The Laplace Transform (Sect. 4.1)

The Laplace Transform (Sect. 4.1).

I The definition of the Laplace Transform.

I Review: Improper integrals.

I Examples of Laplace Transforms.

I A table of Laplace Transforms.

I Properties of the Laplace Transform.

I Laplace Transform and differential equations.








The definition of the Laplace Transform.

DefinitionThe function F : DF → R is the Laplace transform of a functionf : [0,∞) → R iff for all s ∈ DF holds,

F (s) =

∫ ∞

0

e−st f (t) dt,

where DF ⊂ R is the set where the integral converges.

Remark: The domain DF of F depends on the function f .

Notation: We often denote: F (s) = L[f (t)].

I This notation L[ ] emphasizes that the Laplace transformdefines a map from a set of functions into a set of functions.

I Functions are denoted as t 7→ f (t).

I The Laplace transform is also a function: f 7→ L[f ].



F (s) =

∫ ∞

0

e−st f (t) dt,









F (s) =

∫ ∞

0

e−st f (t) dt,









F (s) =

∫ ∞

0

e−st f (t) dt,









F (s) =

∫ ∞

0

e−st f (t) dt,









F (s) =

∫ ∞

0

e−st f (t) dt,














Review: Improper integrals.

Recall: Improper integral are defined as a limit.∫ ∞

t0

g(t) dt = limN→∞

∫ N

t0

g(t) dt.

I The integral converges iff the limit exists.

I The integral diverges iff the limit does not exist.

Example

Compute the improper integral

∫ ∞

0e−at dt, with a > 0.

Solution:

∫ ∞

0

e−at dt = limN→∞

∫ N

0


−1

a

(e−aN − 1

).

Since limN→∞

e−aN = 0 for a > 0, we conclude

∫ ∞

0

e−at dt =1

a. C



t0


∫ N

t0

g(t) dt.



Example


∫ ∞


Solution:

∫ ∞

0


∫ N

0


−1

a

(e−aN − 1

).

Since limN→∞


∫ ∞

0

e−at dt =1

a. C



t0


∫ N

t0

g(t) dt.



Example


∫ ∞


Solution:

∫ ∞

0


∫ N

0


−1

a

(e−aN − 1

).

Since limN→∞


∫ ∞

0

e−at dt =1

a. C



t0


∫ N

t0

g(t) dt.



Example


∫ ∞


Solution:

∫ ∞

0


∫ N

0


−1

a

(e−aN − 1

).

Since limN→∞


∫ ∞

0

e−at dt =1

a. C



t0


∫ N

t0

g(t) dt.



Example


∫ ∞


Solution:

∫ ∞

0


∫ N

0

e−at dt

= limN→∞

−1

a

(e−aN − 1

).

Since limN→∞


∫ ∞

0

e−at dt =1

a. C



t0


∫ N

t0

g(t) dt.



Example


∫ ∞


Solution:

∫ ∞

0


∫ N

0


−1

a

(e−aN − 1

).

Since limN→∞


∫ ∞

0

e−at dt =1

a. C



t0


∫ N

t0

g(t) dt.



Example


∫ ∞


Solution:

∫ ∞

0


∫ N

0


−1

a

(e−aN − 1

).

Since limN→∞

e−aN = 0 for a > 0,

we conclude

∫ ∞

0

e−at dt =1

a. C



t0


∫ N

t0

g(t) dt.



Example


∫ ∞


Solution:

∫ ∞

0


∫ N

0


−1

a

(e−aN − 1

).

Since limN→∞


∫ ∞

0

e−at dt =1

a. C








Examples of Laplace Transforms.

Example

Compute L[1].

Solution: We have to find the Laplace Transform of f (t) = 1.Following the definition we obtain,

L[1] =

∫ ∞

0

e−st 1 dt =

∫ ∞

0

e−st dt

But

∫ ∞

0

e−at dt =1

afor a > 0, and diverges for a 6 0.

Therefore L[1] =1

s, for s > 0, and L[1] does not exists for s 6 0.

In other words, F (s) = L[1] is the function F : DF → R given by

f (t) = 1, F (s) =1

s, DF = (0,∞). C


Example

Compute L[1].

Solution: We have to find the Laplace Transform of f (t) = 1.

Following the definition we obtain,

L[1] =

∫ ∞

0

e−st 1 dt =

∫ ∞

0

e−st dt

But

∫ ∞

0

e−at dt =1


Therefore L[1] =1



f (t) = 1, F (s) =1

s, DF = (0,∞). C


Example

Compute L[1].


L[1] =

∫ ∞

0

e−st 1 dt

=

∫ ∞

0

e−st dt

But

∫ ∞

0

e−at dt =1


Therefore L[1] =1



f (t) = 1, F (s) =1

s, DF = (0,∞). C


Example

Compute L[1].


L[1] =

∫ ∞

0

e−st 1 dt =

∫ ∞

0

e−st dt

But

∫ ∞

0

e−at dt =1


Therefore L[1] =1



f (t) = 1, F (s) =1

s, DF = (0,∞). C


Example

Compute L[1].


L[1] =

∫ ∞

0

e−st 1 dt =

∫ ∞

0

e−st dt

But

∫ ∞

0

e−at dt =1

afor a > 0,

and diverges for a 6 0.

Therefore L[1] =1



f (t) = 1, F (s) =1

s, DF = (0,∞). C


Example

Compute L[1].


L[1] =

∫ ∞

0

e−st 1 dt =

∫ ∞

0

e−st dt

But

∫ ∞

0

e−at dt =1


Therefore L[1] =1



f (t) = 1, F (s) =1

s, DF = (0,∞). C


Example

Compute L[1].


L[1] =

∫ ∞

0

e−st 1 dt =

∫ ∞

0

e−st dt

But

∫ ∞

0

e−at dt =1


Therefore L[1] =1

s, for s > 0,

and L[1] does not exists for s 6 0.


f (t) = 1, F (s) =1

s, DF = (0,∞). C


Example

Compute L[1].


L[1] =

∫ ∞

0

e−st 1 dt =

∫ ∞

0

e−st dt

But

∫ ∞

0

e−at dt =1


Therefore L[1] =1



f (t) = 1, F (s) =1

s, DF = (0,∞). C


Example

Compute L[1].


L[1] =

∫ ∞

0

e−st 1 dt =

∫ ∞

0

e−st dt

But

∫ ∞

0

e−at dt =1


Therefore L[1] =1



f (t) = 1, F (s) =1

s, DF = (0,∞). C


Example

Compute L[eat ], where a ∈ R.

Solution: Following the definition of Laplace Transform,

L[eat ] =

∫ ∞

0

e−steat dt =

∫ ∞

0

e−(s−a)t dt.

We have seen that the improper integral is given by∫ ∞

0e−(s−a)t dt =

1

(s − a)for (s − a) > 0.

We conclude that L[eat ] =1

s − afor s > a. In other words,

f (t) = eat , F (s) =1

(s − a), s > a. C


Example



L[eat ] =

∫ ∞

0

e−steat dt

=

∫ ∞

0

e−(s−a)t dt.


0e−(s−a)t dt =

1

(s − a)for (s − a) > 0.



f (t) = eat , F (s) =1

(s − a), s > a. C


Example



L[eat ] =

∫ ∞

0

e−steat dt =

∫ ∞

0

e−(s−a)t dt.


0e−(s−a)t dt =

1

(s − a)for (s − a) > 0.



f (t) = eat , F (s) =1

(s − a), s > a. C


Example



L[eat ] =

∫ ∞

0

e−steat dt =

∫ ∞

0

e−(s−a)t dt.


0e−(s−a)t dt =

1

(s − a)for (s − a) > 0.



f (t) = eat , F (s) =1

(s − a), s > a. C


Example



L[eat ] =

∫ ∞

0

e−steat dt =

∫ ∞

0

e−(s−a)t dt.


0e−(s−a)t dt =

1

(s − a)for (s − a) > 0.


s − afor s > a.

In other words,

f (t) = eat , F (s) =1

(s − a), s > a. C


Example



L[eat ] =

∫ ∞

0

e−steat dt =

∫ ∞

0

e−(s−a)t dt.


0e−(s−a)t dt =

1

(s − a)for (s − a) > 0.



f (t) = eat , F (s) =1

(s − a), s > a. C

Examples of Laplace Transforms.Example

Compute L[sin(at)], where a ∈ R.

Solution: In this case we need to compute

L[sin(at)] = limN→∞

∫ N

0

e−st sin(at) dt.

Integrating by parts twice it is not difficult to obtain:∫ N

0

e−st sin(at) dt =

−1

s

[e−st sin(at)

]∣∣∣N0

− a

s2

[e−st cos(at)

]∣∣∣N0

− a2

s2

∫ N

0

e−st sin(at) dt.

This identity implies(1 +

a2

s2

) ∫ N

0

e−st sin(at) dt = −1

s

[e−st sin(at)

]∣∣∣N0− a

s2

[e−st cos(at)

]∣∣∣N0.





∫ N

0

e−st sin(at) dt.


0

e−st sin(at) dt =

−1

s

[e−st sin(at)

]∣∣∣N0

− a

s2

[e−st cos(at)

]∣∣∣N0

− a2

s2

∫ N

0

e−st sin(at) dt.


a2

s2

) ∫ N

0


s

[e−st sin(at)

]∣∣∣N0− a

s2

[e−st cos(at)

]∣∣∣N0.





∫ N

0

e−st sin(at) dt.


0

e−st sin(at) dt =

−1

s

[e−st sin(at)

]∣∣∣N0

− a

s2

[e−st cos(at)

]∣∣∣N0

− a2

s2

∫ N

0

e−st sin(at) dt.


a2

s2

) ∫ N

0


s

[e−st sin(at)

]∣∣∣N0− a

s2

[e−st cos(at)

]∣∣∣N0.





∫ N

0

e−st sin(at) dt.


0

e−st sin(at) dt =

−1

s

[e−st sin(at)

]∣∣∣N0

− a

s2

[e−st cos(at)

]∣∣∣N0

− a2

s2

∫ N

0

e−st sin(at) dt.


a2

s2

) ∫ N

0


s

[e−st sin(at)

]∣∣∣N0− a

s2

[e−st cos(at)

]∣∣∣N0.


Example


Solution: Recall the identity:(1 +

a2

s2

) ∫ N

0


s

[e−st sin(at)

]∣∣∣N0− a

s2

[e−st cos(at)

]∣∣∣N0.

Hence, it is not difficult to see that(s2 + a2

s2

) ∫ ∞

0

e−st sin(at) dt =a

s2, s > 0,

which is equivalent to

L[sin(at)] =a

s2 + a2, s > 0. C


Example



a2

s2

) ∫ N

0


s

[e−st sin(at)

]∣∣∣N0− a

s2

[e−st cos(at)

]∣∣∣N0.


s2

) ∫ ∞

0


s2, s > 0,


L[sin(at)] =a

s2 + a2, s > 0. C


Example



a2

s2

) ∫ N

0


s

[e−st sin(at)

]∣∣∣N0− a

s2

[e−st cos(at)

]∣∣∣N0.


s2

) ∫ ∞

0


s2, s > 0,


L[sin(at)] =a

s2 + a2, s > 0. C








A table of Laplace Transforms.

f (t) = 1 F (s) =1

ss > 0,

f (t) = eat F (s) =1

s − as > max{a, 0},

f (t) = tn F (s) =n!

s(n+1)s > 0,

f (t) = sin(at) F (s) =a

s2 + a2s > 0,

f (t) = cos(at) F (s) =s

s2 + a2s > 0,

f (t) = sinh(at) F (s) =a

s2 − a2s > |a|,

f (t) = cosh(at) F (s) =s

s2 − a2s > |a|,

f (t) = tneat F (s) =n!

(s − a)(n+1)s > max{a, 0},

f (t) = eat sin(bt) F (s) =b

(s − a)2 + b2s > max{a, 0}.








Properties of the Laplace Transform.

Theorem (Sufficient conditions)

If the function f : [0,∞) → R is piecewise continuous and thereexist positive constants k and a such that

|f (t)| 6 k eat ,

then the Laplace Transform of f exists for all s > a.

Theorem (Linear combination)

If the L[f ] and L[g ] are well-defined and a, b are constants, then

L[af + bg ] = aL[f ] + bL[g ].

Proof: Integration is a linear operation:∫[a f (t) + b g(t)] dt = a

∫f (t) dt + b

∫g(t) dt.




|f (t)| 6 k eat ,




L[af + bg ] = aL[f ] + bL[g ].


∫f (t) dt + b

∫g(t) dt.




|f (t)| 6 k eat ,




L[af + bg ] = aL[f ] + bL[g ].

Proof: Integration is a linear operation:

∫[a f (t) + b g(t)] dt = a

∫f (t) dt + b

∫g(t) dt.




|f (t)| 6 k eat ,




L[af + bg ] = aL[f ] + bL[g ].


∫f (t) dt + b

∫g(t) dt.


Theorem (Derivatives)

If the L[f ] and L[f ′] are well-defined, then holds,

L[f ′] = s L[f ]− f (0). (1)

Furthermore, if L[f ′′] is well-defined, then it also holds

L[f ′′] = s2 L[f ]− s f (0)− f ′(0). (2)

Proof of Eq (2): Use Eq. (1) twice:

L[f ′′] = L[(f ′)′] = sL[(f ′)]− f ′(0) = s(sL[f ]− f (0)

)− f ′(0),

that is,L[f ′′] = s2 L[f ]− s f (0)− f ′(0).




