My textbook (schaum's outline of ODE's) says that
Proposition A. If $\Lambda[f(x)](s)$ is the laplace transform $\Lambda[f(x)](s)=\int_{0}^\infty e^{-st}f(t)dt$, then
$$\Lambda[\frac {\delta ^{(n)} y}{\delta x^n}](s)=s^n\Lambda[y]-s^{n-1}y(0)-s^{n-2}y'(0)-....-sy^{(n-2)}(0)-y^{n-1}(0)$$
This is then used to solve ODE's.
I have no intuitive idea why proposition A would be true. Is there a simple intuitive proof for this? Do you have an intuitive way of understanding this proposition?
ps. I don't know what this theorem is called, so I couldn't name it in the question title.
Your notation is weird. $$\mathcal{L}[f(t)](s) = \int_0^\infty f(t) e^{-st}dt$$ Integrating by parts : $$\mathcal{L}[f'(t)](s) = \int_0^\infty f'(t) e^{-st}dt=f(t)e^{-st}|_0^\infty+s\int_0^\infty f(t) e^{-st}dt=-f(0)+s\mathcal{L}[f(t)](s)$$ Thus $$\mathcal{L}[f^{(n)}(t)](s) = -f^{(n-1)}(0)+s\mathcal{L}[f^{(n-1)}(t)](s)\\=-f^{(n-1)}(0)-sf^{(n-2)}(0)+s^2\mathcal{L}[f^{(n-2)}(t)](s)\\= \ldots \\=s^n\mathcal{L}[f(t)](s)-\sum_{m=1}^{n}s^{m-1}f^{(n-m)}(0)$$