Solve $x''(t) + tx'(t) - (t+1)x = 0$ using the Laplace transform

Question

$x''(t) + tx'(t) - (t+1)x = 0$, where $ x(0) = x'(0) = 1. $

The Laplace transform of $tx'$ is $-X(s) - sX'(s)$

The Laplace transform of $tx$ is $-X'(s)$

So we get $s^2X-s-1-X(s)-sX'(s) + X'(s) - X(s) = 0.$

$$X'(s) + X(s) \frac{s^2-2}{s-1} = \frac{s+1}{1-s}.$$

I tried solving this differential equation and then inverse Laplace it but it would't work

The answer is too simple for how much I am complicating it : $ x(t) = e^t.$

Answer 1

You got a sign error in your rearrangement: you should have$$X^\prime+\frac{s^2-2}{\color{blue}{1-s}}X=\frac{s+1}{1-s}.$$If that's still hard to solve, use the Ansatz $X=\frac{Y}{s-1}$ so$$(1+s)Y-Y^\prime=s+1,$$which has obvious root$$Y=1\implies X=\frac{1}{s-1}\implies x=e^t.$$The large-$s$ behaviour of a Laplace transform precludes any other choice of $Y$.

Answer 2

J.G's answer is for sure the best answer, but you can also use general method for solving this first DE. $$X^\prime+\frac{s^2-2}{1-s}X=\frac{s+1}{1-s}.$$ The integrating factor $\mu (s):$ $$\mu (s)=\exp \int \frac{s^2-2}{1-s}ds$$ $$\mu (s)=\exp \left (\frac{-s^2}2-s +\ln |{s-1}| \right )$$ $$(X\mu (s))'=-({s+1})\exp \left (\frac{-s^2}2-s \right ).$$ Integrate $$X\mu (s)=- \int ({s+1})\exp \left (\frac{-s^2}2-s \right )ds$$ $$X\mu (s)=\exp \left (\frac{-s^2}2-s \right )+K$$ $$X (s)=\left (\frac 1{s-1} \right )+\frac K {\mu(s)}$$ And using Inverse Laplace's transform you get $x(t)=e^t$