Docsity
Log inSign up
Docsity
Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity

Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips
Guidelines and tips
Sell on Docsity
Sell on Docsity
Log inSign up

Prepare for your exams

Study with the several resources on Docsity

Find documents

Prepare for your exams with the study notes shared by other students like you on Docsity

Search Store documents

The best documents sold by students who completed their studies

Search through all study resources
Docsity AINEW

Summarize your documents, ask them questions, convert them into quizzes and concept maps

Explore questions

Clear up your doubts by reading the answers to questions asked by your fellow students

Earn points to download

Earn points by helping other students or get them with a premium plan

Share documents
download point icon20 Points

For each uploaded document

Answer questions
download point icon5 Points

For each given answer (max 1 per day)

All the ways to get free points
Get points immediately

Choose a premium plan with all the points you need

Study Opportunities

Choose your next study program

Get in touch with the best universities in the world. Search through thousands of universities and official partners

Community

Ask the community

Ask the community for help and clear up your study doubts

University Rankings

Discover the best universities in your country according to Docsity users

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

From our blog

Exams and Study
Go to the blog

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

Tennessee Technological University (TTU)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

koofers-user-muy
koofers-user-muy 🇺🇸

10 documents

1 / 4

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
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

Partial preview of the text

Download Trial-Solution Method for Homogeneous and Nonhomogeneous Differential Equations and more Assignments Mechanical Engineering in PDF only on Docsity!

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

− t − t

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 .