# 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?

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 $$\ddot z-\dot z+\lambda z=0\ .$$

• This certainly is a good solution but the exercise really said it was preferable to use the transformation $f(x):= (e^x)$. How would it look like if you use this transformation or is it impossible? But this method seems to be easier by the way. – Belgium_Physics Dec 13 '17 at 15:00

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. $$f(t)= c_1e^{\frac{1+\sqrt{4\lambda+1}}{2}t}+c_2e^{\frac{1-\sqrt{4\lambda+1}}{2}t}=c_1(e^t)^{\frac{1+\sqrt{4\lambda+1}}{2}}+c_2(e^t)^{\frac{1-\sqrt{4\lambda+1}}{2}}$$

$$y(x)=c_1 x^{\frac{1+\sqrt{4\lambda+1}}{2}}+c_2x^{\frac{1-\sqrt{4\lambda+1}}{2}}$$