L[f ′] = s L[f ]− f (0). (1)


L[f ′′] = s2 L[f ]− s f (0)− f ′(0). (2)

Proof of Eq (2):

Use Eq. (1) twice:

L[f ′′] = L[(f ′)′] = sL[(f ′)]− f ′(0) = s(sL[f ]− f (0)

)− f ′(0),

that is,L[f ′′] = s2 L[f ]− s f (0)− f ′(0).




L[f ′] = s L[f ]− f (0). (1)


L[f ′′] = s2 L[f ]− s f (0)− f ′(0). (2)


L[f ′′] = L[(f ′)′] = sL[(f ′)]− f ′(0) = s(sL[f ]− f (0)

)− f ′(0),

that is,L[f ′′] = s2 L[f ]− s f (0)− f ′(0).




L[f ′] = s L[f ]− f (0). (1)


L[f ′′] = s2 L[f ]− s f (0)− f ′(0). (2)


L[f ′′] = L[(f ′)′]

= sL[(f ′)]− f ′(0) = s(sL[f ]− f (0)

)− f ′(0),

that is,L[f ′′] = s2 L[f ]− s f (0)− f ′(0).




L[f ′] = s L[f ]− f (0). (1)


L[f ′′] = s2 L[f ]− s f (0)− f ′(0). (2)


L[f ′′] = L[(f ′)′] = sL[(f ′)]− f ′(0)

= s(sL[f ]− f (0)

)− f ′(0),

that is,L[f ′′] = s2 L[f ]− s f (0)− f ′(0).




L[f ′] = s L[f ]− f (0). (1)


L[f ′′] = s2 L[f ]− s f (0)− f ′(0). (2)


L[f ′′] = L[(f ′)′] = sL[(f ′)]− f ′(0) = s(sL[f ]− f (0)

)− f ′(0),

that is,L[f ′′] = s2 L[f ]− s f (0)− f ′(0).




L[f ′] = s L[f ]− f (0). (1)


L[f ′′] = s2 L[f ]− s f (0)− f ′(0). (2)


L[f ′′] = L[(f ′)′] = sL[(f ′)]− f ′(0) = s(sL[f ]− f (0)

)− f ′(0),

that is,L[f ′′] = s2 L[f ]− s f (0)− f ′(0).


Proof of Eq (1): Recall the definition of the Laplace Transform,

L[f ′] =

∫ ∞

0

e−st f ′(t) dt

= limn→∞

∫ n

0

e−st f ′(t) dt

Integrating by parts,

limn→∞

∫ n

0

e−st f ′(t) dt = limn→∞

[(e−st f (t)

)∣∣∣n0

−∫ n

0

(−s)e−st f (t) dt]

L[f ′] = limn→∞

[e−snf (n)−f (0)

]+s

∫ ∞

0

e−st f (t) dt = −f (0)+s L[f ],

where we used that limn→∞ e−snf (n) = 0 for s big enough, and wealso used that L[f ] is well-defined.

We then conclude that L[f ′] = s L[f ]− f (0).



L[f ′] =

∫ ∞

0


∫ n

0

e−st f ′(t) dt


limn→∞

∫ n

0


[(e−st f (t)

)∣∣∣n0

−∫ n

0



[e−snf (n)−f (0)

]+s

∫ ∞

0

e−st f (t) dt = −f (0)+s L[f ],





L[f ′] =

∫ ∞

0


∫ n

0

e−st f ′(t) dt


limn→∞

∫ n

0


[(e−st f (t)

)∣∣∣n0

−∫ n

0



[e−snf (n)−f (0)

]+s

∫ ∞

0

e−st f (t) dt = −f (0)+s L[f ],





L[f ′] =

∫ ∞

0


∫ n

0

e−st f ′(t) dt


limn→∞

∫ n

0


[(e−st f (t)

)∣∣∣n0

−∫ n

0



[e−snf (n)−f (0)

]+s

∫ ∞

0

e−st f (t) dt

= −f (0)+s L[f ],





L[f ′] =

∫ ∞

0


∫ n

0

e−st f ′(t) dt


limn→∞

∫ n

0


[(e−st f (t)

)∣∣∣n0

−∫ n

0



[e−snf (n)−f (0)

]+s

∫ ∞

0

e−st f (t) dt = −f (0)+s L[f ],





L[f ′] =

∫ ∞

0


∫ n

0

e−st f ′(t) dt


limn→∞

∫ n

0


[(e−st f (t)

)∣∣∣n0

−∫ n

0



[e−snf (n)−f (0)

]+s

∫ ∞

0

e−st f (t) dt = −f (0)+s L[f ],

where we used that limn→∞ e−snf (n) = 0 for s big enough,

and wealso used that L[f ] is well-defined.




L[f ′] =

∫ ∞

0


∫ n

0

e−st f ′(t) dt


limn→∞

∫ n

0


[(e−st f (t)

)∣∣∣n0

−∫ n

0



[e−snf (n)−f (0)

]+s

∫ ∞

0

e−st f (t) dt = −f (0)+s L[f ],





L[f ′] =

∫ ∞

0


∫ n

0

e−st f ′(t) dt


limn→∞

∫ n

0


[(e−st f (t)

)∣∣∣n0

−∫ n

0



[e−snf (n)−f (0)

]+s

∫ ∞

0

e−st f (t) dt = −f (0)+s L[f ],










Laplace Transform and differential equations.

Remark: Laplace Transforms can be used to find solutions todifferential equations with constant coefficients.

Idea of the method:

L

[Differential Eq.

for y(t).

](1)−→

Algebraic Eq.

for L[y(t)].

(2)−→

(2)−→Solve the

Algebraic Eq.

for L[y(t)].

(3)−→Transform back

to obtain y(t).

(Using the table.)



Idea of the method:

L

[Differential Eq.

for y(t).

](1)−→

Algebraic Eq.

for L[y(t)].

(2)−→

(2)−→Solve the

Algebraic Eq.

for L[y(t)].


to obtain y(t).

(Using the table.)



Idea of the method:

L

[Differential Eq.

for y(t).

]

(1)−→Algebraic Eq.

for L[y(t)].

(2)−→

(2)−→Solve the

Algebraic Eq.

for L[y(t)].


to obtain y(t).

(Using the table.)



Idea of the method:

L

[Differential Eq.

for y(t).

](1)−→

Algebraic Eq.

for L[y(t)].

(2)−→

(2)−→Solve the

Algebraic Eq.

for L[y(t)].


to obtain y(t).

(Using the table.)



Idea of the method:

L

[Differential Eq.

for y(t).

](1)−→

Algebraic Eq.

for L[y(t)].

(2)−→

(2)−→Solve the

Algebraic Eq.

for L[y(t)].


to obtain y(t).

(Using the table.)



Idea of the method:

L

[Differential Eq.

for y(t).

](1)−→

Algebraic Eq.

for L[y(t)].

(2)−→

(2)−→Solve the

Algebraic Eq.

for L[y(t)].


to obtain y(t).

(Using the table.)


Example

Use the Laplace transform to find the solution y(t) to the IVP

y ′ + 2y = 0, y(0) = 3.

Solution: We know the solution: y(t) = 3e−2t .

(1): Compute the Laplace transform of the differential equation,

L[y ′ + 2y ] = L[0] ⇒ L[y ′ + 2y ] = 0.

Find an algebraic equation for L[y ]. Recall linearity:

L[y ′] + 2L[y ] = 0.

Also recall the property: L[y ′] = s L[y ]− y(0), that is,[s L[y ]− y(0)

]+ 2L[y ] = 0 ⇒ (s + 2)L[y ] = y(0).


Example


y ′ + 2y = 0, y(0) = 3.



L[y ′ + 2y ] = L[0] ⇒ L[y ′ + 2y ] = 0.


L[y ′] + 2L[y ] = 0.


]+ 2L[y ] = 0 ⇒ (s + 2)L[y ] = y(0).


Example


y ′ + 2y = 0, y(0) = 3.



L[y ′ + 2y ] = L[0]

⇒ L[y ′ + 2y ] = 0.


L[y ′] + 2L[y ] = 0.


]+ 2L[y ] = 0 ⇒ (s + 2)L[y ] = y(0).


Example


y ′ + 2y = 0, y(0) = 3.



L[y ′ + 2y ] = L[0] ⇒ L[y ′ + 2y ] = 0.


L[y ′] + 2L[y ] = 0.


]+ 2L[y ] = 0 ⇒ (s + 2)L[y ] = y(0).


Example


y ′ + 2y = 0, y(0) = 3.



L[y ′ + 2y ] = L[0] ⇒ L[y ′ + 2y ] = 0.

Find an algebraic equation for L[y ].

Recall linearity:

L[y ′] + 2L[y ] = 0.


]+ 2L[y ] = 0 ⇒ (s + 2)L[y ] = y(0).


Example


y ′ + 2y = 0, y(0) = 3.



L[y ′ + 2y ] = L[0] ⇒ L[y ′ + 2y ] = 0.


L[y ′] + 2L[y ] = 0.


]+ 2L[y ] = 0 ⇒ (s + 2)L[y ] = y(0).


Example


y ′ + 2y = 0, y(0) = 3.



L[y ′ + 2y ] = L[0] ⇒ L[y ′ + 2y ] = 0.


L[y ′] + 2L[y ] = 0.

Also recall the property: L[y ′] = s L[y ]− y(0),

that is,[s L[y ]− y(0)

]+ 2L[y ] = 0 ⇒ (s + 2)L[y ] = y(0).


Example


y ′ + 2y = 0, y(0) = 3.



L[y ′ + 2y ] = L[0] ⇒ L[y ′ + 2y ] = 0.


L[y ′] + 2L[y ] = 0.


]+ 2L[y ] = 0 ⇒ (s + 2)L[y ] = y(0).


Example


y ′ + 2y = 0, y(0) = 3.

Solution: Recall: (s + 2)L[y ] = y(0).

(2): Solve the algebraic equation for L[y ].

L[y ] =y(0)

s + 2, y(0) = 3, ⇒ L[y ] =

3

s + 2.

(3): Transform back to y(t). From the table:

L[eat ] =1

s − a⇒ 3

s + 2= 3L[e−2t ] ⇒ 3

s + 2= L[3e−2t ].

Hence, L[y ] = L[3e−2t ] ⇒ y(t) = 3e−2t . C


Example


y ′ + 2y = 0, y(0) = 3.



L[y ] =y(0)

s + 2, y(0) = 3, ⇒ L[y ] =

3

s + 2.


L[eat ] =1

s − a⇒ 3

s + 2= 3L[e−2t ] ⇒ 3

s + 2= L[3e−2t ].



Example


y ′ + 2y = 0, y(0) = 3.



L[y ] =y(0)

s + 2,

y(0) = 3, ⇒ L[y ] =3

s + 2.


L[eat ] =1

s − a⇒ 3

s + 2= 3L[e−2t ] ⇒ 3

s + 2= L[3e−2t ].



Example


y ′ + 2y = 0, y(0) = 3.



L[y ] =y(0)

s + 2, y(0) = 3,

⇒ L[y ] =3

s + 2.


L[eat ] =1

s − a⇒ 3

s + 2= 3L[e−2t ] ⇒ 3

s + 2= L[3e−2t ].



Example


y ′ + 2y = 0, y(0) = 3.



L[y ] =y(0)

s + 2, y(0) = 3, ⇒ L[y ] =

3

s + 2.


L[eat ] =1

s − a⇒ 3

s + 2= 3L[e−2t ] ⇒ 3

s + 2= L[3e−2t ].



Example


y ′ + 2y = 0, y(0) = 3.



L[y ] =y(0)

s + 2, y(0) = 3, ⇒ L[y ] =

3

s + 2.

(3): Transform back to y(t).

From the table:

L[eat ] =1

s − a⇒ 3

s + 2= 3L[e−2t ] ⇒ 3

s + 2= L[3e−2t ].



Example


y ′ + 2y = 0, y(0) = 3.



L[y ] =y(0)

s + 2, y(0) = 3, ⇒ L[y ] =

3

s + 2.


L[eat ] =1

s − a

⇒ 3

s + 2= 3L[e−2t ] ⇒ 3

s + 2= L[3e−2t ].



Example


y ′ + 2y = 0, y(0) = 3.



L[y ] =y(0)

s + 2, y(0) = 3, ⇒ L[y ] =

3

s + 2.


L[eat ] =1

s − a⇒ 3

s + 2= 3L[e−2t ]

⇒ 3

s + 2= L[3e−2t ].



Example


y ′ + 2y = 0, y(0) = 3.



L[y ] =y(0)

s + 2, y(0) = 3, ⇒ L[y ] =

3

s + 2.


L[eat ] =1

s − a⇒ 3

s + 2= 3L[e−2t ] ⇒ 3

s + 2= L[3e−2t ].



Example


y ′ + 2y = 0, y(0) = 3.



L[y ] =y(0)

s + 2, y(0) = 3, ⇒ L[y ] =

3

s + 2.


L[eat ] =1

s − a⇒ 3

s + 2= 3L[e−2t ] ⇒ 3

s + 2= L[3e−2t ].

Hence, L[y ] = L[3e−2t ]

⇒ y(t) = 3e−2t . C


Example


y ′ + 2y = 0, y(0) = 3.



L[y ] =y(0)

s + 2, y(0) = 3, ⇒ L[y ] =

3

s + 2.


L[eat ] =1

s − a⇒ 3

s + 2= 3L[e−2t ] ⇒ 3

s + 2= L[3e−2t ].


The Laplace Transform and the IVP (Sect. 4.2).

