Reformulate this initial value problem as an equivalent integral equation. $$\dfrac{d^2y}{dx^2}+y=x, \ \ \ \ 0\leq x \leq 1, \ \ \ \ y(0)=1, y'(1)=0 $$
I'm not sure how to do this. I think I start by letting $\phi=\dfrac{d^2y}{dx^2}$ and integrating over $[0,1]$ which gives: $$y'(1)-y'(0)=\int^1_0 \phi(t) \ dt $$ Then using $y'(1)=0$, this gives: $$y'(0)=-\int^1_0 \phi(t) \ dt$$
Do I then integrate this over $[0,x]$?
This is one example of a few inital value problems that I'm not sure how to reformulate so is there a method I follow so that I can apply this to the others?
You can use that by partial integration $$ y(x)=y(0)+y'(0)x+\int_0^x(x-s)y''(s)\,ds $$ or first transform the second order equation into a first order system and use the usual Picard integral equation.