It is known that if $f$ is a function of exponential order $b$, then $F (s) = L (f (t)) (s)$ has a derivative for $s>b$ and satisfies $ \dfrac{d}{ds}F(s)=-\mathcal{L}(t\,f(t))(s) $
Consider the next DE:
$ \left\{ \begin{array}{ll} t\,y''(t)+t\,y'(t)-y=0, & \mbox{} t> 0 \\ y(0)=0,\;y'(0)=1 & \end{array} \right. $
If $Y(s)=\mathcal{L}(y(t))(s)$ Show that $Y'(s)+\dfrac{2}{s}\,Y(s)=0$
I start my proof by applying Laplace to the differential equation and replacing I get this:
$$-\frac{d}{ds}(sY(s)-1)-\frac{d}{ds}(Y(s))-\frac{Y}{s}=0.$$
I differentiate and solve for $Y(s)$:
$$Y(s) = \frac{C}{s},$$
but it does not satisfy the demonstration:
$$-\frac{C}{s^{2}}+\frac{2C}{s^{2}}=0.$$
Someone can help me? I don't quite understand the first paragraph of the question and maybe I'm forgetting something.
In taking the LT, I get \begin{align*} L[t\,y'']+L[t\,y']-Y&=0\\ -\frac{d}{ds}L[y'']-\frac{d}{ds}L[y']-Y&=0\\ -\frac{d}{ds}\left[s^2 Y-sy(0^-)-y'(0^-)\right]-\frac{d}{ds}\left[s Y-y(0^-)\right]-Y&=0\\ -\frac{d}{ds}\left[s^2 Y-1\right]-\frac{d}{ds}\left[s Y\right]-Y&=0\\ \left[2sY+s^2Y'\right]+\left[Y+sY'\right]+Y&=0\\ 2Y(s+1)+sY'(s+1)&=0\\ 2Y+sY'&=0\qquad\text{for } s\not=-1. \end{align*} Surely you can finish. The solution to this DE is $$Y(s)=\frac{C}{s^2}.$$ The Inverse Laplace Transform would then get you $y(t)=C\,t,$ which you can see by inspection solves the original ODE if $C=1.$ That is, the solution is $y(t)=t.$