I Solving differential equations using L[ ].I Homogeneous IVP.I First, second, higher order equations.I Non-homogeneous IVP.

Solving differential equations using L[ ].

Remark: The method works with:

I Constant coefficient equations.

I Homogeneous and non-homogeneous equations.

I First, second, higher order equations.

Idea of the method:

L

[Differential Eq.

for y(t).

](1)−→

Algebraic Eq.

for L[y(t)].

(2)−→

(2)−→Solve the

Algebraic Eq.

for L[y(t)].


to obtain y(t).

(Using the table.)






Idea of the method:

L

[Differential Eq.

for y(t).

](1)−→

Algebraic Eq.

for L[y(t)].

(2)−→

(2)−→Solve the

Algebraic Eq.

for L[y(t)].


to obtain y(t).

(Using the table.)






Idea of the method:

L

[Differential Eq.

for y(t).

](1)−→

Algebraic Eq.

for L[y(t)].

(2)−→

(2)−→Solve the

Algebraic Eq.

for L[y(t)].


to obtain y(t).

(Using the table.)






Idea of the method:

L

[Differential Eq.

for y(t).

](1)−→

Algebraic Eq.

for L[y(t)].

(2)−→

(2)−→Solve the

Algebraic Eq.

for L[y(t)].


to obtain y(t).

(Using the table.)






Idea of the method:

L

[Differential Eq.

for y(t).

](1)−→

Algebraic Eq.

for L[y(t)].

(2)−→

(2)−→Solve the

Algebraic Eq.

for L[y(t)].


to obtain y(t).

(Using the table.)






Idea of the method:

L

[Differential Eq.

for y(t).

]

(1)−→Algebraic Eq.

for L[y(t)].

(2)−→

(2)−→Solve the

Algebraic Eq.

for L[y(t)].


to obtain y(t).

(Using the table.)






Idea of the method:

L

[Differential Eq.

for y(t).

](1)−→

Algebraic Eq.

for L[y(t)].

(2)−→

(2)−→Solve the

Algebraic Eq.

for L[y(t)].


to obtain y(t).

(Using the table.)






Idea of the method:

L

[Differential Eq.

for y(t).

](1)−→

Algebraic Eq.

for L[y(t)].

(2)−→

(2)−→Solve the

Algebraic Eq.

for L[y(t)].


to obtain y(t).

(Using the table.)






Idea of the method:

L

[Differential Eq.

for y(t).

](1)−→

Algebraic Eq.

for L[y(t)].

(2)−→

(2)−→Solve the

Algebraic Eq.

for L[y(t)].


to obtain y(t).

(Using the table.)


Idea of the method:

L

[Differential Eq.

for y(t).

](1)−→

Algebraic Eq.

for L[y(t)].

(2)−→

(2)−→Solve the

Algebraic Eq.

for L[y(t)].


to obtain y(t).

(Using the table.)

Recall:

(a) L[a f (t) + b g(t)] = aL[f (t)] + bL[g(t)];

(b) L[y (n)

]= sn L[y ]− s(n−1) y(0)− s(n−2) y ′(0)−· · ·−y (n−1)(0).


Idea of the method:

L

[Differential Eq.

for y(t).

](1)−→

Algebraic Eq.

for L[y(t)].

(2)−→

(2)−→Solve the

Algebraic Eq.

for L[y(t)].


to obtain y(t).

(Using the table.)

Recall:


(b) L[y (n)

]= sn L[y ]− s(n−1) y(0)− s(n−2) y ′(0)−· · ·−y (n−1)(0).


Idea of the method:

L

[Differential Eq.

for y(t).

](1)−→

Algebraic Eq.

for L[y(t)].

(2)−→

(2)−→Solve the

Algebraic Eq.

for L[y(t)].


to obtain y(t).

(Using the table.)

Recall:


(b) L[y (n)

]= sn L[y ]− s(n−1) y(0)− s(n−2) y ′(0)−· · ·−y (n−1)(0).



Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.

Solution: Compute the L[ ] of the differential equation,

L[y ′′ − y ′ − 2y ] = L[0] ⇒ L[y ′′ − y ′ − 2y ] = 0.

The L[ ] is a linear function, so

L[y ′′]− L[y ′]− 2L[y ] = 0.

Derivatives are transformed into power functions,[s2 L[y ]− s y(0)− y ′(0)

]−

[s L[y ]− y(0)

]− 2L[y ] = 0,

We the obtain (s2 − s − 2)L[y ] = (s − 1) y(0) + y ′(0).

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


L[y ′′ − y ′ − 2y ] = L[0] ⇒ L[y ′′ − y ′ − 2y ] = 0.


L[y ′′]− L[y ′]− 2L[y ] = 0.


]−

[s L[y ]− y(0)

]− 2L[y ] = 0,


Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


L[y ′′ − y ′ − 2y ] = L[0]

⇒ L[y ′′ − y ′ − 2y ] = 0.


L[y ′′]− L[y ′]− 2L[y ] = 0.


]−

[s L[y ]− y(0)

]− 2L[y ] = 0,


Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


L[y ′′ − y ′ − 2y ] = L[0] ⇒ L[y ′′ − y ′ − 2y ] = 0.


L[y ′′]− L[y ′]− 2L[y ] = 0.


]−

[s L[y ]− y(0)

]− 2L[y ] = 0,


Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


L[y ′′ − y ′ − 2y ] = L[0] ⇒ L[y ′′ − y ′ − 2y ] = 0.

The L[ ] is a linear function,

so

L[y ′′]− L[y ′]− 2L[y ] = 0.


]−

[s L[y ]− y(0)

]− 2L[y ] = 0,


Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


L[y ′′ − y ′ − 2y ] = L[0] ⇒ L[y ′′ − y ′ − 2y ] = 0.


L[y ′′]− L[y ′]− 2L[y ] = 0.


]−

[s L[y ]− y(0)

]− 2L[y ] = 0,


Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


L[y ′′ − y ′ − 2y ] = L[0] ⇒ L[y ′′ − y ′ − 2y ] = 0.


L[y ′′]− L[y ′]− 2L[y ] = 0.


]−

[s L[y ]− y(0)

]− 2L[y ] = 0,


Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


L[y ′′ − y ′ − 2y ] = L[0] ⇒ L[y ′′ − y ′ − 2y ] = 0.


L[y ′′]− L[y ′]− 2L[y ] = 0.


]−

[s L[y ]− y(0)

]− 2L[y ] = 0,


Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.

Solution: Recall: (s2 − s − 2)L[y ] = (s − 1) y(0) + y ′(0).

Differential equation for yL[ ]−→ Algebraic equation for L[y ].

Introduce the initial condition,

(s2 − s − 2)L[y ] = (s − 1).

We can solve for the unknown L[y ] as follows,

L[y ] =(s − 1)

(s2 − s − 2).

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.




(s2 − s − 2)L[y ] = (s − 1).


L[y ] =(s − 1)

(s2 − s − 2).

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.




(s2 − s − 2)L[y ] = (s − 1).


L[y ] =(s − 1)

(s2 − s − 2).

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.




(s2 − s − 2)L[y ] = (s − 1).


L[y ] =(s − 1)

(s2 − s − 2).

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.




(s2 − s − 2)L[y ] = (s − 1).


L[y ] =(s − 1)

(s2 − s − 2).

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.




(s2 − s − 2)L[y ] = (s − 1).


L[y ] =(s − 1)

(s2 − s − 2).

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.

Solution: Recall: L[y ] =(s − 1)

(s2 − s − 2).

The partial fraction method: Find the zeros of the denominator,

s2 − s − 2 = 0 ⇒ s± =1

2

[1±

√1 + 8

]⇒

{s+ = 2,

s− = −1,

Therefore, we rewrite: L[y ] =(s − 1)

(s − 2)(s + 1).

Find constants a and b such that

(s − 1)

(s − 2)(s + 1)=

a

s − 2+

b

s + 1.

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


(s2 − s − 2).


s2 − s − 2 = 0

⇒ s± =1

2

[1±

√1 + 8

]⇒

{s+ = 2,

s− = −1,


(s − 2)(s + 1).


(s − 1)

(s − 2)(s + 1)=

a

s − 2+

b

s + 1.

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


(s2 − s − 2).


s2 − s − 2 = 0 ⇒ s± =1

2

[1±

√1 + 8

]

⇒

{s+ = 2,

s− = −1,


(s − 2)(s + 1).


(s − 1)

(s − 2)(s + 1)=

a

s − 2+

b

s + 1.

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


(s2 − s − 2).


s2 − s − 2 = 0 ⇒ s± =1

2

[1±

√1 + 8

]⇒

{s+ = 2,

s− = −1,


(s − 2)(s + 1).


(s − 1)

(s − 2)(s + 1)=

a

s − 2+

b

s + 1.

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


(s2 − s − 2).


s2 − s − 2 = 0 ⇒ s± =1

2

[1±

√1 + 8

]⇒

{s+ = 2,

s− = −1,


(s − 2)(s + 1).


(s − 1)

(s − 2)(s + 1)=

a

s − 2+

b

s + 1.

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


(s2 − s − 2).


s2 − s − 2 = 0 ⇒ s± =1

2

[1±

√1 + 8

]⇒

{s+ = 2,

s− = −1,


(s − 2)(s + 1).


(s − 1)

(s − 2)(s + 1)=

a

s − 2+

b

s + 1.

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.

Solution: Recall:(s − 1)

(s − 2)(s + 1)=

a

s − 2+

b

s + 1.

A simple calculation shows

(s − 1)

(s − 2)(s + 1)=

a

s − 2+

b

s + 1=

a(s + 1) + b(s − 2)

(s − 2)(s + 1)

(s − 1) = s(a + b) + (a− 2b) ⇒

{a + b = 1,

a− 2b = −1

Hence, a =1

3and b =

2

3. Then, L[y ] =

1

3

1

(s − 2)+

2

3

1

(s + 1).

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


(s − 2)(s + 1)=

a

s − 2+

b

s + 1.


(s − 1)

(s − 2)(s + 1)=

a

s − 2+

b

s + 1

=a(s + 1) + b(s − 2)

(s − 2)(s + 1)

(s − 1) = s(a + b) + (a− 2b) ⇒

{a + b = 1,

a− 2b = −1

Hence, a =1

3and b =

2

3. Then, L[y ] =

1

3

1

(s − 2)+

2

3

1

(s + 1).

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


(s − 2)(s + 1)=

a

s − 2+

b

s + 1.


(s − 1)

(s − 2)(s + 1)=

a

s − 2+

b

s + 1=

a(s + 1) + b(s − 2)

(s − 2)(s + 1)

(s − 1) = s(a + b) + (a− 2b) ⇒

{a + b = 1,

a− 2b = −1

Hence, a =1

3and b =

2

3. Then, L[y ] =

1

3

1

(s − 2)+

2

3

1

(s + 1).

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


(s − 2)(s + 1)=

a

s − 2+

b

s + 1.


(s − 1)

(s − 2)(s + 1)=

a

s − 2+

b

s + 1=

a(s + 1) + b(s − 2)

(s − 2)(s + 1)

(s − 1) = s(a + b) + (a− 2b)

⇒

{a + b = 1,

a− 2b = −1

Hence, a =1

3and b =

2

3. Then, L[y ] =

1

3

1

(s − 2)+

2

3

1

(s + 1).

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


(s − 2)(s + 1)=

a

s − 2+

b

s + 1.


(s − 1)

(s − 2)(s + 1)=

a

s − 2+

b

s + 1=

a(s + 1) + b(s − 2)

(s − 2)(s + 1)

(s − 1) = s(a + b) + (a− 2b) ⇒

{a + b = 1,

a− 2b = −1

Hence, a =1

3and b =

2

3. Then, L[y ] =

1

3

1

(s − 2)+

2

3

1

(s + 1).

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


(s − 2)(s + 1)=

a

s − 2+

b

s + 1.


(s − 1)

(s − 2)(s + 1)=

a

s − 2+

b

s + 1=

a(s + 1) + b(s − 2)

(s − 2)(s + 1)

(s − 1) = s(a + b) + (a− 2b) ⇒

{a + b = 1,

a− 2b = −1

Hence, a =1

3and b =

2

3.

Then, L[y ] =1

3

1

(s − 2)+

2

3

1

(s + 1).

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


(s − 2)(s + 1)=

a

s − 2+

b

s + 1.


(s − 1)

(s − 2)(s + 1)=

a

s − 2+

b

s + 1=

a(s + 1) + b(s − 2)

(s − 2)(s + 1)

(s − 1) = s(a + b) + (a− 2b) ⇒

{a + b = 1,

a− 2b = −1

Hence, a =1

3and b =

2

3. Then, L[y ] =

1

3

1

(s − 2)+

2

3

1

(s + 1).

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.

Solution: Recall: L[y ] =1

3

1

(s − 2)+

2

3

1

(s + 1).

From the table:

L[eat ] =1

s − a⇒ 1

s − 2= L[e2t ],

1

s + 1= L[e−t ].

So we arrive at the equation

L[y ] =1

3L[e2t ] +

2

3L[e−t ] = L

[1

3

(e2t + 2e−t

)]We conclude that: y(t) =

1

3

(e2t + 2e−t

). C

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


3

1

(s − 2)+

2

3

1

(s + 1). From the table:

L[eat ] =1

s − a

⇒ 1

s − 2= L[e2t ],

1

s + 1= L[e−t ].


L[y ] =1

3L[e2t ] +

2

3L[e−t ] = L

[1

3

(e2t + 2e−t


1

3

(e2t + 2e−t

). C

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


3

1

(s − 2)+

2

3

1


L[eat ] =1

s − a⇒ 1

s − 2= L[e2t ],

1

s + 1= L[e−t ].


