Is it possible to say that $L(f^n)=s^nL(f)$ when the differential equation is not in the rest condition?

Question

Question

Use the Laplace transform to solve the following equation: $y'+2y=\cos(3t)$ ; where $y(0)=1$

---

In class our teacher wrote that "When in rest condition: $L(f^n)=s^nL(f)$", but I want to use this on this problem even though the rest condition is not satisfied here.

Note - we were taught that the rest condition is when $$y^{(n-1)}(0)=0, y^{(n-2)}(0)=0,\ldots,y^{(0)}(0)=0$$

Answer 1

In general case we have: $$L(f^n)=s^nL(f)-s^{n-1}f(0)-\dots-sf^{n-2}(0)-f^{n-1}(0)$$
so $$L(y')=sL(y)-y(0)=sL(y)-1$$
then $$sL(y)-1+2L(y)=\frac{s}{s^2+9}$$ $$L(y)=\frac{s^2+s+9}{(s+2)(s^2+9)}$$ $$L(y)=\frac{\frac{2}{13}s+\frac{9}{13}}{s^2+9}+\frac{\frac{11}{13}}{s+2}$$ $$L(y)=\frac{2}{13}\frac{s}{s^2+9}+\frac{9}{13}\frac{1}{3}\frac{3}{s^2+9}+\frac{11}{13}\frac{1}{s+2}$$ $$y(t)=\frac{2}{13}\cos 3t+\frac{9}{39}\sin 3t+\frac{11}{13}e^{-2t}$$

Answer 2

The general formula is $$ L(f^{(n)})(s) = s^n L(f)(s) - \sum_{k=1}^n s^{n-k} f^{(k-1)}(0) $$

In particular, $L(y')(s) = s L(y)(s) - y(0)$.