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Trial-Solution Method for Homogeneous and Nonhomogeneous Differential Equations, Assignments of Mechanical Engineering

Step-by-step solutions to various differential equations using the trial-solution method. The equations are homogeneous and nohomogeneous, and the solutions involve finding the values of constants and trigonometric functions. The document also includes initial conditions.

Typology: Assignments

Pre 2010

Uploaded on 07/30/2009

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koofers-user-muy 🇺🇸

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ME 3050 – DMC (SP/09)
1
HW #1: Differential Equations
Solve the following differential equations using the Trial-Solution method.
1. 13)0(4)0(045
=
==++ xxxxx &&&&
Homogeneous eq.
045
=
++ xxx &&&
Let st
ex =, then we get 0)45( 2=++ st
ess Î 045
2=++ ss
0)4)(1(45
2=++=++ ssss Î 4or 1
=
s Î tt eex 4
or
=
tt
cBeeAxx 4 +==
From the I.C.’s,
=
=
==
=+=
3
1
134)0(
4)0(
B
A
BAx
BAx
&
Therefore, tt eetx 4
3)( += .
2. 3)0(1)0(096
=
==++ xxxxx &&&&
Homogeneous eq.
096
=
++ xxx &&&
Let st
ex =, then we get 0)96( 2=++ st
ess Î 096
2=++ ss
0)3(96 22 =+=++ sss Î 3or 3
=
s Î tt etex 33 or
=
tt
cetBeAxx 33 +==
From the I.C.’s,
=
=
=+=
==
6
1
33)0(
1)0(
B
A
BAx
Ax
&
Therefore, tt etex 33 6 += .
3. 1)0(0)0(816
=
==+ xxxx &&&
Homogeneous eq.
016 =+ xx
&&
Let st
ex =, then we get 0)16( 2=+ st
es Î 016
2=+s
016
2=+s Î js 4±= Î tjtj eex 44 or
=Î )4sin(or )4cos( ttx
=
)4sin()4cos( tBtAxc+=
Nohomogeneous eq.
816 =+ xx
&&
Try a constant solution, Cx
=
, then 816
=
C Î 2
1
=C
5.0=
p
x
Therefore, 5.0)4sin()4cos(
+
+
=+= tBtAxxx pc
From the I.C.’s,
pf3
pf4

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HW #1: Differential Equations

Solve the following differential equations using the Trial-Solution method.

  1. & x & + 5 x &+ 4 x = 0 x ( 0 )= 4 x &( 0 )=− 13
    • Homogeneous eq.

x && + 5 x &+ 4 x = 0

Let

st x = e , then we get ( 5 4 ) 0

2

    • =

st s s e Î 5 4 0

2 s + s + =

2 s + s + = s + s + = Î s =− 1 or− 4 Î

t t x e e

4 or

− −

t t x xc Ae Be

− − 4 = = +

  • From the I.C.’s,

B

A

x A B

x A B

  • Therefore,

t t x t e e

4 ( ) 3

− − = +.

  1. & x & + 6 x &+ 9 x = 0 x ( 0 )= 1 x &( 0 )= 3
    • Homogeneous eq.

x && + 6 x &+ 9 x = 0

Let

st x = e , then we get ( 6 9 ) 0

2

    • =

st s s e Î 6 9 0

2 s + s + =

2 2 s + s + = s + = Î s =− 3 or− 3 Î

t t x e te

3 3 or

− −

t t x xc Ae Bte

− 3 − 3 = = +

  • From the I.C.’s,

B

A

x A B

x A

  • Therefore,

t t x e te

3 3 6

− − = +.

  1. & x & + 16 x = 8 x ( 0 )= 0 x &( 0 )= 1
    • Homogeneous eq.

x &&+ 16 x = 0

Let

st x = e , then we get ( 16 ) 0

2

  • =

st s e Î 16 0

2 s + =

2 s + = Î s = ± 4 j Î

j t jt x e e

4 4 or

− = Î x =cos( 4 t )orsin( 4 t )

x (^) c = A cos( 4 t )+ B sin( 4 t )

  • Nohomogeneous eq.

x &&+ 16 x = 8

Try a constant solution, x = C , then 16 C = 8 Î 2

C =

xp = 0. 5

Therefore, x = xc + xp = A cos( 4 t )+ B sin( 4 t )+ 0. 5

  • From the I.C.’s,

B

A

x B

x A

  • Therefore, 2

sin( 4 ) 4

cos( 4 ) 2

x ( t )= − t + t +.

  1. & x & + 6 x &+ 34 x = 68 x ( 0 )= 1 x &( 0 )= 0
    • Homogeneous eq.

x && + 6 x &+ 34 x = 0

Let

st x = e , then we get ( 6 34 ) 0

2

    • =

st s s e Î 6 34 0

2 s + s + =

2 s + s + = Î s = − 3 ± j 5 Î

t jt t jt x e e e e

3 5 3 5 or

− − − = Î

cos( 5 )or sin( 5 )

3 3 x e t e t

tt

cos( 5 ) sin( 5 )

3 3 x Ae t Be t

t t c

− − = +

  • Nohomogeneous eq.

x && + 6 x &+ 34 x = 68

Try a constant solution, x = C , then 34 C = 68 Î C = 2

xp = 2

Therefore, cos( 5 ) sin( 5 ) 2

3 3 = + = + +

− − x x x Ae t Be t

t t c p

  • From the I.C.’s,

B

A

x A B

x A

  • Therefore, sin( 5 ) 2 5

( ) cos( 5 )

3 3 = − − +

− − x t e t e t

t t .

2 & x & + 5 x &+ 2 x = 5 x ( 0 )= x & =

  • Homogeneous eq.

2 x && + 5 x &+ 2 x = 0

Let

st x = e , then we get ( 2 5 2 ) 0

2

    • =

st s s e Î 2 5 2 0

2 s + s + =

2 s + s + = Î ( 2 s + 1 )( s + 2 )= 0 Î s =− 1 2 or− 2 Î

t

t x e e 2 2

1

or

t

t xc Ae Be 2 2

1 −

− = +

  • Nohomogeneous eq.

2 x && + 5 x &+ 2 x = 5

Try a constant solution, x = C , then 2 C = 5 Î C = 52

xp =

Therefore, 2

1

= + = + +

t

t x xc xp Ae Be

  • From the I.C.’s,

xp =

Therefore, 4

= + = cos( )+ sin()+

− − x x x Ae t Be t

t t c p

  • From the I.C.’s,

B

A

x A B

x A

  • Therefore,

4

sin() 4

cos() 4

− − x t e t e t

t t .