Solve equation by shooting method I've got homework to numerical methods to solve this equation using shooting method
$ y'' - (1-e^{-x})y = 0, \quad y(0) = 1, \quad y(x\to+\infty) = 0 $
Hints I've got:

*

*Easily you can find asymptotic solution, using this solution you can use functional substition, that will reduce functional values of solution


*You can use substition of independent variable to make integration domain finite interval.


*Hints above also say: Graph of function is unbounded in $x$ and $y$ axis. Using suitable  substitution you can make graph bounded in $\mathbb{R}^2$
Can anyone help me with transformations? Thanks!
 A: For part 1, they want you to note that if $y$ is unbounded, it grows asymptotically as $e^x$. Therefore, $e^{-x}y(x)$ will be bounded. Letting $z(x) = e^{-x}y(x)$, we find that
$$
z''(x) +2 z'(x) +e^{-x}z(x)= 0\;\;\;;\;\;\;z(0) = 1\;\;\;;\;\;\;z(\infty) = 0.
$$
For part 2, the infinite interval on $x$ needs to be reduced to a finite one. The equation itself suggests something like $u = e^{-x}$. Let's use $u = 1-e^{-x}$ so that the transform is increasing. Letting $z(x) =  w(1-e^{-x})$ gives
$$
(1-u) w''(u) + w'(u) +w(u) = 0\;\;\;;\;\;\; w(0) = 1\;\;\;;\;\;\;w(1) = 0.
$$
Now you can try different initial conditions for $w'(0)$ and see which ones get $w(1)$ close to $0$. Of course, you can't actually do calculations at $u=1$ because of the singular point, but you can force $w(1-\epsilon) = 0$ for some small $\epsilon$. You can then recover $y'(0)$ using $y'(0) = w'(0)+w(0)$.
A: You can not integrate over infinite intervals. So you have to set an upper limit $b$ and establish some boundary condition there.
The limit point is a saddle point. There are exactly 2 paths that lead to it in the phase portrait, and any deviation from these will at some point exponentially diverge.
For $x>b$ the factor $e^{-x}$ becomes very small, for a suitably large $b$. The reduced equation $y''-y=0$ has known solutions, aka "far-field solutions". The one that remains bounded satisfies $y'+y=0$. Thus you can take $y'(b)+y(b)=0$ as upper boundary condition.

To get higher accuracy, or the same accuracy with a smaller $b$, compute the next perturbation term in $y=c_0e^{-x}+c_1e^{-2x}+...$ to get
$$
-3c_1e^{-2x}+...=c_0e^{-2x}+c_1e^{-3x}+...
$$
Thus an improved boundary condition is $$(3-e^{-b})y'(b)+(3+2e^{-b})y(b)=0.$$
