First Order Differential

Equations

National Chiao Tung University

Chun-Jen Tsai

9/26/2011

2/50

Separable Equations (1/2)

� A first order DE of the form

is said to be separable or to have separable variables.Divide both side by h(y), the DE becomes

Integrating both sides w.r.t. x, we have

( ) ( ) ( ) ( ) )(/1)(where, yhyfyfxgyhxgdx

dy===

( ) ( )xgdx

dyyf =

( ) ( ) .)( Cdxxgdxdx

dyxyf += ∫∫

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Separable Equations (2/2)

Cancelling the differential term dx, we have

If the two anti-derivatives

can be found, we have the family of equations

F(y(x)) = G(x) + C

that conforms to the differential equation.

( ) ( ) .Cdxxgdyyf += ∫∫

( ) ( )∫∫ == dxxgxGdyyfyF )(and)(

4/50

Example: dy/dx ＝ –6xy, y(0) = 7

� Rearranging the equation, we have dy/y = –6xdx,

therefore,

Since y(0) = 7, we have

ln y = –3x2 + C, or y = e–3x eC .

The particular solution becomes

.3ln

6/

2Cxy

xdxydy

+−=→

−=∫ ∫

2

.723x

ey−=

5/50

Example: dy/dx ＝ –x/y, y(4) = –3

� Since ∫ ydy = –∫ xdx, we have y2/2 = –x2/2 + c1.

The solution must pass (4, –3), thus, c1 = 25/2.

→ the solution is the lower half-circle of radius 5centered at (0, 0).

x

y

(4, –3)

family of solutions:x2 + y2 = C

A particular solution:x2 + y2 = 25

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Losing a Solution

� Some care should be exercised when separating variables, since the variable divisors could be zero in

some cases.

� If r is a zero of h(y), then y = r is a constant solution of

the DE. However, y = r may not show up in the family

of solutions. Recall that this is called a singular solution.

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Example: dy/dx = y2－4

� Since y2－4 is separable

∫ ∫=−

dxy

dy

42 ∫ ∫=

+−

−dxdy

yy 2

4/1

2

4/1

12ln4

12ln

4

1cxyy +=+−−

24

22

2or,4

2

2ln

cxe

y

ycx

y

y +=+

−+=

+

−

x

x

ce

cey

4

4

1

12

−

+=

Note: The solutions y = ±2 have been excluded in the first step!

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Example: dy/dx = 6x(y – 1)2/3, y(1) = 1

� Separation of variables gives

Note that C = –1 fulfills the initial condition. In addition, a singular solution y = 1 is lost in the first step of

separation of variables.

→ The IVP has non-unique solutions.

( )

( )

( ) .1)(

1

213

1

32

23/1

3/2

Cxxy

Cxy

xdxdyy

++=→

+=−→

=−

∫ ∫

9/50

Natural Growth and Decay Models

� The differential equation

is a widely used model for natural phenomena whose rate of change over time is proportional to its current

size.

� For example, if a population has birth and death rates

β and δ, respectively. The differential change in size P(t) of the population changes is

constant.aiswhere, kkxdt

dx=

.)()()(

lim0

Pt

ttPttP

dt

dP

tδβ

δβ−=

∆

∆−∆=

→∆

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Solution to the Natural Growth Eq.

� The DE dx/dt = kx is a easily separable equation:

If the initial condition is x(0) = x0, we have the

particular solution

.0,ln

1

1

kteCxxCktx

kdtdxx

=→≠+=→

= ∫∫

.0

ktexx =

Natural growth Natural decay

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

1

10

2

3

4

5

6

7

8

9

k > 0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

1.5

2

2.5

3

k < 0

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Other Mathematical Models

� Cooling and Heating Models (by Newton):

� Torricelli’s Law of draining tank:

).( TAkdt

dT−=

.2gyacdt

dV−=

hole area a

y(t)V(t)

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Linear First Order DE

� A first-order differential equation of the form:

(1)

is said to be a linear equation. When g(x) = 0, the

linear equation is said to be homogeneous, otherwise it is non-homogeneous.

