Undetermined Coefficients problem

  • #1
DanielJackins
40
0

Homework Statement



Find the unique solution to the differential equation

y''+4y'−5y=−435sin(2t)

satisfying the initial conditions y(0)=29 and y'(0)=47

The Attempt at a Solution



So (I think) I found the correct general solution. I let y(t)=Acos(2t)+Bsin(2t) and eventually worked it out to be A = -24 and B = 27, giving me y(t) = -24cos(2t)+27sin(2t). I found the auxiliary equation as well, y = c1e^t+c2e^(-5t), and then if I'm not mistaken I add them together.

This gives me y = c1e^t+c2e^(-5t)-24cos(2t)+27sin(2t), and y' = c1e^t-5c2e^(-5t)+48sin(2t)+54cos(2t). I subbed in the initial values, and got c1 = 43 and c2 = 10. So finally, I had y(t) = 43e^t+10e^(-5t)-24cos(2t)+27sin(2t), which was incorrect. Anybody see what I'm doing wrong?

Thanks!
 

Answers and Replies

  • #2
shiri
85
0

Homework Statement



Find the unique solution to the differential equation

y''+4y'−5y=−435sin(2t)

satisfying the initial conditions y(0)=29 and y'(0)=47

The Attempt at a Solution



So (I think) I found the correct general solution. I let y(t)=Acos(2t)+Bsin(2t) and eventually worked it out to be A = -24 and B = 27, giving me y(t) = -24cos(2t)+27sin(2t). I found the auxiliary equation as well, y = c1e^t+c2e^(-5t), and then if I'm not mistaken I add them together.

This gives me y = c1e^t+c2e^(-5t)-24cos(2t)+27sin(2t), and y' = c1e^t-5c2e^(-5t)+48sin(2t)+54cos(2t). I subbed in the initial values, and got c1 = 43 and c2 = 10. So finally, I had y(t) = 43e^t+10e^(-5t)-24cos(2t)+27sin(2t), which was incorrect. Anybody see what I'm doing wrong?

Thanks!

It appears you didn't do a characteristic equation.

eg. y'' + 0y' -1y = -578sin(4t)

r^2 - 1 = 0
 
  • #3
DanielJackins
40
0
I thought I did? I said r^2+4r-5 = 0, which gave me r = 1 and r = -5, giving me the equation y = c1e^t+c2e^(-5t)
 
  • #4
Lunat1c
66
0

Homework Statement



Find the unique solution to the differential equation

y''+4y'−5y=−435sin(2t)

satisfying the initial conditions y(0)=29 and y'(0)=47

The Attempt at a Solution



So (I think) I found the correct general solution. I let y(t)=Acos(2t)+Bsin(2t) and eventually worked it out to be A = -24 and B = 27, giving me y(t) = -24cos(2t)+27sin(2t). I found the auxiliary equation as well, y = c1e^t+c2e^(-5t), and then if I'm not mistaken I add them together.

This gives me y = c1e^t+c2e^(-5t)-24cos(2t)+27sin(2t), and y' = c1e^t-5c2e^(-5t)+48sin(2t)+54cos(2t). I subbed in the initial values, and got c1 = 43 and c2 = 10. So finally, I had y(t) = 43e^t+10e^(-5t)-24cos(2t)+27sin(2t), which was incorrect. Anybody see what I'm doing wrong?

Thanks!

Your auxiliar equation is r^2 + 4r - 5 = 0, factorising that you get r=1 and r=-5 just like you said. The complementary function should be y(t) = c1e^(r1t) + c2e^(r2t) where r1 and r2 are the roots of the auxiliar equation.

Regarding A and B of your trial solution, I'd check your working again because one of the signs might be wrong.

Once you find both the constants of the trial solution, you add it to the complementary function, differentiate as needed to find the particular solution.
 
  • #5
shiri
85
0
I thought I did? I said r^2+4r-5 = 0, which gave me r = 1 and r = -5, giving me the equation y = c1e^t+c2e^(-5t)

Yup, my mistake

didn't you do a second derivative for yp? furthermore, both yp and y'p are wrong.
 
Last edited:
  • #6
DanielJackins
40
0
Sorry, can you point out where I went wrong?
 
  • #7
Lunat1c
66
0
Sorry, I misread the first part. You made a mistake when calculating A and B most probably. I got A=24 and B=27
 
  • #8
shiri
85
0
Sorry, can you point out where I went wrong?

derive yp = Asin(2t) + Bcos(2t) until you solve the second derivative, y''p
 
  • #9
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,585
2,223
You just made an algebraic mistake somewhere. As Lunat1c said, A should equal +24, not -24.
 

Suggested for: Undetermined Coefficients problem

Replies
7
Views
1K
Replies
5
Views
740
  • Last Post
Replies
14
Views
2K
Replies
3
Views
664
  • Last Post
Replies
1
Views
1K
Replies
2
Views
206
  • Last Post
Replies
5
Views
2K
Replies
2
Views
193
  • Last Post
Replies
7
Views
683
Top