Differential Equations *RESOLVED

[kevinthenerd]

I'm working on studying for my differentials equation exam, and I'm wondering if the back of my textbook is wrong.

Use Laplace Transformations to find the solution to this IVP:

y'' + 3y' + 2y = 4t^2

y(0)=0

y'(0)=0

I started with something that looks like this:

s^2 L[y] - sy'(0) - y(0) + 3sL[y] - 3y(0) + 2L[y] = 4 * (2 / s^3)

Am I braindead? Help me out here. Thanks. The final exam is 12.5 hours from now, and I'm running low on sleep from last night.

 

[hypn0tik]

Originally posted by: kevinthenerd
I'm working on studying for my differentials equation exam, and I'm wondering if the back of my textbook is wrong.

Use Laplace Transformations to find the solution to this IVP:

y'' + 3y' + 2y = 4t^2

y(0)=0

y'(0)=0

I started with something that looks like this:

s^2 L[y] - sy'(0) - y(0) + 3sL[y] - 3y(0) + 2L[y] = 4 * (2 / s^3)

Am I braindead? Help me out here. Thanks. The final exam is 12.5 hours from now, and I'm running low on sleep from last night.

What exactly is the problem?

 

[kevinthenerd]

Originally posted by: hypn0tik
What exactly is the problem?

Find y by itself.

Originally posted by: kevinthenerd
Use Laplace Transformations to find the solution to this IVP:

y'' + 3y' + 2y = 4t^2

y(0)=0

y'(0)=0

 

[hypn0tik]

Well, this line looks right to me:

s^2 L[y] - sy'(0) - y(0) + 3sL[y] - 3y(0) + 2L[y] = 4 * (2 / s^3)


you know that both y(0) and y'(0) = 0, so simplify and isolate L[y]

L[y] (s^2 + 3s + 2) = 8/s^3

--> L[y] = 8/(s^3)(s^2 + 3s + 2)

L[y] = 8/[s^3 (s+2)(s+1)]

Find the partial fraction expansion and you should have your result.

 

[esun]

Originally posted by: kevinthenerd
I'm working on studying for my differentials equation exam, and I'm wondering if the back of my textbook is wrong.

Use Laplace Transformations to find the solution to this IVP:

y'' + 3y' + 2y = 4t^2

y(0)=0

y'(0)=0

I started with something that looks like this:

s^2 L[y] - sy'(0) - y(0) + 3sL[y] - 3y(0) + 2L[y] = 4 * (2 / s^3)

Am I braindead? Help me out here. Thanks. The final exam is 12.5 hours from now, and I'm running low on sleep from last night.

If I may assume you took the Laplace transform correctly (don't wanna go look up the terms in tables right now), then substitude in the initial conditions and get

s^2 L[y] + 3s L[y] + 2 L[y] = 8 / s^3

So you'll get

L[y] = 8 / [(s^3 * (s^2 + 3s + 2))]

Then just find a way to take the inverse transform of that sucker (probably have to factor and do a partial fraction expansion).

 

[MSCoder610]

s^2 L[y] - sy'(0) - y(0) + 3sL[y] - 3y(0) + 2L[y] = 4 * (2 / s^3)

Doesn't quite look right to me-
L[y']=sL[y]-y(0)
L[y'']=sL[y']-y'(0)
L[y'']=s(L[y]-y(0))-y'(0) = s^2 L[y] - s y(0) - y'(0)

 

[kevinthenerd]

I got it. The partial fractions messed me up. Thanks.

 

[hypn0tik]

Good luck.

Just out of curiosity, what does your exam cover?