I am trying to solve this ODE using symmetry methods.
$$y''(x)=ay'(x)$$
With the boundary condition $y(0)=0$ and $y(\infty)=y_\infty $.
The base solution is $y(x)=y_{\infty}(1-e^{ax})$.
I found 8 one parameter symmetry groups. For the first four symmetries I was able to calculate the explicit form of the symmetry solution. But the fifth group is quiet hard.
For the first group $G_1 (x,y+\varepsilon)$ I did the following:
$$\hat{x}=x$$ $$\hat{y}=y(x)+\varepsilon$$
Then I solved for $x$ in the first equation and plugged the result into the second equation to obtain:
$$\hat{y}=y(\hat{x})+\varepsilon.$$
It is easy to see that this does not satisfy the b.c. if $y(x)=y_\infty(1-e^{ax})$, as $$\hat{y}(\hat{x}=0)=y(\hat{x}=0)+\varepsilon=0+\varepsilon\neq0.$$
I was able to use this method up to group $G_4$, but failed at $G_5$.
$$G_5: \left[a\left( 2\pi ik_1+\ln\left(\exp\left(ax-2\pi ik_1\right) +a\varepsilon y \right) \right),y\right]$$
I tried the following:
$$\hat{x}=a\left( 2\pi ik_1+\ln\left(\exp\left(ax-2\pi ik_1\right) +a\varepsilon y \right) \right)$$ $$\hat{y}=y(x)$$
I solved for $x$ but it still contains $y$ in it. How can I check the boundary conditions without the explicit symmetry solution? Is there maybe a way to even use the infinitesimal generators to so?
Thank you for reading my questions.