nonlinear ODE shooting method using Newton $y'' = 2y'-2y+2$
with $y(0)=1 $ and $ y(\frac{\pi}{2})=2$
I have to solve this using shooting method (Newton).
First thing I need to do it replace the right boundary problem with a specified slope at left boundary. I am told in the question that $y'(0) = 0$. Then the note tells me to use Newton's method for $y(\frac{\pi}{2};0)-2$ ?
I am quite confused as to how to proceed and do this question. I cannot find any similar examples.
Thanks.

I'll write out the idea of the shooting method in your problem, although the idea is pretty general. Any solution to $y''=2y'-2y+2,y(0)=1,y(\pi/2)=2$ is also a solution to $y''=2y'-2y+2,y(0)=1,y'(0)=s$ for some unknown number $s$. Denoting the solutions to this family of IVPs by $y(x;s)$, we can define $F(s)=y(\pi/2;s)$. $F$ can be approximately numerically evaluated using an IVP solver. 
We then want to solve the equation $F(s)=2$, which can be done using a numerical method for 1D root finding, such as bisection, Newton's method, or the secant method. Newton's method is not easy to implement in this situation, because it is not easy to compute $F'(s)$. But bisection and the secant method are both easy to implement in this situation.