L[y ] =1

3L[e2t ] +

2

3L[e−t ] = L

[1

3

(e2t + 2e−t


1

3

(e2t + 2e−t

). C

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


3

1

(s − 2)+

2

3

1


L[eat ] =1

s − a⇒ 1

s − 2= L[e2t ],

1

s + 1= L[e−t ].


L[y ] =1

3L[e2t ] +

2

3L[e−t ] = L

[1

3

(e2t + 2e−t


1

3

(e2t + 2e−t

). C

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


3

1

(s − 2)+

2

3

1


L[eat ] =1

s − a⇒ 1

s − 2= L[e2t ],

1

s + 1= L[e−t ].


L[y ] =1

3L[e2t ] +

2

3L[e−t ]

= L[1

3

(e2t + 2e−t


1

3

(e2t + 2e−t

). C

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


3

1

(s − 2)+

2

3

1


L[eat ] =1

s − a⇒ 1

s − 2= L[e2t ],

1

s + 1= L[e−t ].


L[y ] =1

3L[e2t ] +

2

3L[e−t ] = L

[1

3

(e2t + 2e−t

)]

We conclude that: y(t) =1

3

(e2t + 2e−t

). C

Homogeneous IVP.

Example


y ′′ − y ′ − 2y = 0, y(0) = 1, y ′(0) = 0.


3

1

(s − 2)+

2

3

1


L[eat ] =1

s − a⇒ 1

s − 2= L[e2t ],

1

s + 1= L[e−t ].


L[y ] =1

3L[e2t ] +

2

3L[e−t ] = L

[1

3

(e2t + 2e−t


1

3

(e2t + 2e−t

). C

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.


L[y ′′ − 4y ′ + 4y ] = L[0] = 0.


L[y ′′]− 4L[y ′] + 4L[y ] = 0.


]− 4

[s L[y ]− y(0)

]+ 4L[y ] = 0,

Therefore, (s2 − 4s + 4)L[y ] = (s − 4) y(0) + y ′(0).

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.


L[y ′′ − 4y ′ + 4y ] = L[0] = 0.


L[y ′′]− 4L[y ′] + 4L[y ] = 0.


]− 4

[s L[y ]− y(0)

]+ 4L[y ] = 0,


Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.


L[y ′′ − 4y ′ + 4y ] = L[0] = 0.


L[y ′′]− 4L[y ′] + 4L[y ] = 0.


]− 4

[s L[y ]− y(0)

]+ 4L[y ] = 0,


Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.


L[y ′′ − 4y ′ + 4y ] = L[0] = 0.


L[y ′′]− 4L[y ′] + 4L[y ] = 0.


]− 4

[s L[y ]− y(0)

]+ 4L[y ] = 0,


Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.


L[y ′′ − 4y ′ + 4y ] = L[0] = 0.


L[y ′′]− 4L[y ′] + 4L[y ] = 0.


]− 4

[s L[y ]− y(0)

]+ 4L[y ] = 0,


Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.


L[y ′′ − 4y ′ + 4y ] = L[0] = 0.


L[y ′′]− 4L[y ′] + 4L[y ] = 0.


]− 4

[s L[y ]− y(0)

]+ 4L[y ] = 0,


Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.

Solution: Recall: (s2 − 4s + 4)L[y ] = (s − 4) y(0) + y ′(0).

Introduce the initial conditions, (s2 − 4s + 4)L[y ] = s − 3.

Solve for L[y ] as follows: L[y ] =(s − 3)

(s2 − 4s + 4).

The partial fraction method: Find the roots of the denominator,

s2−4s+4 = 0 ⇒ s± =1

2

[4±

√16− 16

]⇒ s+ = s− = 2.

We obtain: L[y ] =(s − 3)

(s − 2)2.

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.


Introduce the initial conditions,

(s2 − 4s + 4)L[y ] = s − 3.


(s2 − 4s + 4).


s2−4s+4 = 0 ⇒ s± =1

2

[4±

√16− 16

]⇒ s+ = s− = 2.


(s − 2)2.

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.




(s2 − 4s + 4).


s2−4s+4 = 0 ⇒ s± =1

2

[4±

√16− 16

]⇒ s+ = s− = 2.


(s − 2)2.

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.



Solve for L[y ] as follows:

L[y ] =(s − 3)

(s2 − 4s + 4).


s2−4s+4 = 0 ⇒ s± =1

2

[4±

√16− 16

]⇒ s+ = s− = 2.


(s − 2)2.

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.




(s2 − 4s + 4).


s2−4s+4 = 0 ⇒ s± =1

2

[4±

√16− 16

]⇒ s+ = s− = 2.


(s − 2)2.

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.




(s2 − 4s + 4).

The partial fraction method:

Find the roots of the denominator,

s2−4s+4 = 0 ⇒ s± =1

2

[4±

√16− 16

]⇒ s+ = s− = 2.


(s − 2)2.

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.




(s2 − 4s + 4).


s2−4s+4 = 0

⇒ s± =1

2

[4±

√16− 16

]⇒ s+ = s− = 2.


(s − 2)2.

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.




(s2 − 4s + 4).


s2−4s+4 = 0 ⇒ s± =1

2

[4±

√16− 16

]

⇒ s+ = s− = 2.


(s − 2)2.

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.




(s2 − 4s + 4).


s2−4s+4 = 0 ⇒ s± =1

2

[4±

√16− 16

]⇒ s+ = s− = 2.


(s − 2)2.

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.




(s2 − 4s + 4).


s2−4s+4 = 0 ⇒ s± =1

2

[4±

√16− 16

]⇒ s+ = s− = 2.


(s − 2)2.

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.


(s − 2)2.

We find the partial fraction,

(s − 3)

(s − 2)2=

a

(s − 2)+

b

(s − 2)2⇒ s − 3 = a(s − 2) + b

If s = 2, then b = −1. If s = 3, then a = 1. Hence

L[y ] =1

s − 2− 1

(s − 2)2.

From the Laplace transforms table:

L[eat ] =1

s − a⇒ 1

s − 2= L[e2t ],

L[tneat ] =n!

(s − a)(n+1)⇒ 1

(s − 2)2= L[te2t ].

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.


(s − 2)2. We find the partial fraction,

(s − 3)

(s − 2)2=

a

(s − 2)+

b

(s − 2)2

⇒ s − 3 = a(s − 2) + b


L[y ] =1

s − 2− 1

(s − 2)2.


L[eat ] =1

s − a⇒ 1

s − 2= L[e2t ],

L[tneat ] =n!

(s − a)(n+1)⇒ 1

(s − 2)2= L[te2t ].

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.



(s − 3)

(s − 2)2=

a

(s − 2)+

b

(s − 2)2⇒ s − 3 = a(s − 2) + b


L[y ] =1

s − 2− 1

(s − 2)2.


L[eat ] =1

s − a⇒ 1

s − 2= L[e2t ],

L[tneat ] =n!

(s − a)(n+1)⇒ 1

(s − 2)2= L[te2t ].

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.



(s − 3)

(s − 2)2=

a

(s − 2)+

b

(s − 2)2⇒ s − 3 = a(s − 2) + b

If s = 2,

then b = −1. If s = 3, then a = 1. Hence

L[y ] =1

s − 2− 1

(s − 2)2.


L[eat ] =1

s − a⇒ 1

s − 2= L[e2t ],

L[tneat ] =n!

(s − a)(n+1)⇒ 1

(s − 2)2= L[te2t ].

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.



(s − 3)

(s − 2)2=

a

(s − 2)+

b

(s − 2)2⇒ s − 3 = a(s − 2) + b

If s = 2, then b = −1.

If s = 3, then a = 1. Hence

L[y ] =1

s − 2− 1

(s − 2)2.


L[eat ] =1

s − a⇒ 1

s − 2= L[e2t ],

L[tneat ] =n!

(s − a)(n+1)⇒ 1

(s − 2)2= L[te2t ].

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.



(s − 3)

(s − 2)2=

a

(s − 2)+

b

(s − 2)2⇒ s − 3 = a(s − 2) + b

If s = 2, then b = −1. If s = 3,

then a = 1. Hence

L[y ] =1

s − 2− 1

(s − 2)2.


L[eat ] =1

s − a⇒ 1

s − 2= L[e2t ],

L[tneat ] =n!

(s − a)(n+1)⇒ 1

(s − 2)2= L[te2t ].

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.



(s − 3)

(s − 2)2=

a

(s − 2)+

b

(s − 2)2⇒ s − 3 = a(s − 2) + b

If s = 2, then b = −1. If s = 3, then a = 1.

Hence

L[y ] =1

s − 2− 1

(s − 2)2.


L[eat ] =1

s − a⇒ 1

s − 2= L[e2t ],

L[tneat ] =n!

(s − a)(n+1)⇒ 1

(s − 2)2= L[te2t ].

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.



(s − 3)

(s − 2)2=

a

(s − 2)+

b

(s − 2)2⇒ s − 3 = a(s − 2) + b


L[y ] =1

s − 2− 1

(s − 2)2.


L[eat ] =1

s − a⇒ 1

s − 2= L[e2t ],

L[tneat ] =n!

(s − a)(n+1)⇒ 1

(s − 2)2= L[te2t ].

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.



(s − 3)

(s − 2)2=

a

(s − 2)+

b

(s − 2)2⇒ s − 3 = a(s − 2) + b


L[y ] =1

s − 2− 1

(s − 2)2.


L[eat ] =1

s − a

⇒ 1

s − 2= L[e2t ],

L[tneat ] =n!

(s − a)(n+1)⇒ 1

(s − 2)2= L[te2t ].

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.



(s − 3)

(s − 2)2=

a

(s − 2)+

b

(s − 2)2⇒ s − 3 = a(s − 2) + b


L[y ] =1

s − 2− 1

(s − 2)2.


L[eat ] =1

s − a⇒ 1

s − 2= L[e2t ],

L[tneat ] =n!

(s − a)(n+1)⇒ 1

(s − 2)2= L[te2t ].

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.



(s − 3)

(s − 2)2=

a

(s − 2)+

b

(s − 2)2⇒ s − 3 = a(s − 2) + b


L[y ] =1

s − 2− 1

(s − 2)2.


L[eat ] =1

s − a⇒ 1

s − 2= L[e2t ],

L[tneat ] =n!

(s − a)(n+1)

⇒ 1

(s − 2)2= L[te2t ].

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.



(s − 3)

(s − 2)2=

a

(s − 2)+

b

(s − 2)2⇒ s − 3 = a(s − 2) + b


L[y ] =1

s − 2− 1

(s − 2)2.


L[eat ] =1

s − a⇒ 1

s − 2= L[e2t ],

L[tneat ] =n!

(s − a)(n+1)⇒ 1

(s − 2)2= L[te2t ].

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.


s − 2− 1

(s − 2)2and

1

s − 2= L[e2t ],

1

(s − 2)2= L[te2t ].


L[y ] = L[e2t ]− L[te2t ] = L[e2t − te2t

].

We conclude that y(t) = e2t − te2t . C

Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.


s − 2− 1

(s − 2)2and

1

s − 2= L[e2t ],

1

(s − 2)2= L[te2t ].


L[y ] = L[e2t ]− L[te2t ]

= L[e2t − te2t

].


Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.


s − 2− 1

(s − 2)2and

1

s − 2= L[e2t ],

1

(s − 2)2= L[te2t ].



].


Homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 0, y(0) = 1, y ′(0) = 1.


s − 2− 1

(s − 2)2and

1

s − 2= L[e2t ],

1

(s − 2)2= L[te2t ].



].




First, second, higher order equations.

Example

Use the Laplace Transform to find the solution of y (4) − 4y = 0,

y(0) = 1, y ′(0) = 1, y ′′(0) = −2, y ′′′(0) = 0.

Solution: Compute the L[ ] of the equation,

L[y (4)

]− 4L[y ] = 0.

[s4 L[y ]− s3 y(0)− s2 y ′(0)− s y ′′(0)− y ′′′(0)

]− 4L[y ] = 0.

[s4 L[y ]− s3 + 2s

]− 4L[y ] = 0 ⇒ (s4 − 4)L[y ] = s3 − 2s,

We obtain, L[y ] =s3 − 2s

(s4 − 4).


Example


y(0) = 1, y ′(0) = 1, y ′′(0) = −2, y ′′′(0) = 0.


L[y (4)

]− 4L[y ] = 0.

[s4 L[y ]− s3 y(0)− s2 y ′(0)− s y ′′(0)− y ′′′(0)

]− 4L[y ] = 0.

[s4 L[y ]− s3 + 2s

]− 4L[y ] = 0 ⇒ (s4 − 4)L[y ] = s3 − 2s,


(s4 − 4).


Example


y(0) = 1, y ′(0) = 1, y ′′(0) = −2, y ′′′(0) = 0.


L[y (4)

]− 4L[y ] = 0.

[s4 L[y ]− s3 y(0)− s2 y ′(0)− s y ′′(0)− y ′′′(0)

]− 4L[y ] = 0.

[s4 L[y ]− s3 + 2s

]− 4L[y ] = 0 ⇒ (s4 − 4)L[y ] = s3 − 2s,


(s4 − 4).


Example


y(0) = 1, y ′(0) = 1, y ′′(0) = −2, y ′′′(0) = 0.


L[y (4)

]− 4L[y ] = 0.

[s4 L[y ]− s3 y(0)− s2 y ′(0)− s y ′′(0)− y ′′′(0)

]− 4L[y ] = 0.

