Use Laplace transform to solve $xy”+(1-x)y’+my=0$

Question

Use Laplace transform to solve $xy''+(1-x)y'+my=0$.

(a) $y=\displaystyle\frac{e^t}{k!} \frac{d^k}{dt^k}(t^{-k}e^{-t})$

(b) $y=\displaystyle\frac{e^t}{k} \frac{d^k}{dt^k}(t^k e^{-t})$

(c) $y=\displaystyle\frac{e^t}{k!} \frac{d^k}{dt^k}(t^k e^{-t})$

Answer 1

Hint.

$$ \mathcal{L}\left(x f(x)\right) = -\frac{d}{ds}\mathcal{L}\left(f(x)\right) $$

then

$$ -\frac{d}{ds}\mathcal{L}\left(\ddot y\right)+\frac{d}{ds}\mathcal{L}\left(\dot y\right)+\mathcal{L}\left(\dot y\right)+m\mathcal{L}\left(y\right)=0 $$

with

$$ \mathcal{L}\left(\ddot y\right) = s^2Y(s)-\dot y(0)-s y(0)\\ \mathcal{L}\left(\dot y\right) = sY(s)-y(0)\\ $$

and then solve the $Y(s)$ DE.

NOTE

After deriving we get at

$$ Y'(s)(s-s^2) + (m+1-s)Y(s) = 0 $$

with solution

$$ Y(s) = C_0 \frac{(1-s)^m}{s^{m+1}} $$