How to use the Laplace Transform on a DE with a time coefficent

[sliperyfrog]

Homework Statement

ty'' + y' = 2t², y(0) = 0

Homework Equations

laplace(f''(t)) = s²laplace(f(t)) -sf(0) - f'(0)
(-1) (d/ds) (F(s))

The Attempt at a Solution

I know how to solve the problem except for the ty'' part. I tried using the equation and I got -d/ds(s²Y(s) - 0 - f'(0)) which becomes -2sY(s). But this is wrong because it does not lead to the correct answer y = (2/3)t³+c₁t²

 

[Ray Vickson]

Homework Statement

ty'' + y' = 2t², y(0) = 0

Homework Equations

laplace(f''(t)) = s²laplace(f(t)) -sf(0) - f'(0)
(-1) (d/ds) (F(s))

The Attempt at a Solution

I know how to solve the problem except for the ty'' part. I tried using the equation and I got -d/ds(s²Y(s) - 0 - f'(0)) which becomes -2sY(s). But this is wrong because it does not lead to the correct answer y = (2/3)t³+c₁t²

Your equation for $Y(s)$ is a first-order differential equation, since it involves both $Y(s)$ and $dY(s)/ds$.

 

[sliperyfrog]

So in the problem when I get the dY(s)/ds after using the product rule what am I supposed to do? Is it the same as having the laplace of y'(t)?

 

[LCKurtz]

I got -d/ds(s²Y(s) - 0 - f'(0)) which becomes -2sY(s).

No. $\frac d {ds} (s^2Y(s) \ne -2sY(s))$. You need the product rule.

So in the problem when I get the dY(s)/ds after using the product rule what am I supposed to do? Is it the same as having the LaPlace of y'(t)?

No, it isn't the same as the LaPlace of $y'(t)$. You already know the LaPlace of that is $sY(s)-y(0)$. Assuming you do correctly use the product rule, as has already been pointed out to you, you will get a differential equation in $Y(s)$ which you have to solve for $Y(s)$ before you try to take the inverse.

 

[sliperyfrog]

Thanks for the help I got the answer!