# Use Laplace transform to find a solution for $tx''+x'+tx = 0$

I was hoping I could get some help checking my work through.

$\mathcal{L}\{x'(t)\} = sX(s)-1$ and $\mathcal{L}\{x''(t)\} = s^2X(s)-s$

Using the equality $\mathcal{L}\{-tf(t)\} = F'(s)$ we can rewrite the original system as:

$$-\frac{d}{ds}[s^2X(s)-s]+[sX(s)-1]-\frac{d}{ds}[X(s)] = 0$$

$$(s^2+1)X'(s)+sX(s)=0$$

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

Now I'm not sure where to go from here. How can I make this differentiation easier to understand so I can solve the problem? My textbook says the answers should be:

$$X(s) = \frac{C}{\sqrt{s^2+1}}$$

but I don't see how I can get that from my above formulation. Any help?

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Hint: $$\frac{X'}{X}=\frac{-s}{s^2+1}\implies \log X=-\frac{1}{2}\log (s^2+1)+\log C\implies \ldots$$
Just take your equation $\displaystyle{\frac{X'(s)}{X(s)} = -\frac{s}{s^2+1}}$ a little further: write $X'(s)=\frac{dX}{ds}$ then multiply both sides by $ds$ to get $$\frac{dX}{X} = -\frac{s}{s^2+1}\,ds.$$
Integrate both sides, noting that the left-hand side is of the form $\int \frac{du}{u}$. Solve the result for $X(s)$ and you are done.