� Dividing both side of (1) by the leading coefficient a1(x), we have the standard form:

)()()( 01 xgyxadx

dyxa =+

)()( xQyxPdx

dy=+

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Solution by Integrating Factors

� We can solve for the solution by multiply both sides of the standard form by e∫P(x)dx, thus:

� e∫P(x)dx is called the integrating factor.

∫=∫+∫ dxxPdxxPdxxP

exQyexPdx

dye

)()()(

)()(

∫=

∫ dxxPdxxP

exQyedx

d )()(

)(

CdxexQyedxxPdxxP

+∫=∫∫

)()(

)(

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The Property of Standard Form

� The solution of dy/dx + P(x)y = Q(x) is the sum of two solutions, namely, y = yc + yp, where yc is a solution of

the associated homogeneous equation

and yp is a particular solution of the original DE.

� Verification:

0)( =+ yxPdx

dy

[ ] [ ] )()()()(

)(0

xQyxPdx

dyyxP

dx

dyyyxPyy

dx

dy

xf

p

p

cc

pcpc =

++

+=+++

44 344 2144 344 21

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Homogeneous Solution yc

� The homogeneous equation is separable:

Integrating both sides and solving for y gives

yc = ce–∫P(x)dx,

� For convenience, let yc = cy1(x), where y1 = e–∫P(x)dx.

0)( =+ yxPdx

dy.0)( =+ dxxP

y

dy

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Particular Solution yp (1/2)

� A particular solution can be found using a procedure called variation of parameters:

Assume that yp = u(x)y1(x) = u(x)e–∫P(x)dx, substitute yp

into dy/dx + P(x)y = Q(x), we have

)()( 111 xQuyxP

dx

duy

dx

dyu =++

)()( 1

0

11 xQ

dx

duyyxP

dx

dyu =+

+

44 344 21∫= dx

xy

xQu

)(

)(

1

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Particular Solution yp (2/2)

Therefore,

The general solution form of a 1-st order linear DE is:

.)()(

)( )()()(

1

1 ∫∫ ∫∫=∫

==

−−dxxQeeedx

xy

xQuyy

dxxPdxxPdxxP

p

.)()()()(

∫∫∫+∫=+=

−−dxxQeeceyyy

dxxPdxxPdxxP

pc

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Dropping Integrating Factor Constant

� Note that you do not need to keep the constant when computing the anti-derivative of the integrating factor.Assume that G(x) is the anti-derivative of P(x), since

e∫P(x)dx = eG(x) + C = C1eG(x),

The constant C1 = eC will simply be cancelled out on

both side of the differential equation.

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Example: Solve dy/dx – 3y = 6

� Solution:

xdx

ee3)3( −−

=∫

[ ] xxeye

dx

d 33 6 −− =

xxxeye

dx

dye

333 63 −−− =−

ceyexx +−= −− 33 2

∞<<∞−+−= xceyx ,2 3

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General Solution on I

� If P(x) and Q(x) in the standard form are continuous on an open interval I, then

(2)

is a general solution of dy/dx + P(x)y = Q(x). That is, every solutions on I has the form of (2). In another

words, there is no singular solution for the linear first order differential equation on I.

∫∫∫+∫=+=

−−dxxQeeceyyy

dxxPdxxPdxxP

pc )()()()(

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Particular Solution on I

� Given an initial condition y(x0) = y0 to the linear first order DE dy/dx + P(x)y = Q(x) on I where P(x) and Q(x)

are continuous, the particular solution of the DE has

the form:

Note that it is easy to verify that y(x0) = y0.