[s4 L[y ]− s3 + 2s

]− 4L[y ] = 0

⇒ (s4 − 4)L[y ] = s3 − 2s,


(s4 − 4).


Example


y(0) = 1, y ′(0) = 1, y ′′(0) = −2, y ′′′(0) = 0.


L[y (4)

]− 4L[y ] = 0.

[s4 L[y ]− s3 y(0)− s2 y ′(0)− s y ′′(0)− y ′′′(0)

]− 4L[y ] = 0.

[s4 L[y ]− s3 + 2s

]− 4L[y ] = 0 ⇒ (s4 − 4)L[y ] = s3 − 2s,


(s4 − 4).


Example


y(0) = 1, y ′(0) = 1, y ′′(0) = −2, y ′′′(0) = 0.


L[y (4)

]− 4L[y ] = 0.

[s4 L[y ]− s3 y(0)− s2 y ′(0)− s y ′′(0)− y ′′′(0)

]− 4L[y ] = 0.

[s4 L[y ]− s3 + 2s

]− 4L[y ] = 0 ⇒ (s4 − 4)L[y ] = s3 − 2s,


(s4 − 4).


Example


y(0) = 1, y ′(0) = 1, y ′′(0) = −2, y ′′′(0) = 0.

Solution: Recall: L[y ] =s3 − 2s

(s4 − 4).

L[y ] =s(s2 − 2)

(s2 − 2)(s2 + 2)⇒ L[y ] =

s

(s2 + 2).

The last expression is in the table of Laplace Transforms,

L[y ] =s(

s2 +[√

2]2) = L

[cos(

√2 t)

].

We conclude that y(t) = cos(√

2 t). C


Example


y(0) = 1, y ′(0) = 1, y ′′(0) = −2, y ′′′(0) = 0.


(s4 − 4).

L[y ] =s(s2 − 2)

(s2 − 2)(s2 + 2)⇒ L[y ] =

s

(s2 + 2).


L[y ] =s(

s2 +[√

2]2) = L

[cos(

√2 t)

].


2 t). C


Example


y(0) = 1, y ′(0) = 1, y ′′(0) = −2, y ′′′(0) = 0.


(s4 − 4).

L[y ] =s(s2 − 2)

(s2 − 2)(s2 + 2)

⇒ L[y ] =s

(s2 + 2).


L[y ] =s(

s2 +[√

2]2) = L

[cos(

√2 t)

].


2 t). C


Example


y(0) = 1, y ′(0) = 1, y ′′(0) = −2, y ′′′(0) = 0.


(s4 − 4).

L[y ] =s(s2 − 2)

(s2 − 2)(s2 + 2)⇒ L[y ] =

s

(s2 + 2).


L[y ] =s(

s2 +[√

2]2) = L

[cos(

√2 t)

].


2 t). C


Example


y(0) = 1, y ′(0) = 1, y ′′(0) = −2, y ′′′(0) = 0.


(s4 − 4).

L[y ] =s(s2 − 2)

(s2 − 2)(s2 + 2)⇒ L[y ] =

s

(s2 + 2).


L[y ] =s(

s2 +[√

2]2) = L

[cos(

√2 t)

].


2 t). C


Example


y(0) = 1, y ′(0) = 1, y ′′(0) = −2, y ′′′(0) = 0.


(s4 − 4).

L[y ] =s(s2 − 2)

(s2 − 2)(s2 + 2)⇒ L[y ] =

s

(s2 + 2).


L[y ] =s(

s2 +[√

2]2)

= L[cos(

√2 t)

].


2 t). C


Example


y(0) = 1, y ′(0) = 1, y ′′(0) = −2, y ′′′(0) = 0.


(s4 − 4).

L[y ] =s(s2 − 2)

(s2 − 2)(s2 + 2)⇒ L[y ] =

s

(s2 + 2).


L[y ] =s(

s2 +[√

2]2) = L

[cos(

√2 t)

].


2 t). C


Example


y(0) = 1, y ′(0) = 1, y ′′(0) = −2, y ′′′(0) = 0.


(s4 − 4).

L[y ] =s(s2 − 2)

(s2 − 2)(s2 + 2)⇒ L[y ] =

s

(s2 + 2).


L[y ] =s(

s2 +[√

2]2) = L

[cos(

√2 t)

].


2 t). C



Non-homogeneous IVP.

Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.

Solution: Compute the Laplace transform of the equation,

L[y ′′ − 4y ′ + 4y ] = L[3 sin(2t)].

The right-hand side above can be expressed as follows,

L[3 sin(2t)] = 3L[sin(2t)] = 32

s2 + 22=

6

s2 + 4.

Introduce this source term in the differential equation,

L[y ′′]− 4L[y ′] + 4L[y ] =6

s2 + 4.


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


L[y ′′ − 4y ′ + 4y ] = L[3 sin(2t)].


L[3 sin(2t)] = 3L[sin(2t)] = 32

s2 + 22=

6

s2 + 4.


L[y ′′]− 4L[y ′] + 4L[y ] =6

s2 + 4.


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


L[y ′′ − 4y ′ + 4y ] = L[3 sin(2t)].


L[3 sin(2t)] = 3L[sin(2t)] = 32

s2 + 22=

6

s2 + 4.


L[y ′′]− 4L[y ′] + 4L[y ] =6

s2 + 4.


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


L[y ′′ − 4y ′ + 4y ] = L[3 sin(2t)].


L[3 sin(2t)] = 3L[sin(2t)] = 32

s2 + 22=

6

s2 + 4.


L[y ′′]− 4L[y ′] + 4L[y ] =6

s2 + 4.


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


L[y ′′ − 4y ′ + 4y ] = L[3 sin(2t)].


L[3 sin(2t)] = 3L[sin(2t)]

= 32

s2 + 22=

6

s2 + 4.


L[y ′′]− 4L[y ′] + 4L[y ] =6

s2 + 4.


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


L[y ′′ − 4y ′ + 4y ] = L[3 sin(2t)].


L[3 sin(2t)] = 3L[sin(2t)] = 32

s2 + 22

=6

s2 + 4.


L[y ′′]− 4L[y ′] + 4L[y ] =6

s2 + 4.


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


L[y ′′ − 4y ′ + 4y ] = L[3 sin(2t)].


L[3 sin(2t)] = 3L[sin(2t)] = 32

s2 + 22=

6

s2 + 4.


L[y ′′]− 4L[y ′] + 4L[y ] =6

s2 + 4.


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


L[y ′′ − 4y ′ + 4y ] = L[3 sin(2t)].


L[3 sin(2t)] = 3L[sin(2t)] = 32

s2 + 22=

6

s2 + 4.


L[y ′′]− 4L[y ′] + 4L[y ] =6

s2 + 4.


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


L[y ′′ − 4y ′ + 4y ] = L[3 sin(2t)].


L[3 sin(2t)] = 3L[sin(2t)] = 32

s2 + 22=

6

s2 + 4.


L[y ′′]− 4L[y ′] + 4L[y ] =6

s2 + 4.


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.

Solution: Recall: L[y ′′]− 4L[y ′] + 4L[y ] =6

s2 + 4.


]− 4

[s L[y ]− y(0)

]+ 4L[y ] =

6

s2 + 4.

Rewrite the above equation,

(s2 − 4s + 4)L[y ] = (s − 4) y(0) + y ′(0) +6

s2 + 4.


(s2 − 4s + 4)L[y ] = s − 3 +6

s2 + 4.


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


s2 + 4.


]− 4

[s L[y ]− y(0)

]+ 4L[y ] =

6

s2 + 4.


(s2 − 4s + 4)L[y ] = (s − 4) y(0) + y ′(0) +6

s2 + 4.


(s2 − 4s + 4)L[y ] = s − 3 +6

s2 + 4.


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


s2 + 4.


]− 4

[s L[y ]− y(0)

]+ 4L[y ] =

6

s2 + 4.


(s2 − 4s + 4)L[y ] = (s − 4) y(0) + y ′(0) +6

s2 + 4.


(s2 − 4s + 4)L[y ] = s − 3 +6

s2 + 4.


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


s2 + 4.


]− 4

[s L[y ]− y(0)

]+ 4L[y ] =

6

s2 + 4.


(s2 − 4s + 4)L[y ] = (s − 4) y(0) + y ′(0) +6

s2 + 4.


(s2 − 4s + 4)L[y ] = s − 3 +6

s2 + 4.

Non-homogeneous IVP.Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.

Solution: Recall: (s2 − 4s + 4)L[y ] = s − 3 +6

s2 + 4.

Therefore, L[y ] =(s − 3)

(s2 − 4s + 4)+

6

(s2 − 4 + 4)(s2 + 4).

From an Example above: s2 − 4s + 4 = (s − 2)2,

L[y ] =1

s − 2− 1

(s − 2)2+

6

(s − 2)2(s2 + 4).

From an Example above we know that

L[e2t − te2t ] =1

s − 2− 1

(s − 2)2.



y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


s2 + 4.


(s2 − 4s + 4)+

6

(s2 − 4 + 4)(s2 + 4).


L[y ] =1

s − 2− 1

(s − 2)2+

6

(s − 2)2(s2 + 4).


L[e2t − te2t ] =1

s − 2− 1

(s − 2)2.



y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


s2 + 4.


(s2 − 4s + 4)+

6

(s2 − 4 + 4)(s2 + 4).


L[y ] =1

s − 2− 1

(s − 2)2+

6

(s − 2)2(s2 + 4).


L[e2t − te2t ] =1

s − 2− 1

(s − 2)2.



y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


s2 + 4.


(s2 − 4s + 4)+

6

(s2 − 4 + 4)(s2 + 4).


L[y ] =1

s − 2− 1

(s − 2)2+

6

(s − 2)2(s2 + 4).


L[e2t − te2t ] =1

s − 2− 1

(s − 2)2.



y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


s2 + 4.


(s2 − 4s + 4)+

6

(s2 − 4 + 4)(s2 + 4).


L[y ] =1

s − 2− 1

(s − 2)2+

6

(s − 2)2(s2 + 4).


L[e2t − te2t ] =1

s − 2− 1

(s − 2)2.


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.

Solution: Recall: L[y ] = L[e2t − te2t ] +6

(s − 2)2(s2 + 4).

Use Partial fractions to simplify the last term above.

Find constants a, b, c , d , such that

6

(s − 2)2 (s2 + 4)=

as + b

s2 + 4+

c

(s − 2)+

d

(s − 2)2

6

(s − 2)2 (s2 + 4)=

(as + b)(s − 2)2 + c(s − 2)(s2 + 4) + d(s2 + 4)

(s2 + 4)(s − 2)2

6 = (as + b)(s − 2)2 + c(s − 2)(s2 + 4) + d(s2 + 4).


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


(s − 2)2(s2 + 4).



6

(s − 2)2 (s2 + 4)=

as + b

s2 + 4+

c

(s − 2)+

d

(s − 2)2

6

(s − 2)2 (s2 + 4)=

(as + b)(s − 2)2 + c(s − 2)(s2 + 4) + d(s2 + 4)

(s2 + 4)(s − 2)2

6 = (as + b)(s − 2)2 + c(s − 2)(s2 + 4) + d(s2 + 4).


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


(s − 2)2(s2 + 4).



6

(s − 2)2 (s2 + 4)=

as + b

s2 + 4+

c

(s − 2)+

d

(s − 2)2

6

(s − 2)2 (s2 + 4)=

(as + b)(s − 2)2 + c(s − 2)(s2 + 4) + d(s2 + 4)

(s2 + 4)(s − 2)2

6 = (as + b)(s − 2)2 + c(s − 2)(s2 + 4) + d(s2 + 4).


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


(s − 2)2(s2 + 4).



6

(s − 2)2 (s2 + 4)=

as + b

s2 + 4+

c

(s − 2)+

d

(s − 2)2

6

(s − 2)2 (s2 + 4)=

(as + b)(s − 2)2 + c(s − 2)(s2 + 4) + d(s2 + 4)

(s2 + 4)(s − 2)2

6 = (as + b)(s − 2)2 + c(s − 2)(s2 + 4) + d(s2 + 4).


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


(s − 2)2(s2 + 4).



6

(s − 2)2 (s2 + 4)=

as + b

s2 + 4+

c

(s − 2)+

d

(s − 2)2

6

(s − 2)2 (s2 + 4)=

(as + b)(s − 2)2 + c(s − 2)(s2 + 4) + d(s2 + 4)

(s2 + 4)(s − 2)2

6 = (as + b)(s − 2)2 + c(s − 2)(s2 + 4) + d(s2 + 4).


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.

Solution: 6 = (as + b)(s − 2)2 + c(s − 2)(s2 + 4) + d(s2 + 4).

6 = (as + b)(s2 − 4s + 4) + c(s3 + 4s − 2s2 − 8) + d(s2 + 4)

6 = a(s3−4s2+4s)+b(s2−4s+4)+c(s3+4s−2s2−8)+d(s2+4).

6 = (a + c)s3 + (−4a + b − 2c + d)s2

+ (4a− 4b + 4c)s + (4b − 8c + 4d).

We obtain the system

a + c= 0, −4a + b − 2c + d= 0,

4a− 4b + 4c= 0, 4b − 8c + 4d = 6.


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


6 = (as + b)(s2 − 4s + 4) + c(s3 + 4s − 2s2 − 8) + d(s2 + 4)

6 = a(s3−4s2+4s)+b(s2−4s+4)+c(s3+4s−2s2−8)+d(s2+4).

