Solve $y''-xy'+y = 1 , y(0)=1, y'(0) = 2 $ with Laplace transform

Question

What's making me get stuck is the Laplace transform of $xy'$. I'm aware of different methods of solving this, but it's asking specifically for Laplace transform.

Answer 1

Note that if $y(x)$ has Laplace transform $\bar y (s)$, then

• $y'$ has Laplace transform $s\bar y - y(0)$
• $y''$ has Laplace transform $s^2\bar y - sy(0)-y'(0)$
• $xy'$ has Laplace transform $-\frac{d}{ds}\bar y$

So, $y''-xy + y = 1, y(0)=1, y'(0)=2$ becomes:

$$(s^2\bar y-s-2)+\frac{d}{ds}(s\bar y-1)+\bar y=\frac{1}{s}$$

Rearranging gives

$$s\bar y'+(s^2+2)\bar y=\frac{1}{s}+s+2$$

From which we see:

$$s^2\bar y'+(s^3+2s)\bar y=s^2+2s+1$$

Letting $u=s^2\bar y$, this becomes:

$$u'+sy=(s+1)^2$$

Using the integrating factor of $e^{s^2/2}$, we see:

$$\frac{d}{ds}(ue^{s^2/2})=(s+1)^2e^{s^2/2}$$

A bit of careful integration by parts gets us that $ue^{s^2/2}=(s+2)e^{s^2/2}+C$. We now have $u(s)=s+2+Ce^{-s^2/2}$, whence $\bar y=\frac{1}{s}+\frac{2}{s^2}+CL^{-1}(e^{-s^2/2})$, where $L^{-1}$ represents the Laplace transform.

Noting that $2x+1$ has Laplace transform $\frac{1}{s}+\frac{2}{s^2}$, we find this to be the solution for $C=0$.

I'm not entirely sure how to reconcile/deal with the case for $C\neq 0$. We've found a solution, in any case.