∫+

∫= ∫

− x

x

duuPdttP

dttQeyexy

t

x

x

x

0

00 )()()(

0

)(

22/50

Example: (x2–9)dy/dx + xy = 0

� Solution

P(x) is continuous on (–∞, –3), (–3, 3), and (3, ∞), but solution only defined on the first and third intervals. The integrating factor is:

)9()(,0

9 22 −=∴=

−+

x

xxPy

x

x

dx

dy

929ln2/1)9/(22/1)9/( 222

−==∫=∫ −−−xeee

xxxdxxxdx

[ ] 092 =− yxdx

d

3or3,92 −<>=− xxCyx

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Example: IVP y' + y = x, y(0) = 4

� Since P(x) = 1 and Q(x) = x are continuous on (–∞, ∞), we have integrating factor e∫dx = ex:

[ ] xxxeye

dx

d=

∞<<∞−+−= −xcexy

x ,)1(

4

2

0

- 4

- 2

- 4 - 2 0 2 4

x

c = 0

y

c > 0

c < 0

0

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Example: Discontinuous f(x)

� Find a continuous function satisfying

and y(0) = 0.

Solution:

→ find c2 so that

>

≤≤==+

1,0

10,1)(),(

x

xxfxfy

dx

dy

>

≤≤−=

−

−

1,

10,1

2 xec

xey

x

x

)1()(lim1

yxyx

=+→

f(x)

x

y

x1

25/50

Solution by Substitutions

� If the DE has the form

and assume the solution has the form y = β(x, v), where v is a sub-function of f(x, y).

By chain rule:

we have:

The solution of dv/dx = g(x, v), when substituted into the equation

y = β(x, v), is a solution to the original DE dy/dx = f(x, y).

),,( yxfdx

dy=

),,(),(),( yxfdx

dvvx

dx

dxvx

dx

dyvx =+= ββ

[ ]).,(

),(),()),(,(

vxgvx

vxvxxf

dx

dv

v

x =−

=β

ββ

26/50

Example: dy/dx = (x + y + 3)2

� Let v = x + y + 3, thus, y = v – x – 3. Then

and

The solution can be obtained by separating variables:

So, v = tan(x – C) and the family of solutions of the DE

is:y = tan(x – C) – x – 3.

,1−=dx

dv

dx

dy

.tan1

1

2Cv

v

dvx +=

+= −

∫

.1 2v

dx

dv+=

27/50

Principles of Solution by Substitution

� There are many cases of DE which can be solved by substitution:

� Part of the DE has the form f(ax + by + c)

� The DE has homogeneous equation form

� Bernoulli equations

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Special Form of f(x, y)

� A differential equation of the form

where b≠0 can always be reduced to an equation

with separable variables by means of the substitutionv = ax＋by＋c.

),( cbyaxfdx

dy++=

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Example: dy/dx = (–2x + y)2–7, y(0) = 0

� Solution:Let u =－2x＋y, then du/dx =－2＋dy/dx.

The DE can be reduced to du/dx = u2－9.

dxduuu

dxuu

du=

+−

−→=

+−→

3

1

3

1

6

1

)3)(3(

166

1 ,3

3

3

3ln

6

1 cxecce

u

ucx

u

u==

+

−→+=

+

−→

x

x

ce

cexy

6

6

1

)1(32

−

++=→

.1,0)0( −==→ cyx

y

30/50

Homogeneous Equations (1/2)

� If f(tx, ty) = tαf(x, y) for some real number α, then f is said to be a homogeneous equation of degree α.

Example: f (x, y) = x3＋y3 is a homogeneous equation of degree 3.

� Similarly, a first-order DE in differential form

M(x, y)dx＋N(x, y)dy = 0

is said to be homogeneous if both M and N are homogeneous functions of the same degree.

31/50

Homogeneous Equations (2/2)

� A homogeneous first-order differential equation can also be expressed in the form:

Therefore, if v = y/x , we have y = vx and

→ the DE becomes a separable one.

),(x

yF

dx

dy=

,dx

dvxv

dx

dy+=

,)( vvFdx

dvx −=

32/50

Example: (x2＋y2)dx＋(x2－xy)dy = 0

� Solution:M and N are homogeneous with degree 2.

