Having trouble solving this Laplace transform:

y''−8y'+16y=t, y(0)=0, y'(0)=1

Here's what I have so far:

y(s)(s-4)^2=1/s+1

y(s)=1/[(s^2)(s-4)^2]+1/(s-4)^2

partial decomp of 1/[(s^2)(s-4)^2]: 1/16(1/s^2)-1/8[1/(s-4)]+1/16[1/(s-4)^2]

y(s)=1/16(1/s^2)-1/8[1/(s-4)]+17/16[1/(s-4)^2]

y(t)=t/16-[e^(4t)]/8+[17te^(4t)]/16

But I get the wrong answer. Help?

Question

Having trouble solving this Laplace transform:

y''−8y'+16y=t, y(0)=0, y'(0)=1

Here's what I have so far:

y(s)(s-4)^2=1/s+1

y(s)=1/[(s^2)(s-4)^2]+1/(s-4)^2

partial decomp of 1/[(s^2)(s-4)^2]: 1/16(1/s^2)-1/8[1/(s-4)]+1/16[1/(s-4)^2]

y(s)=1/16(1/s^2)-1/8[1/(s-4)]+17/16[1/(s-4)^2]

y(t)=t/16-[e^(4t)]/8+[17te^(4t)]/16

But I get the wrong answer. Help?

kainui · Answer

Give me some time to refresh and work through this. It's been a while for me but I think I can help you out.

anonymous · Answer

Alright, thank you.

kainui · Answer

I think I found your problem. When you did partial fractions you needed to add an extra term:

$$\Large \frac{1}{s^2(s-4)^2}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s-4}+\frac{D}{(s-4)^2}$$

anonymous · Answer

Yeah, I did that. I got A=0. Maybe I did the partial fraction decomp wrong?

kainui · Answer

Yeah I think so, try it again. I think that's it.

anonymous · Answer

Alright, give me 2 min.

kainui · Answer

No problem, there's a lot going on in that mess haha.

anonymous · Answer

Still getting A=0.

anonymous · Answer

And yeah, there's way too much arithmetic in differential equations.

kainui · Answer

Hmm, I'm getting A=1/32. Ok, I guess it won't be very pleasant for me to type out how I got there.

kainui · Answer

How do you solve partial fractions?

kainui · Answer

I basically distribute everything out, collect terms of the same powers of s to create 4 equations. Then you can either put it in a matrix to solve it or whatever.

anonymous · Answer

Ahh, wait a minute. I put in Cs(s-4) instead of Cs^2(s-4). Let me rework it.

kainui · Answer

Cool =)

anonymous · Answer

You get A=1/32, B=1/16,C=-1/32,D=1/16?

anonymous · Answer

Just submitted answer. It's right!! Second time in two days partial fraction decomp has tripped me up. Anyways, thanks for the help. I really appreciate it.

kainui · Answer

Awesome, glad I could help. It wasn't even really the transform part that's hard haha.

anonymous · Answer

Yeah, I feel like this class tests me more on attention to detail than anything else.

kainui · Answer

Yeah, in a way these classes sort of battle hardened me to calculating things faster and more precisely, which is kind of good in a way. It's like you have to go through calculus to master algebra haha.