Is the Laplace Transform Correct for the Differential Equation y-8y'+20y=te^t?

[Jeff12341234]

y"-8y'+20y=te^t, y(0)=0, y'(0)=0

I need to know if I made a mistake in getting to the step below:

L^-1{ 1/[(s-1)²(s²-8s+20)] }

because when I solve that, I get something pretty gnarly..

 

[Mark44]

y"-8y'+20y=te^t, y(0)=0, y'(0)=0

I need to know if I made a mistake in getting to the step below:

L^-1{ 1/[(s-1)²(s²-8s+20)] }

because when I solve that, I get something pretty gnarly..

Show us how you got to that step.

 

[Jeff12341234]

 

[Mark44]

So y(t) = L^-1(Y(s)) = L^-1(1 /[(s - 1)²(s² -8s + 20)]

Break up the right side using partial fractions, and then you can take inverse Laplace transforms of them separately. This is how you should decompose the right side.

$$ \frac{1}{(s - 1)^2(s^2 - 8s + 20)} = \frac{A}{s - 1} + \frac{B}{(s - 1)^2} + \frac{Cs + D}{s^2 - 8s + 20}$$

When you figure out A, B, C, and D, check your work to make sure you haven't made an error. Then we can talk about the final step.

 

[Jeff12341234]

So I was doing it right.. The answer was kinda big so I thought I made a mistake somewhere.

 

[Jeff12341234]

So is that right?

 

[Dick]

So is that right?

Yes, it is. Well done.

 

[Mark44]

So is that right?

In your post just before the one I'm quoting, you showed the solution you found. I assumed that you had checked your solution to verify that it works.