Let y = vx and dy = v dx＋x dv, we have

→ (x2＋v2x2)dx＋(x2－vx2)[v dx＋x dv] = 0

→ x2 (1＋v)dx＋x3 (1－v) dv = 0

Therefore

01

210

1

1=+

++−→=+

+

−

x

dxdv

vx

dxdv

v

v

cxvv lnln1ln2 =+++− y = . . .

33/50

Bernoulli’s Equation

� The differential equation

where n is any real number, is called Bernoulli’s

equation. Note that for n = 0 and n = 1, it is linear. For any other n, the substitution v = y1–n reduces any

equation of this form into a linear equation.

nyxQyxP

dx

dy)()( =+

).()1()()1( xQnvxPndx

dv−=−+

34/50

Example: x dy/dx + y = x2y2

� Rewrite the equation in Bernoulli’s form:

substitute with y = v–1 and dy/dx = –v–2dv/dx.

, the integrating factor on (0, ∞)

is e–∫dx/x = x–1, we have

x–1v = –x + c → y = 1/(– x2 + cx).

,1 2

xyyxdx

dy=+

xvxdx

dv−=−→

1

[ ] 11 −=−vx

dx

d

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Total Differential of a Function

� In calculus, if z = f (x, y) is a function of two variables with continuous first partial derivatives in a region R

of the xy-plane, the (total) derivative of z w.r.t. the

independent variable x can be written as:

In differential form, we have

dz is called the total differential of z.

.dx

dy

y

f

x

f

dx

dz

∂

∂+

∂

∂=

.dyy

fdx

x

fdz

∂

∂+

∂

∂=

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Exact Differential

� If f (x, y) = c, we then have:

→ The one-parameter family of curves f (x, y) = c,is a set of solutions to the first order DE.

� A differential expression M(x, y) dx + N(x, y) dy is an

exact differential in a region R of the xy-plane if it corresponds to the differential of some function f(x, y).

That is,M(x, y) = fx , N(x, y) = fy.

.0=∂

∂+

∂

∂dy

y

fdx

x

f

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Exact Differential Equations

� A first-order differential equation of the form

M(x, y) dx + N(x, y) dy = 0

is said to be an exact differential equation if the

expression on the left side is an exact differential.

� Example: f (x, y) = xy3 = c, then y3dx + 3xy2dy = 0

is an exact differential equation. However, the equivalent differential equation ydx + 3xdy = 0 is not

exact.

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Criterion for an Exact Differential

� Theorem: Let M(x, y) and N(x, y) be continuous and

have continuous first partial derivatives in a rectangular region R defined by a < x < b, c < y < d.

Then a necessary and sufficient condition thatM(x, y) dx + N(x, y) dy be an exact differential is

.x

N

y

M

∂

∂=

∂

∂

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Proof of the Sufficiency

� If M(x, y) dx + N(x, y) dy is exact, there exists some function f such that for all x in R,

Therefore, M(x, y) = ∂f/∂x, and N(x, y) = ∂f/∂y, and

.),(),( dyy

fdx

x

fdyyxNdxyxM

∂

∂+

∂

∂=+

.2

x

N

y

f

xxy

f

x

f

yy

M

∂

∂=

∂

∂

∂

∂=

∂∂

∂=

∂

∂

∂

∂=

∂

∂

40/50

Proof of the Necessity (1/2)

� Note that we have

where g(y), is some function of y. Since we want

therefore,

If g'(y) is a function of y alone, integrating g'(y) w.r.t. y,

gives us the solution.

),(),(),(),( ygdxyxMyxfyxMx

f+=→=

∂

∂∫

),(),(),(),( ygdxyxMy

yxNyxNy

f′+

∂

∂=→=

∂

∂∫

∫∂

∂−=′ dxyxM

yyxNyg ),(),()(

41/50

Proof of the Necessity (2/2)

� Since

g'(y) would be a function of y alone if ∂M/∂y = ∂N/∂x is

true because ∂/∂x[N(x, y) – (∂/∂ y)∫ M(x, y)dx] = 0.

