First Order Eq Ns

8/12/2019 First Order Eq Ns

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Examples for obtaining a set of first-order ODE from a higher order ODE

1) )2cos(3)1()1( 2 x x y x y x ⋅⋅⋅⋅=⋅++′′′⋅− π

First rewrite the equation and re-label the derivatives as follows:

)2cos(3)1(00)1( 2 x x y x y y y x ⋅⋅⋅⋅=⋅++′⋅+′′⋅+′′′⋅− π

Note: the higher order derivative is labeled differently.

The set of first-order differential equations is:

20

2

21

10

)1(

)1()2cos(3

−

⋅+−⋅⋅⋅⋅=

′

=′

=′

x

z x x x z

z z

z z

π

with initial conditions

)0()0(

)0()0(

)0()0(

2

1

0

y z

y z

y z

′′=

′=

=

Note: the number of first-order differential equations is three because this is a third orderdifferential equation.

Note: there will be three initial conditions because this is a third order differentialequation.

In Mathcad notation we can write:

−

⋅+−⋅⋅⋅⋅

=

20

2

1

)1(

)1()2cos(3

:),(

x

z x x x

z

z

z x D

π

and

′′

′=

)0(

)0(

)0(

:

y

y

y

zinitial

Note: actual numbers should be substituted in the expression for the initial conditions.

z2’ z2 z1 z0



2) )cos(2120)sin(5

t e

t x xt

t ⋅⋅++=⋅+⋅

⋅ ω

This equation is already a first order differential equation ! Solving for the first order

derivative, xD :

)sin(

20)cos(215

t

xt e

t

xt

⋅−⋅⋅++

=⋅

ω

D with initial condition )0( x

Note: the number of first-order differential equations is one because this is a first order

differential equation.

Note: there is one initial condition because this is a first order differential equation.


)sin(

20)cos(21

:),(5

t

xt e

t

xt Dt

⋅−⋅⋅++

=⋅

ω

and )0(: x xinitial =

Note: actual numbers must be substituted in the expression for the initial condition.



3) t t y y yt yt ⋅+=+⋅+⋅⋅+⋅+ 9)cos()sin(5)3sin()2( 2

First rewrite the equation and re-label the derivatives as follows:

t t y y yt y yt ⋅+=+⋅+⋅⋅+⋅+⋅+

9)cos()sin(5)3sin(0)2(

2

Note: )sin( y becomes )sin( 0w .

Note: the higher order derivative is labeled differently.


)2(

9)cos()sin(5)3sin(

2

0012

21

10

+

⋅++−⋅−⋅⋅−=

=

=

t

t t wwwt w

ww

ww

D

D

D

with initial conds.

)0()0(

)0()0(

)0()0(

2

1

0

yw

yw

yw

DD

D

=

=

=

Note: the number of first-order differential equations is three because this is a third order

differential equation.

Note: there will be three initial conditions because this is a third order differential

equation.


+

⋅++−⋅−⋅⋅−

=

)2(

9)cos()sin(5)3sin(

:),(

2

001

2

1

t

t t wwwt

w

w

wt D and

=

)0(

)0(

)0(

:

y

y

y

winitial


2w w2 w1 w0 w0



4) Here we have two differential equations that are “coupled” (connected):

0)(

)sin()(

2

211

=−−⋅

⋅⋅=⋅+−+⋅+⋅

y xk ym

t F xk y xk xb xm

ω

First rewrite the first equation and re-label the derivatives as follows:

)sin()( 211 t F yk xk k xb xm ⋅⋅=⋅−⋅++⋅+⋅ ω

Now rewrite the second equation and re-label the derivatives.

002 =⋅−⋅+⋅+⋅ xk yk y ym

Note: notice how the labels for the second equation are a continuation of the labels for the

first equation.


[ ]

[ ]022

3

32

2021111

10

1

)sin()(1

zk zk m

z

z z

t F zk zk k zbm

z

z z

⋅+⋅−⋅=

=

⋅⋅+⋅+⋅+−⋅−⋅=

=

ω

and initial conds

)0()0(

)0()0(

)0()0(

)0()0(

3

2

1

0

y z

y z

x z

x z

=

=

=

=

Note: the number of first-order differential equations is four because two second order

differential equations are used.

Note: there will be four initial conditions. Two for each second order differential

equation.

1 z z1 z0

z3 3 z z2




[ ]

[ ]

⋅+⋅−⋅

⋅⋅+⋅+⋅+−⋅−⋅

=

022

3

202111

1

1

)sin()(1

:),(

zk zk m

z

t F zk zk k zbm

z

zt D

ω

and

=

)0(

)0()0(

)0(

:

y

y x

x

zinitial


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First Order Eq Ns