Undetermined Coefficients problem

[DanielJackins]

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!

 

[shiri]

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

 

[DanielJackins]

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)

 

[Lunat1c]

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.

 

[shiri]

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.

 

[DanielJackins]

Sorry, can you point out where I went wrong?

 

[Lunat1c]

Sorry, I misread the first part. You made a mistake when calculating A and B most probably. I got A=24 and B=27

 

[shiri]

Sorry, can you point out where I went wrong?

derive y_p = Asin(2t) + Bcos(2t) until you solve the second derivative, y''_p

 

[vela]

You just made an algebraic mistake somewhere. As Lunat1c said, A should equal +24, not -24.