6 = (a + c)s3 + (−4a + b − 2c + d)s2

+ (4a− 4b + 4c)s + (4b − 8c + 4d).


a + c= 0, −4a + b − 2c + d= 0,

4a− 4b + 4c= 0, 4b − 8c + 4d = 6.


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


6 = (as + b)(s2 − 4s + 4) + c(s3 + 4s − 2s2 − 8) + d(s2 + 4)

6 = a(s3−4s2+4s)+b(s2−4s+4)+c(s3+4s−2s2−8)+d(s2+4).

6 = (a + c)s3 + (−4a + b − 2c + d)s2

+ (4a− 4b + 4c)s + (4b − 8c + 4d).


a + c= 0, −4a + b − 2c + d= 0,

4a− 4b + 4c= 0, 4b − 8c + 4d = 6.


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


6 = (as + b)(s2 − 4s + 4) + c(s3 + 4s − 2s2 − 8) + d(s2 + 4)

6 = a(s3−4s2+4s)+b(s2−4s+4)+c(s3+4s−2s2−8)+d(s2+4).

6 = (a + c)s3 + (−4a + b − 2c + d)s2

+ (4a− 4b + 4c)s + (4b − 8c + 4d).


a + c= 0, −4a + b − 2c + d= 0,

4a− 4b + 4c= 0, 4b − 8c + 4d = 6.


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


6 = (as + b)(s2 − 4s + 4) + c(s3 + 4s − 2s2 − 8) + d(s2 + 4)

6 = a(s3−4s2+4s)+b(s2−4s+4)+c(s3+4s−2s2−8)+d(s2+4).

6 = (a + c)s3 + (−4a + b − 2c + d)s2

+ (4a− 4b + 4c)s + (4b − 8c + 4d).


a + c= 0, −4a + b − 2c + d= 0,

4a− 4b + 4c= 0, 4b − 8c + 4d = 6.


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.

Solution: The solution for this linear system is

a =3

8, b = 0, c = −3

8, d =

3

4.

6

(s − 2)2 (s2 + 4)=

3

8

s

s2 + 4− 3

8

1

(s − 2)+

3

4

1

(s − 2)2.

Use the table of Laplace Transforms

6

(s − 2)2 (s2 + 4)=

3

8L[cos(2t)]− 3

8L[e2t ] +

3

4L[te2t ].

6

(s − 2)2 (s2 + 4)= L

[3

8cos(2t)− 3

8e2t +

3

4te2t

].


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


a =3

8, b = 0, c = −3

8, d =

3

4.

6

(s − 2)2 (s2 + 4)=

3

8

s

s2 + 4− 3

8

1

(s − 2)+

3

4

1

(s − 2)2.


6

(s − 2)2 (s2 + 4)=

3

8L[cos(2t)]− 3

8L[e2t ] +

3

4L[te2t ].

6

(s − 2)2 (s2 + 4)= L

[3

8cos(2t)− 3

8e2t +

3

4te2t

].


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


a =3

8, b = 0, c = −3

8, d =

3

4.

6

(s − 2)2 (s2 + 4)=

3

8

s

s2 + 4− 3

8

1

(s − 2)+

3

4

1

(s − 2)2.


6

(s − 2)2 (s2 + 4)=

3

8L[cos(2t)]− 3

8L[e2t ] +

3

4L[te2t ].

6

(s − 2)2 (s2 + 4)= L

[3

8cos(2t)− 3

8e2t +

3

4te2t

].


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


a =3

8, b = 0, c = −3

8, d =

3

4.

6

(s − 2)2 (s2 + 4)=

3

8

s

s2 + 4− 3

8

1

(s − 2)+

3

4

1

(s − 2)2.


6

(s − 2)2 (s2 + 4)=

3

8L[cos(2t)]− 3

8L[e2t ] +

3

4L[te2t ].

6

(s − 2)2 (s2 + 4)= L

[3

8cos(2t)− 3

8e2t +

3

4te2t

].


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.

Solution: Summary: L[y ] = L[e2t − te2t ] +6

(s − 2)2(s2 + 4),

6

(s − 2)2 (s2 + 4)= L

[3

8cos(2t)− 3

8e2t +

3

4te2t

].

L[y(t)] = L[(1− t) e2t +

3

8(−1 + 2t) e2t +

3

8cos(2t)

].

We conclude that

y(t) = (1− t) e2t +3

8(2t − 1) e2t +

3

8cos(2t). C


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


(s − 2)2(s2 + 4),

6

(s − 2)2 (s2 + 4)= L

[3

8cos(2t)− 3

8e2t +

3

4te2t

].

L[y(t)] = L[(1− t) e2t +

3

8(−1 + 2t) e2t +

3

8cos(2t)

].

We conclude that

y(t) = (1− t) e2t +3

8(2t − 1) e2t +

3

8cos(2t). C


Example


y ′′ − 4y ′ + 4y = 3 sin(2t), y(0) = 1, y ′(0) = 1.


(s − 2)2(s2 + 4),

6

(s − 2)2 (s2 + 4)= L

[3

8cos(2t)− 3

8e2t +

3

4te2t

].

L[y(t)] = L[(1− t) e2t +

3

8(−1 + 2t) e2t +

3

8cos(2t)

].

We conclude that

y(t) = (1− t) e2t +3

8(2t − 1) e2t +

3

8cos(2t). C

The Laplace Transform of step functions (Sect. 4.3).

I Overview and notation.

I The definition of a step function.

I Piecewise discontinuous functions.

I The Laplace Transform of discontinuous functions.


Overview and notation.

Overview: The Laplace Transform method can be used to solveconstant coefficients differential equations with discontinuoussource functions.

Notation:If L[f (t)] = F (s), then we denote L−1[F (s)] = f (t).

Remark: One can show that for a particular type of functions f ,that includes all functions we work with in this Section, thenotation above is well-defined.

Example

From the Laplace Transform table we know that L[eat

]=

1

s − a.

Then also holds that L−1[ 1

s − a

]= eat . C





Example


]=

1

s − a.


s − a

]= eat . C





Example


]=

1

s − a.


s − a

]= eat . C





Example


]=

1

s − a.


s − a

]= eat . C





Example


]=

1

s − a.


s − a

]= eat . C







The definition of a step function.

DefinitionA function u is called a step function at t = 0 iff holds

u(t) =

{0 for t < 0,

1 for t > 0.

Example

Graph the step function values u(t) above, and the translationsu(t − c) and u(t + c) with c > 0.

Solution:u(t)

t

1

0

u(t − c)

t

1

0 c

u(t + c)

0

1

t− c

C



u(t) =

{0 for t < 0,

1 for t > 0.

Example


Solution:u(t)

t

1

0

u(t − c)

t

1

0 c

u(t + c)

0

1

t− c

C



u(t) =

{0 for t < 0,

1 for t > 0.

Example


Solution:u(t)

t

1

0

u(t − c)

t

1

0 c

u(t + c)

0

1

t− c

C



u(t) =

{0 for t < 0,

1 for t > 0.

Example


Solution:u(t)

t

1

0

u(t − c)

t

1

0 c

u(t + c)

0

1

t− c

C



u(t) =

{0 for t < 0,

1 for t > 0.

Example


Solution:u(t)

t

1

0

u(t − c)

t

1

0 c

u(t + c)

0

1

t− c

C


Remark: Given any function values f (t) and c > 0, then f (t − c)is a right translation of f and f (t + c) is a left translation of f .

Example

f ( t ) = e

0 t

a t

1

f ( t ) f ( t ) = e

0 t

1

c

a ( t − c )f ( t )

a t

0 t

f ( t )

1

f ( t ) = u ( t ) e f ( t ) = u ( t − c ) e

0 t

1

c

a ( t − c )f ( t )



Example

f ( t ) = e

0 t

a t

1

f ( t )

f ( t ) = e

0 t

1

c

a ( t − c )f ( t )

a t

0 t

f ( t )

1

f ( t ) = u ( t ) e f ( t ) = u ( t − c ) e

0 t

1

c

a ( t − c )f ( t )



Example

f ( t ) = e

0 t

a t

1

f ( t ) f ( t ) = e

0 t

1

c

a ( t − c )f ( t )

a t

0 t

f ( t )

1

f ( t ) = u ( t ) e f ( t ) = u ( t − c ) e

0 t

1

c

a ( t − c )f ( t )



Example

f ( t ) = e

0 t

a t

1

f ( t ) f ( t ) = e

0 t

1

c

a ( t − c )f ( t )

a t

0 t

f ( t )

1

f ( t ) = u ( t ) e

f ( t ) = u ( t − c ) e

0 t

1

c

a ( t − c )f ( t )



Example

f ( t ) = e

0 t

a t

1

f ( t ) f ( t ) = e

0 t

1

c

a ( t − c )f ( t )

a t

0 t

f ( t )

1

f ( t ) = u ( t ) e f ( t ) = u ( t − c ) e

0 t

1

c

a ( t − c )f ( t )







Piecewise discontinuous functions.

Example

Graph of the function b(t) = u(t − a)− u(t − b), with 0 < a < b.

Solution: The bump function b can be graphed as follows:

u(t −a)

0 a b

1

t

u(t −b)

0 a b

1

t

t0 a b

b(t)

1

C


Example



u(t −a)

0 a b

1

t

u(t −b)

0 a b

1

t

t0 a b

b(t)

1

C


Example



u(t −a)

0 a b

1

t

u(t −b)

0 a b

1

t

t0 a b

b(t)

1

C


Example



u(t −a)

0 a b

1

t

u(t −b)

0 a b

1

t

t0 a b

b(t)

1

C


Example

Graph of the function f (t) = eat[u(t − 1)− u(t − 2)

].

Solution:

a t

t1 2

1

a t

f ( t ) = e [ u ( t −1 ) − u ( t −2 ) ]

[ u ( t −1 ) − u ( t −2 ) ]

e

y

Notation: It is common in the literature to denote the functionvalues u(t − c) as uc(t).


Example


].

Solution:

a t

t1 2

1

a t

f ( t ) = e [ u ( t −1 ) − u ( t −2 ) ]

[ u ( t −1 ) − u ( t −2 ) ]

e

y



Example


].

Solution:

a t

t1 2

1

a t

f ( t ) = e [ u ( t −1 ) − u ( t −2 ) ]

[ u ( t −1 ) − u ( t −2 ) ]

e

y








The Laplace Transform of discontinuous functions.

TheoremGiven any real number c > 0, the following equation holds,

L[u(t − c)] =e−cs

s, s > 0.

Proof:

L[u(t − c)] =

∫ ∞

0

e−stu(t − c) dt =

∫ ∞

ce−st dt,

L[u(t − c)] = limN→∞

−1

s

(e−Ns − e−cs

)=

e−cs

s, s > 0.

We conclude that L[u(t − c)] =e−cs

s.




s, s > 0.

Proof:

L[u(t − c)] =

∫ ∞

0

e−stu(t − c) dt

=

∫ ∞

ce−st dt,


−1

s

(e−Ns − e−cs

)=

e−cs

s, s > 0.


s.




s, s > 0.

Proof:

L[u(t − c)] =

∫ ∞

0


∫ ∞

ce−st dt,


−1

s

(e−Ns − e−cs

)=

e−cs

s, s > 0.


s.




s, s > 0.

Proof:

L[u(t − c)] =

∫ ∞

0


∫ ∞

ce−st dt,


−1

s

(e−Ns − e−cs

)

=e−cs

s, s > 0.


s.




s, s > 0.

Proof:

L[u(t − c)] =

∫ ∞

0


∫ ∞

ce−st dt,


−1

s

(e−Ns − e−cs

)=

e−cs

s, s > 0.


s.


Example

Compute L[3u(t − 2)].

Solution: L[3u(t − 2)] = 3L[u(t − 2)] = 3e−2s

s.

We conclude: L[3u(t − 2)] =3e−2s

s. C

Example

Compute L−1[2e−3s

s

].

Solution: Since L[u(t − c)] =e−cs

s, for c = 3 we get

L−1[e−3s

s

]= u(t − 3). Therefore, L−1

[2e−3s

s

]= 2u(t − 3). C


Example


Solution: L[3u(t − 2)] = 3L[u(t − 2)]

= 3e−2s

s.


s. C

Example


s

].


s, for c = 3 we get

L−1[e−3s

s


[2e−3s

s

]= 2u(t − 3). C


Example



s.


s. C

Example


s

].


s, for c = 3 we get

L−1[e−3s

s


[2e−3s

s

]= 2u(t − 3). C


Example



s.


s. C

Example


s

].


s, for c = 3 we get

L−1[e−3s

s


[2e−3s

s

]= 2u(t − 3). C


Example



s.


s. C

Example


s

].


s, for c = 3 we get

L−1[e−3s

s


[2e−3s

s

]= 2u(t − 3). C


Example



s.


s. C

Example


s

].


s,

for c = 3 we get

L−1[e−3s

s


[2e−3s

s

]= 2u(t − 3). C


Example



s.


s. C

Example


s

].


s, for c = 3 we get

L−1[e−3s

s

]

= u(t − 3). Therefore, L−1[2e−3s

s

]= 2u(t − 3). C


Example



s.


s. C

Example


s

].


s, for c = 3 we get

L−1[e−3s

s

]= u(t − 3).

Therefore, L−1[2e−3s

s

]= 2u(t − 3). C


Example



s.


s. C

Example


s

].


s, for c = 3 we get

L−1[e−3s

s


[2e−3s

s

]= 2u(t − 3). C








Theorem (Translations)

If F (s) = L[f (t)] exists for s > a > 0 and c > 0, then holds

L[u(t − c)f (t − c)] = e−cs F (s), s > a.

