Second order ODE with exponential coefficient, $Y''(x)+aY'(x)+b^2e^{2ax}Y(x)=0$ How to find the solution of the given equation?. Can it be transformed to a Bessel equation or equation with constant coefficients?.
$$Y''(x)+aY'(x)+b^2e^{2ax}Y(x)=0$$
Thank you

We first assume $a\neq 0$. We start by substituting $y(x)=e^{-ax}u(x)$. Note that
$$y'(x)=e^{-ax} u'(x)-ae^{-ax}u(x),$$
$$y''(x)=e^{-ax}u''(x)-2ae^{-ax}u'(x)+a^2 e^{-ax}u(x).$$
Therefore, one obtains the ODE
$$e^{-ax}u''(x)-ae^{-ax}u'(x)+b^2 e^{ax}u(x)=0.$$
Since $e^{-ax}\neq 0$, one can divide by $e^{-ax}$ to obtain
$$u''(x)-au'(x)+b^2 e^{2ax}u(x)=0.\tag{1}$$
We now substitute $z=e^{ax}$. One finds that $z'(x)=az$, thus by the chain rule one obtains
$$u'(x)=azu'(z),$$
$$u''(x)=a^2 z^2 u''(z)+a^2 zu'(z).$$
Substituting this into $(1)$ gives you
$$a^2 z^2u''(z)+b^2 z^2 u(z)=0.$$
One can divide by $a^2 z^2=a^2 e^{2ax}\neq 0$ therefore one obtains a simple ODE with constant coefficients
$$u''(z)+\left(\frac{b}{a}\right)^2u(z)=0.$$
If $a=0$ the ODE is already in a similar form in $y(x)$, so you don't need to do any substitutions.
