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!

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 $y'(0) = w'(0)+w(0)$.