Furthermore,

L[ect f (t)] = F (s − c), s > a + c .

Remark:

I L[translation (uf )

]= (exp)

(L[f ]

).

I L[(exp) (f )

]= translation

(L[f ]

).

Equivalent notation:

I L[u(t − c)f (t − c)] = e−cs L[f (t)],

I L[ect f (t)] = L[f ](s − c).




L[u(t − c)f (t − c)] = e−cs F (s), s > a.

Furthermore,

L[ect f (t)] = F (s − c), s > a + c .

Remark:


]= (exp)

(L[f ]

).

I L[(exp) (f )

]= translation

(L[f ]

).


I L[u(t − c)f (t − c)] = e−cs L[f (t)],

I L[ect f (t)] = L[f ](s − c).




L[u(t − c)f (t − c)] = e−cs F (s), s > a.

Furthermore,

L[ect f (t)] = F (s − c), s > a + c .

Remark:


]= (exp)

(L[f ]

).

I L[(exp) (f )

]= translation

(L[f ]

).


I L[u(t − c)f (t − c)] = e−cs L[f (t)],

I L[ect f (t)] = L[f ](s − c).




L[u(t − c)f (t − c)] = e−cs F (s), s > a.

Furthermore,

L[ect f (t)] = F (s − c), s > a + c .

Remark:


]= (exp)

(L[f ]

).

I L[(exp) (f )

]= translation

(L[f ]

).


I L[u(t − c)f (t − c)] = e−cs L[f (t)],

I L[ect f (t)] = L[f ](s − c).




L[u(t − c)f (t − c)] = e−cs F (s), s > a.

Furthermore,

L[ect f (t)] = F (s − c), s > a + c .

Remark:


]= (exp)

(L[f ]

).

I L[(exp) (f )

]= translation

(L[f ]

).


I L[u(t − c)f (t − c)] = e−cs L[f (t)],

I L[ect f (t)] = L[f ](s − c).


Example

Compute L[u(t − 2) sin(a(t − 2))

].

Solution: L[sin(at)] =a

s2 + a2, L[u(t − c)f (t − c)] = e−cs L[f (t)].

L[u(t − 2) sin(a(t − 2))

]= e−2s L[sin(at)] = e−2s a

s2 + a2.

We conclude: L[u(t − 2) sin(a(t − 2))

]= e−2s a

s2 + a2. C

Example

Compute L[e3t sin(at)

].

Solution: Recall: L[ect f (t)] = L[f ](s − c), L[sin(at)] =a

s2 + a2.

We conclude: L[e3t sin(at)

]=

a

(s − 3)2 + a2, with s > 3. C


Example


].


s2 + a2,

L[u(t − c)f (t − c)] = e−cs L[f (t)].

L[u(t − 2) sin(a(t − 2))


s2 + a2.


]= e−2s a

s2 + a2. C

Example


].


s2 + a2.


]=

a

(s − 3)2 + a2, with s > 3. C


Example


].



L[u(t − 2) sin(a(t − 2))


s2 + a2.


]= e−2s a

s2 + a2. C

Example


].


s2 + a2.


]=

a

(s − 3)2 + a2, with s > 3. C


Example


].



L[u(t − 2) sin(a(t − 2))

]= e−2s L[sin(at)]

= e−2s a

s2 + a2.


]= e−2s a

s2 + a2. C

Example


].


s2 + a2.


]=

a

(s − 3)2 + a2, with s > 3. C


Example


].



L[u(t − 2) sin(a(t − 2))


s2 + a2.


]= e−2s a

s2 + a2. C

Example


].


s2 + a2.


]=

a

(s − 3)2 + a2, with s > 3. C


Example


].



L[u(t − 2) sin(a(t − 2))


s2 + a2.


]= e−2s a

s2 + a2. C

Example


].


s2 + a2.


]=

a

(s − 3)2 + a2, with s > 3. C


Example


].



L[u(t − 2) sin(a(t − 2))


s2 + a2.


]= e−2s a

s2 + a2. C

Example


].


s2 + a2.


]=

a

(s − 3)2 + a2, with s > 3. C


Example


].



L[u(t − 2) sin(a(t − 2))


s2 + a2.


]= e−2s a

s2 + a2. C

Example


].

Solution: Recall: L[ect f (t)] = L[f ](s − c),

L[sin(at)] =a

s2 + a2.


]=

a

(s − 3)2 + a2, with s > 3. C


Example


].



L[u(t − 2) sin(a(t − 2))


s2 + a2.


]= e−2s a

s2 + a2. C

Example


].


s2 + a2.


]=

a

(s − 3)2 + a2, with s > 3. C


Example


].



L[u(t − 2) sin(a(t − 2))


s2 + a2.


]= e−2s a

s2 + a2. C

Example


].


s2 + a2.


]=

a

(s − 3)2 + a2, with s > 3. C


Example

Find the Laplace transform of f (t) =

{0, t < 1,

(t2 − 2t + 2), t > 1.

Solution: Using step function notation,

f (t) = u(t − 1)(t2 − 2t + 2).

Completing the square we obtain,

t2 − 2t + 2 = (t2 − 2t + 1)− 1 + 2 = (t − 1)2 + 1.

This is a parabola t2 translated to theright by 1 and up by one. Because ofthe step function, this is a discontinuousfunction at t = 1.

1

10

f ( t )

t


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.


f (t) = u(t − 1)(t2 − 2t + 2).


t2 − 2t + 2 = (t2 − 2t + 1)− 1 + 2 = (t − 1)2 + 1.


1

10

f ( t )

t


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.


f (t) = u(t − 1)(t2 − 2t + 2).


t2 − 2t + 2 = (t2 − 2t + 1)− 1 + 2 = (t − 1)2 + 1.


1

10

f ( t )

t


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.


f (t) = u(t − 1)(t2 − 2t + 2).


t2 − 2t + 2 = (t2 − 2t + 1)− 1 + 2

= (t − 1)2 + 1.


1

10

f ( t )

t


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.


f (t) = u(t − 1)(t2 − 2t + 2).


t2 − 2t + 2 = (t2 − 2t + 1)− 1 + 2 = (t − 1)2 + 1.


1

10

f ( t )

t


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.


f (t) = u(t − 1)(t2 − 2t + 2).


t2 − 2t + 2 = (t2 − 2t + 1)− 1 + 2 = (t − 1)2 + 1.

This is a parabola t2 translated to theright by 1 and up by one.

Because ofthe step function, this is a discontinuousfunction at t = 1.

1

10

f ( t )

t


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.


f (t) = u(t − 1)(t2 − 2t + 2).


t2 − 2t + 2 = (t2 − 2t + 1)− 1 + 2 = (t − 1)2 + 1.


1

10

f ( t )

t


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.


f (t) = u(t − 1)(t2 − 2t + 2).


t2 − 2t + 2 = (t2 − 2t + 1)− 1 + 2 = (t − 1)2 + 1.


1

10

f ( t )

t


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.

Solution: Recall: f (t) = u(t − 1)[(t − 1)2 + 1

].

This is equivalent to

f (t) = u(t − 1) (t − 1)2 + u(t − 1).

Since L[t2] = 2/s3, and L[u(t − c)g(t − c)] = e−cs L[g(t)], then

L[f (t)] = L[u(t − 1) (t − 1)2] + L[u(t − 1)] = e−s 2

s3+ e−s 1

s.

We conclude: L[f (t)] =e−s

s3

(2 + s2

). C


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.


].


f (t) = u(t − 1) (t − 1)2 + u(t − 1).


L[f (t)] = L[u(t − 1) (t − 1)2] + L[u(t − 1)] = e−s 2

s3+ e−s 1

s.


s3

(2 + s2

). C


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.


].


f (t) = u(t − 1) (t − 1)2 + u(t − 1).

Since L[t2] = 2/s3,

and L[u(t − c)g(t − c)] = e−cs L[g(t)], then

L[f (t)] = L[u(t − 1) (t − 1)2] + L[u(t − 1)] = e−s 2

s3+ e−s 1

s.


s3

(2 + s2

). C


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.


].


f (t) = u(t − 1) (t − 1)2 + u(t − 1).

Since L[t2] = 2/s3, and L[u(t − c)g(t − c)] = e−cs L[g(t)],

then

L[f (t)] = L[u(t − 1) (t − 1)2] + L[u(t − 1)] = e−s 2

s3+ e−s 1

s.


s3

(2 + s2

). C


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.


].


f (t) = u(t − 1) (t − 1)2 + u(t − 1).


L[f (t)] = L[u(t − 1) (t − 1)2] + L[u(t − 1)]

= e−s 2

s3+ e−s 1

s.


s3

(2 + s2

). C


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.


].


f (t) = u(t − 1) (t − 1)2 + u(t − 1).


L[f (t)] = L[u(t − 1) (t − 1)2] + L[u(t − 1)] = e−s 2

s3+ e−s 1

s.


s3

(2 + s2

). C


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.


].


f (t) = u(t − 1) (t − 1)2 + u(t − 1).


L[f (t)] = L[u(t − 1) (t − 1)2] + L[u(t − 1)] = e−s 2

s3+ e−s 1

s.


s3

(2 + s2

). C


Remark: The inverse of the formulas in the Theorem above are:

L−1[e−cs F (s)

]= u(t − c) f (t − c),

L−1[F (s − c)

]= ect f (t).

Example

Find L−1[ e−4s

s2 + 9

].

Solution: L−1[ e−4s

s2 + 9

]=

1

3L−1

[e−4s 3

s2 + 9

].

Recall: L−1[ a

s2 + a2

]= sin(at). Then, we conclude that

L−1[ e−4s

s2 + 9

]=

1

3u(t − 4) sin

(3(t − 4)

). C



L−1[e−cs F (s)

]= u(t − c) f (t − c),

L−1[F (s − c)

]= ect f (t).

Example

Find L−1[ e−4s

s2 + 9

].


s2 + 9

]=

1

3L−1

[e−4s 3

s2 + 9

].

Recall: L−1[ a

s2 + a2


L−1[ e−4s

s2 + 9

]=

1

3u(t − 4) sin

(3(t − 4)

). C



L−1[e−cs F (s)

]= u(t − c) f (t − c),

L−1[F (s − c)

]= ect f (t).

Example

Find L−1[ e−4s

s2 + 9

].


s2 + 9

]=

1

3L−1

[e−4s 3

s2 + 9

].

Recall: L−1[ a

s2 + a2


L−1[ e−4s

s2 + 9

]=

1

3u(t − 4) sin

(3(t − 4)

). C



L−1[e−cs F (s)

]= u(t − c) f (t − c),

L−1[F (s − c)

]= ect f (t).

Example

Find L−1[ e−4s

s2 + 9

].


s2 + 9

]=

1

3L−1

[e−4s 3

s2 + 9

].

Recall: L−1[ a

s2 + a2


L−1[ e−4s

s2 + 9

]=

1

3u(t − 4) sin

(3(t − 4)

). C



L−1[e−cs F (s)

]= u(t − c) f (t − c),

L−1[F (s − c)

]= ect f (t).

Example

Find L−1[ e−4s

s2 + 9

].


s2 + 9

]=

1

3L−1

[e−4s 3

s2 + 9

].

Recall: L−1[ a

s2 + a2

]= sin(at).

Then, we conclude that

L−1[ e−4s

s2 + 9

]=

1

3u(t − 4) sin

(3(t − 4)

). C



L−1[e−cs F (s)

]= u(t − c) f (t − c),

L−1[F (s − c)

]= ect f (t).

Example

Find L−1[ e−4s

s2 + 9

].


s2 + 9

]=

1

3L−1

[e−4s 3

s2 + 9

].

Recall: L−1[ a

s2 + a2


L−1[ e−4s

s2 + 9

]=

1

3u(t − 4) sin

(3(t − 4)

). C


Example

Find L−1[ (s − 2)

(s − 2)2 + 9

].

Solution: L−1[ s

s2 + a2

]= cos(at), L−1

[F (s − c)

]= ect f (t).

We conclude: L−1[ (s − 2)

(s − 2)2 + 9

]= e2t cos(3t). C

Example

Find L−1[ 2e−3s

s2 − 4

].

Solution: Recall: L−1[ a

s2 − a2

]= sinh(at)

and L−1[e−cs F (s)

]= u(t − c) f (t − c).


Example

Find L−1[ (s − 2)

(s − 2)2 + 9

].

Solution: L−1[ s

s2 + a2

]= cos(at),

L−1[F (s − c)

]= ect f (t).


(s − 2)2 + 9

]= e2t cos(3t). C

Example

Find L−1[ 2e−3s

s2 − 4

].


s2 − a2

]= sinh(at)


]= u(t − c) f (t − c).


Example

Find L−1[ (s − 2)

(s − 2)2 + 9

].

Solution: L−1[ s

s2 + a2

]= cos(at), L−1

[F (s − c)

]= ect f (t).


(s − 2)2 + 9

]= e2t cos(3t). C

Example

Find L−1[ 2e−3s

s2 − 4

].


s2 − a2

]= sinh(at)


]= u(t − c) f (t − c).


Example

Find L−1[ (s − 2)

(s − 2)2 + 9

].

Solution: L−1[ s

s2 + a2

]= cos(at), L−1

[F (s − c)

]= ect f (t).


(s − 2)2 + 9

]= e2t cos(3t). C

Example

Find L−1[ 2e−3s

s2 − 4

].


s2 − a2

]= sinh(at)


]= u(t − c) f (t − c).


Example

Find L−1[ (s − 2)

(s − 2)2 + 9

].

Solution: L−1[ s

s2 + a2

]= cos(at), L−1

[F (s − c)

]= ect f (t).


