Solve the differential equation: $-x^2 y'' = \lambda y$ I need to solve the differential equation  $-x^2 y'' = \lambda y$ by transforming the differential equation to a equation with constant coefficients. I need to do this by using $f(x) = y(e^x)$. If I do this, I become the equation:
$y''(e^x) + \frac{y'(e^x)}{e^x}+ \frac{\lambda y(e^x)}{(e^x)^2} = 0$
The I find a general solation of: $$y(e^x) = C \exp\left(\frac{-1+\sqrt{1- \frac{4\lambda}{x^2}}}{2(e^x)^2}\right) + C \exp\left(\frac{-1-\sqrt{1- \frac{4\lambda}{x^2}}}{2(e^x)^2}\right) $$
But Wolfram alpha gives the solution:
$$y(x) = c x^{\frac{1}{2} + \frac{1}{2}*\sqrt{1-4\lambda}} + c x^{\frac{1}{2} - \frac{1}{2}*\sqrt{1-4\lambda}}  $$
Can someone help me figure this out?
 A: It is true that the given ODE can be transformed into an equation with constant coefficients by substituting $x:=e^t$. But the necessary computations are  and not at all straightforward, hence error prone.
Consider your equation as a Eulerian ODE instead, and use the "Ansatz" $y(x):=x^\alpha$, with $\alpha\in{\mathbb C}$ to be determined. This leads to
$$-x^2\cdot \alpha(\alpha-1)x^{\alpha-2}=\lambda x^{\alpha}\ ,$$
hence to the "characteristic equation"
$$\alpha^2-\alpha+\lambda=0\ .$$
This equation has the two solutions ${1\over2}\bigl(1\pm\sqrt{1-4\lambda}\bigr)$ and leads to the following general solution of the given ODE:
$$y(x)=C_1 x^{(1+\sqrt{1-4\lambda})/2}+C_2x^{(1-\sqrt{1-4\lambda})/2}\ .$$
If these are real, fine. Otherwise addditional measures (via complex exponentials) are necessary.
For the desired transformation define
$$z(t):=y(e^t)$$ and compute, using the chain rule,
$$\dot z(t)=y'(e^t)\>e^t,\qquad \ddot z(t)=y''(e^t)\>e^{2t}+y'(e^t)\>e^t=-\lambda z(t)+\dot z(t)\ .$$
It folows that $t\mapsto z(t)$ satisfies the constant coefficient equation
$$\ddot z-\dot z+\lambda z=0\ .$$
A: It is true that using $f(X)=y(e^X)$ allows to transform the ODE into an ODE with constant coefficients, but with a little trick at first beginning :
$f(X)=y(e^X)$ is valid any symbol of variable, for example  $f(t)=y(e^t)$
Let $e^t=x \quad\to\quad f(t)=y(x)$
$dx=e^tdt=xdt \quad\to\quad \frac{dt}{dx}=\frac{1}{x}$
$\frac{dy}{dx}=\frac{df}{dt}\frac{dt}{dx}=\frac{1}{x}\frac{df}{dt}$
$\frac{d^2y}{dx^2}=\frac{d\left(\frac{1}{x}\frac{df}{dt}\right)}{dx} =
-\frac{1}{x^2}\frac{df}{dt}+\frac{1}{x}\frac{d^2f}{dt^2}\frac{dt}{dx}=
-\frac{1}{x^2}\frac{df}{dt}+\frac{1}{x^2}\frac{d^2f}{dt^2}$
$-x^2\frac{d^2y}{dx^2}=\lambda y=x^2\left(-\frac{1}{x^2}\frac{df}{dt}+\frac{1}{x^2}\frac{d^2f}{dt^2}\right)= -\frac{df}{dt}+\frac{d^2f}{dt^2}$
$$\frac{d^2f}{dt^2}-\frac{df}{dt}-\lambda f(t)=0$$
This is a linear ODE with constant coefficients easy to solve.
$$f(t)= c_1e^{\frac{1+\sqrt{4\lambda+1}}{2}t}+c_2e^{\frac{1-\sqrt{4\lambda+1}}{2}t}=c_1(e^t)^{\frac{1+\sqrt{4\lambda+1}}{2}}+c_2(e^t)^{\frac{1-\sqrt{4\lambda+1}}{2}} $$
$$y(x)=c_1 x^{\frac{1+\sqrt{4\lambda+1}}{2}}+c_2x^{\frac{1-\sqrt{4\lambda+1}}{2}}$$
