I have to solve the following ODEs:
- $y''+ y = e^{\epsilon \cos x}$, $y(0,\epsilon)=y(1,\epsilon)=0$;
- $y''+ y = \epsilon y^2$, $y(0,\epsilon)=1$, $y'(0,\epsilon)=0$.
I am having trouble with these two questions.
For the first one, I got: \begin{align} y''_0 + y_0 &=1,& y_0(0)&=y_0(1)=0\\ y''_1 + y_1 &= \cos x,& y_1(0)&=y_1(1)=0 \end{align} Not sure where the $\cos x$ is from.
For the second question, unsure how $y_0(x)=\cos x$ and $y_1(x)=1/2-1/3(\cos x)-1/6(\cos 2x)$
) You can write the right side as series expansion in $ϵ$, that is, insert the series for the exponential function $$ e^{ϵ\cos x}=1+ϵ\cos x+\frac{ϵ^2}2\cos^2 x+\frac{ϵ^3}6\cos^2 x+... $$
) The initial equation is the harmonic oscillator $$ y_0''+y_0=0,~~y_0(0)=1,~~y_0'(0)=0 $$ which has $y_0(x)=\cos(x)$ as solution. The next equation of the terms of first order in $ϵ$ is $$ y_1''+y_1=y_0^2,~~y_1(0)=1,~~y_1'(0)=0 $$ and $\cos^2x$ is as is well-known equal to $\frac12(1+\cos(2x))$. Apply the method of undetermined coefficients to find the corresponding particular solution.