differentiation of Laplace transform solution

Question

I am wondering if there is a solution to the differential equation (of sorts): $$\frac{d}{ds}\mathcal{L}\left[y(t)\right]-\mathcal{L}\left[\frac{d}{dt}y(t)\right]=0$$ Using the fact that: $$\frac{d}{ds}\mathcal{L}\left[y(t)\right]=-\mathcal{L}\left[ty(t)\right]$$ we can show that the equation is the same as: $$\mathcal{L}\left[ty(t)\right]+\mathcal{L}\left[y'(t)\right]=0$$ One case that would clearly work is: $$y'+ty=0$$ Which would give: $$y=e^{\frac{t^2}{2}+C_1}$$ Are there any other solutions?

Answer 1

No because the only solution to $\mathcal{L}(f) = 0$ is the function $f \equiv 0$.

Then you are able to use linearity of the laplacian operator to get

$\mathcal{L}(ty+y') = 0 \iff ty+y' = 0$.