In this case, the solution is

,),(),(y

M

x

NdxyxM

yyxN

x ∂

∂−

∂

∂=

∂

∂−

∂

∂∫

.),(),(),(),( ∫ ∫∫

∂

∂−+= dydxyxM

yyxNdxyxMyxf

42/50

Observations

� The method of solution can start from ∂f/∂y = N(x, y)

as well. Then, we have

∫ += )(),(),( yhdyyxNyxf

∫∂

∂−= dyyxN

xyxMyh ),(),()('

43/50

Example: 2xy dx＋(x2－1)dy = 0

� Solution:Since M(x, y) = 2xy, N(x, y) = x2 – 1, we have

∂M/∂y = 2x = ∂N/∂x so the equation is exact and

there exists f(x, y) such that ∂f/∂x = 2xy and∂f/∂y = x2 – 1.

Integrating the first equation→ f(x, y) = x2y + g(y)

Take the partial derivative of y, equate it with N(x, y),we have x2 + g'(y) = x2 – 1. Therefore, g'(y) = –1, and

f (x, y) = x2y – y. The implicit solution is x2y – y = c.

44/50

Reducible 2nd-Order Equations

� Recall that the general form of a 2nd-order DE of y is

F(x, y, y', y") = 0.

If either y or x is missing from F(), then the equation

can be reduced to a 1st-order DE by substitution.

� If F(x, y', y") = 0, the substitution p = y', y" = dp/dx results in

the 1st-order DE F(x, p, p′) = 0. If the solution is p(x, C1), then

� If F(y, y', y") = 0, the substitution p = y', y" = p·dp/dy results in the 1st-order DE F(y, p, p·dp/dy) = 0. If the solution is p(y, C1),

then

.),(

)( 2

1

CCyp

dydy

dy

dxyx +== ∫∫

.),()()( 21 CdxCxpdxxyxy +=′= ∫∫

45/50

Example: xy″+2y′=6x

� Substitute by p = y′, we have

The integrating factor is e∫(2/x)dx = x2, and hence

p = 2x + C1/x2.

Therefore,y(x) = x2 – C1/x + C2.

.62

62 =+→=+ pxdx

dpxp

dx

dpx

46/50

Example: yy″ = (y′)2

� Let p = y′ and assume y and y′ are non-negative,we have

By separation of variables,

Hence

The family of solutions is y(x) = e(C1x – C2) = AeBx.

Note that the solution holds even for A < 0.

.2p

dy

dpyp =

.lnln 1yCpCypy

dy

p

dp=→+=→= ∫∫

.ln11

21

1

∫ +==→== Cyy

dyxC

yCpdy

dx

47/50

� Growth and Decayx(t) is the growth and decay

of a population.

� Half LifeA(t) is the amount of plutonium remaining at any time.

Linear Models (1/2)

00 )(, xtxkxdt

dx==

0)0(, AAkAdt

dA==

t = 2.71t

P

3P0

P0

P(t) = P0e0.4055t

48/50

Linear Models (2/2)

� Newton’s Law of Cooling

where T(t) is the temperature of the object at t > 0, Tm

is the ambient temperature.

� Mixtures

),( mTTkdt

dT−=

21)leaving of rate()entering of rate( RRdt

dA−=−=

A

t500

A=600

49/50

Nonlinear Models (1/2)

� Population Dynamics

If P(t) denotes the size of a population at time t, the

model for exponential growth begins with the assumption that dP/dt = kP for some k > 0.

If the rate of growth only depends on the number

present, the model becomes

or)(/

PfP

dtdP= )(PPf

dt

dP=

50/50

Nonlinear Models (2/2)

� Logistic Equations

If the carrying capacity of the environment is K, we

have f(K) = 0. If the initial condition is f(0) = r, we

have

or

−= P

K

rrP

dt

dP

)( bPaPdt

dP−=

r

K P

f (P)