(s − 2)2 + 9

]= e2t cos(3t). C

Example

Find L−1[ 2e−3s

s2 − 4

].


s2 − a2

]= sinh(at)


]= u(t − c) f (t − c).


Example

Find L−1[ (s − 2)

(s − 2)2 + 9

].

Solution: L−1[ s

s2 + a2

]= cos(at), L−1

[F (s − c)

]= ect f (t).


(s − 2)2 + 9

]= e2t cos(3t). C

Example

Find L−1[ 2e−3s

s2 − 4

].


s2 − a2

]= sinh(at)


]= u(t − c) f (t − c).


Example

Find L−1[ (s − 2)

(s − 2)2 + 9

].

Solution: L−1[ s

s2 + a2

]= cos(at), L−1

[F (s − c)

]= ect f (t).


(s − 2)2 + 9

]= e2t cos(3t). C

Example

Find L−1[ 2e−3s

s2 − 4

].


s2 − a2

]= sinh(at)


]= u(t − c) f (t − c).


Example

Find L−1[ 2e−3s

s2 − 4

].

Solution: Recall:

L−1[ a

s2 − a2

]= sinh(at), L−1

[e−cs F (s)

]= u(t − c) f (t − c).

L−1[ 2e−3s

s2 − 4

]= L−1

[e−3s 2

s2 − 4

].

We conclude: L−1[ 2e−3s

s2 − 4

]= u(t − 3) sinh

(2(t − 3)

). C


Example

Find L−1[ 2e−3s

s2 − 4

].

Solution: Recall:

L−1[ a

s2 − a2

]= sinh(at), L−1

[e−cs F (s)

]= u(t − c) f (t − c).

L−1[ 2e−3s

s2 − 4

]= L−1

[e−3s 2

s2 − 4

].


s2 − 4

]= u(t − 3) sinh

(2(t − 3)

). C


Example

Find L−1[ 2e−3s

s2 − 4

].

Solution: Recall:

L−1[ a

s2 − a2

]= sinh(at), L−1

[e−cs F (s)

]= u(t − c) f (t − c).

L−1[ 2e−3s

s2 − 4

]= L−1

[e−3s 2

s2 − 4

].


s2 − 4

]= u(t − 3) sinh

(2(t − 3)

). C


Example

Find L−1[ e−2s

s2 + s − 2

].

Solution: Find the roots of the denominator:

s± =1

2

[−1±

√1 + 8

]⇒

{s+ = 1,

s− = −2.

Therefore, s2 + s − 2 = (s − 1) (s + 2).

Use partial fractions to simplify the rational function:

1

s2 + s − 2=

1

(s − 1) (s + 2)=

a

(s − 1)+

b

(s + 2),

1

s2 + s − 2= a(s + 2) + b(s − 1) =

(a + b) s + (2a− b)

(s − 1) (s + 2).


Example

Find L−1[ e−2s

s2 + s − 2

].


s± =1

2

[−1±

√1 + 8

]

⇒

{s+ = 1,

s− = −2.

Therefore, s2 + s − 2 = (s − 1) (s + 2).


1

s2 + s − 2=

1

(s − 1) (s + 2)=

a

(s − 1)+

b

(s + 2),

1

s2 + s − 2= a(s + 2) + b(s − 1) =

(a + b) s + (2a− b)

(s − 1) (s + 2).


Example

Find L−1[ e−2s

s2 + s − 2

].


s± =1

2

[−1±

√1 + 8

]⇒

{s+ = 1,

s− = −2.

Therefore, s2 + s − 2 = (s − 1) (s + 2).


1

s2 + s − 2=

1

(s − 1) (s + 2)=

a

(s − 1)+

b

(s + 2),

1

s2 + s − 2= a(s + 2) + b(s − 1) =

(a + b) s + (2a− b)

(s − 1) (s + 2).


Example

Find L−1[ e−2s

s2 + s − 2

].


s± =1

2

[−1±

√1 + 8

]⇒

{s+ = 1,

s− = −2.

Therefore, s2 + s − 2 = (s − 1) (s + 2).


1

s2 + s − 2=

1

(s − 1) (s + 2)=

a

(s − 1)+

b

(s + 2),

1

s2 + s − 2= a(s + 2) + b(s − 1) =

(a + b) s + (2a− b)

(s − 1) (s + 2).


Example

Find L−1[ e−2s

s2 + s − 2

].


s± =1

2

[−1±

√1 + 8

]⇒

{s+ = 1,

s− = −2.

Therefore, s2 + s − 2 = (s − 1) (s + 2).


1

s2 + s − 2=

1

(s − 1) (s + 2)

=a

(s − 1)+

b

(s + 2),

1

s2 + s − 2= a(s + 2) + b(s − 1) =

(a + b) s + (2a− b)

(s − 1) (s + 2).


Example

Find L−1[ e−2s

s2 + s − 2

].


s± =1

2

[−1±

√1 + 8

]⇒

{s+ = 1,

s− = −2.

Therefore, s2 + s − 2 = (s − 1) (s + 2).


1

s2 + s − 2=

1

(s − 1) (s + 2)=

a

(s − 1)+

b

(s + 2),

1

s2 + s − 2= a(s + 2) + b(s − 1) =

(a + b) s + (2a− b)

(s − 1) (s + 2).


Example

Find L−1[ e−2s

s2 + s − 2

].


s± =1

2

[−1±

√1 + 8

]⇒

{s+ = 1,

s− = −2.

Therefore, s2 + s − 2 = (s − 1) (s + 2).


1

s2 + s − 2=

1

(s − 1) (s + 2)=

a

(s − 1)+

b

(s + 2),

1

s2 + s − 2= a(s + 2) + b(s − 1)

=(a + b) s + (2a− b)

(s − 1) (s + 2).


Example

Find L−1[ e−2s

s2 + s − 2

].


s± =1

2

[−1±

√1 + 8

]⇒

{s+ = 1,

s− = −2.

Therefore, s2 + s − 2 = (s − 1) (s + 2).


1

s2 + s − 2=

1

(s − 1) (s + 2)=

a

(s − 1)+

b

(s + 2),

1

s2 + s − 2= a(s + 2) + b(s − 1) =

(a + b) s + (2a− b)

(s − 1) (s + 2).


Example

Find L−1[ e−2s

s2 + s − 2

].

Solution: Recall:1

s2 + s − 2=

(a + b) s + (2a− b)

(s − 1) (s + 2)

a + b = 0, 2a− b = 1, ⇒ a =1

3, b = −1

3.

L−1[ e−2s

s2 + s − 2

]=

1

3L−1

[e−2s 1

s − 1

]− 1

3L−1

[e−2s 1

s + 2

].

Recall: L−1[ 1

s − a

]= eat , L−1

[e−cs F (s)

]= u(t − c) f (t − c),

L−1[ e−2s

s2 + s − 2

]=

1

3u(t − 2) e(t−2) − 1

3u(t − 2) e−2(t−2).

Hence: L−1[ e−2s

s2 + s − 2

]=

1

3u(t − 2)

[e(t−2) − e−2(t−2)

]. C


Example

Find L−1[ e−2s

s2 + s − 2

].

Solution: Recall:1

s2 + s − 2=

(a + b) s + (2a− b)

(s − 1) (s + 2)

a + b = 0,

2a− b = 1, ⇒ a =1

3, b = −1

3.

L−1[ e−2s

s2 + s − 2

]=

1

3L−1

[e−2s 1

s − 1

]− 1

3L−1

[e−2s 1

s + 2

].

Recall: L−1[ 1

s − a

]= eat , L−1

[e−cs F (s)

]= u(t − c) f (t − c),

L−1[ e−2s

s2 + s − 2

]=

1

3u(t − 2) e(t−2) − 1

3u(t − 2) e−2(t−2).


s2 + s − 2

]=

1

3u(t − 2)

[e(t−2) − e−2(t−2)

]. C


Example

Find L−1[ e−2s

s2 + s − 2

].

Solution: Recall:1

s2 + s − 2=

(a + b) s + (2a− b)

(s − 1) (s + 2)

a + b = 0, 2a− b = 1,

⇒ a =1

3, b = −1

3.

L−1[ e−2s

s2 + s − 2

]=

1

3L−1

[e−2s 1

s − 1

]− 1

3L−1

[e−2s 1

s + 2

].

Recall: L−1[ 1

s − a

]= eat , L−1

[e−cs F (s)

]= u(t − c) f (t − c),

L−1[ e−2s

s2 + s − 2

]=

1

3u(t − 2) e(t−2) − 1

3u(t − 2) e−2(t−2).


s2 + s − 2

]=

1

3u(t − 2)

[e(t−2) − e−2(t−2)

]. C


Example

Find L−1[ e−2s

s2 + s − 2

].

Solution: Recall:1

s2 + s − 2=

(a + b) s + (2a− b)

(s − 1) (s + 2)

a + b = 0, 2a− b = 1, ⇒ a =1

3, b = −1

3.

L−1[ e−2s

s2 + s − 2

]=

1

3L−1

[e−2s 1

s − 1

]− 1

3L−1

[e−2s 1

s + 2

].

Recall: L−1[ 1

s − a

]= eat , L−1

[e−cs F (s)

]= u(t − c) f (t − c),

L−1[ e−2s

s2 + s − 2

]=

1

3u(t − 2) e(t−2) − 1

3u(t − 2) e−2(t−2).


s2 + s − 2

]=

1

3u(t − 2)

[e(t−2) − e−2(t−2)

]. C


Example

Find L−1[ e−2s

s2 + s − 2

].

Solution: Recall:1

s2 + s − 2=

(a + b) s + (2a− b)

(s − 1) (s + 2)

a + b = 0, 2a− b = 1, ⇒ a =1

3, b = −1

3.

L−1[ e−2s

s2 + s − 2

]=

1

3L−1

[e−2s 1

s − 1

]− 1

3L−1

[e−2s 1

s + 2

].

Recall: L−1[ 1

s − a

]= eat , L−1

[e−cs F (s)

]= u(t − c) f (t − c),

L−1[ e−2s

s2 + s − 2

]=

1

3u(t − 2) e(t−2) − 1

3u(t − 2) e−2(t−2).


s2 + s − 2

]=

1

3u(t − 2)

[e(t−2) − e−2(t−2)

]. C


Example

Find L−1[ e−2s

s2 + s − 2

].

Solution: Recall:1

s2 + s − 2=

(a + b) s + (2a− b)

(s − 1) (s + 2)

a + b = 0, 2a− b = 1, ⇒ a =1

3, b = −1

3.

L−1[ e−2s

s2 + s − 2

]=

1

3L−1

[e−2s 1

s − 1

]− 1

3L−1

[e−2s 1

s + 2

].

Recall: L−1[ 1

s − a

]= eat ,

L−1[e−cs F (s)

]= u(t − c) f (t − c),

L−1[ e−2s

s2 + s − 2

]=

1

3u(t − 2) e(t−2) − 1

3u(t − 2) e−2(t−2).


s2 + s − 2

]=

1

3u(t − 2)

[e(t−2) − e−2(t−2)

]. C


Example

Find L−1[ e−2s

s2 + s − 2

].

Solution: Recall:1

s2 + s − 2=

(a + b) s + (2a− b)

(s − 1) (s + 2)

a + b = 0, 2a− b = 1, ⇒ a =1

3, b = −1

3.

L−1[ e−2s

s2 + s − 2

]=

1

3L−1

[e−2s 1

s − 1

]− 1

3L−1

[e−2s 1

s + 2

].

Recall: L−1[ 1

s − a

]= eat , L−1

[e−cs F (s)

]= u(t − c) f (t − c),

L−1[ e−2s

s2 + s − 2

]=

1

3u(t − 2) e(t−2) − 1

3u(t − 2) e−2(t−2).


s2 + s − 2

]=

1

3u(t − 2)

[e(t−2) − e−2(t−2)

]. C


Example

Find L−1[ e−2s

s2 + s − 2

].

Solution: Recall:1

s2 + s − 2=

(a + b) s + (2a− b)

(s − 1) (s + 2)

a + b = 0, 2a− b = 1, ⇒ a =1

3, b = −1

3.

L−1[ e−2s

s2 + s − 2

]=

1

3L−1

[e−2s 1

s − 1

]− 1

3L−1

[e−2s 1

s + 2

].

Recall: L−1[ 1

s − a

]= eat , L−1

[e−cs F (s)

]= u(t − c) f (t − c),

L−1[ e−2s

s2 + s − 2

]=

1

3u(t − 2) e(t−2) − 1

3u(t − 2) e−2(t−2).


s2 + s − 2

]=

1

3u(t − 2)

[e(t−2) − e−2(t−2)

]. C


Example

Find L−1[ e−2s

s2 + s − 2

].

Solution: Recall:1

s2 + s − 2=

(a + b) s + (2a− b)

(s − 1) (s + 2)

a + b = 0, 2a− b = 1, ⇒ a =1

3, b = −1

3.

L−1[ e−2s

s2 + s − 2

]=

1

3L−1

[e−2s 1

s − 1

]− 1

3L−1

[e−2s 1

s + 2

].

Recall: L−1[ 1

s − a

]= eat , L−1

[e−cs F (s)

]= u(t − c) f (t − c),

L−1[ e−2s

s2 + s − 2

]=

1

3u(t − 2) e(t−2) − 1

3u(t − 2) e−2(t−2).


s2 + s − 2

]=

1

3u(t − 2)

[e(t−2) − e−2(t−2)

]. C

Documents

The Laplace Transform (Sect. 4.1)