# Thread: Solve DE using Laplace transform

1. ## Solve DE using Laplace transform

Transform the given differential equation to find a nontrivial solution such that $\displaystyle x(0)=0$.

$\displaystyle tx''+(4t-2)x'+(13t-4)x=0$

My attempt:

$\displaystyle \mathcal{L}\{x'(t)\}=sX(s)$ and $\displaystyle \mathcal{L}\{x''(t)\}=s^2X(s)-x'(0)$

$\displaystyle -\frac{d}{ds}[s^2X(s)-x'(0)]-4\frac{d}{ds}[sX(s)]-2sX(s)-13\frac{d}{ds}[X(s)]-4X(s)=0$

$\displaystyle -[s^2X'(s)+2sX(s)]-4[sX'(s)+X(s)]-2sX(s)-13X'(s)-4X(s)=0$

$\displaystyle (-s^2-4s-13)X'(s)-(2s+8)X(s)=0$

$\displaystyle (s^2+4s+13)X'(s)+(2s+8)X(s)=0$

$\displaystyle \frac{X'(s)}{X(s)}=-\frac{2s+8}{s^2+4s+13}$

$\displaystyle \frac{dX}{X}=-\frac{2s+8}{s^2+4s+13}ds$

$\displaystyle \ln X=\int -\frac{2s+8}{s^2+4s+13}ds$

$\displaystyle =\int -\frac{2s+8}{(s+2)^2+9}ds$

$\displaystyle =-2\int \frac{s+4}{(s+2)^2+9}ds$

$\displaystyle =-2\int \left[\frac{s+2}{(s+2)^2+9}+\frac{2}{(s+2)^2+9}\right]ds$

$\displaystyle =-2\left[\frac{1}{2}\ln [(s+2)^2+9]+\frac{2}{3}\tan^{-1}\left(\frac{s+2}{3}\right)\right]+K$

$\displaystyle =-\ln [(s+2)^2+9]-\frac{4}{3}\tan^{-1}\left(\frac{s+2}{3}\right)+K$

The answer given in the textbook is $\displaystyle x(t)=Ce^{-2t}(\sin 3t-3t\cos 3t)$ but I don't seem to be arriving at it. How do I proceed?

2. ## Re: Solve DE using Laplace transform

Originally Posted by alexmahone
Transform the given differential equation to find a nontrivial solution such that $\displaystyle x(0)=0$.

$\displaystyle tx''+(4t-2)x'+(13t-4)x=0$

My attempt:

$\displaystyle \mathcal{L}\{x'(t)\}=sX(s)$ and $\displaystyle \mathcal{L}\{x''(t)\}=s^2X(s)-x'(0)$

$\displaystyle -\frac{d}{ds}[s^2X(s)-x'(0)]-4\frac{d}{ds}[sX(s)]-2sX(s)-13\frac{d}{ds}[X(s)]-4X(s)=0$

$\displaystyle -[s^2X'(s)+2sX(s)]-4[sX'(s)+X(s)]-2sX(s)-13X'(s)-4X(s)=0$
Good so far, but you make an algebra mistake.

$\displaystyle (-s^2-4s-13)X'(s)-(2s+8)X(s)=0$
Should be

$\displaystyle (-s^{2}-4s-13)X'-(4s+8)X=0.$

$\displaystyle (s^2+4s+13)X'(s)+(2s+8)X(s)=0$

$\displaystyle \frac{X'(s)}{X(s)}=-\frac{2s+8}{s^2+4s+13}$

$\displaystyle \frac{dX}{X}=-\frac{2s+8}{s^2+4s+13}ds$

$\displaystyle \ln X=\int -\frac{2s+8}{s^2+4s+13}ds$

$\displaystyle =\int -\frac{2s+8}{(s+2)^2+9}ds$

$\displaystyle =-2\int \frac{s+4}{(s+2)^2+9}ds$

$\displaystyle =-2\int \left[\frac{s+2}{(s+2)^2+9}+\frac{2}{(s+2)^2+9}\right]ds$

$\displaystyle =-2\left[\frac{1}{2}\ln [(s+2)^2+9]+\frac{2}{3}\tan^{-1}\left(\frac{s+2}{3}\right)\right]+K$

$\displaystyle =-\ln [(s+2)^2+9]-\frac{4}{3}\tan^{-1}\left(\frac{s+2}{3}\right)+K$

The answer given in the textbook is $\displaystyle x(t)=Ce^{-2t}(\sin 3t-3t\cos 3t)$ but I don't seem to be arriving at it. How do I proceed?
Carry through the correction above. It will make a fairly big difference, I think.