# Second order differential equation given the form of the solution

Consider the following second order differential equation: $$\frac{d^2 u}{dr} + \frac{N-1}{r}\frac{du}{dr} + \frac{\lambda}{r^2}u = 0$$ in $$(0,1)$$, where $$\lambda = 1+\frac{1}{4}(N-2)^2$$ and $$N = 1,2,3$$. Find all the solutions having the form $$u(r) = r^{-\frac{N-2}{2}}v(log(r))$$. Hint: make the change of variable $$t = log(r)$$ for $$r \in (0,1)$$ and find the ODE that $$v(t) = v(log(r))$$ solves.

I don't understand if I have to actually find the solutions by hand as if I didn't know their form or if I have to use the information of the form of the solution and substitute inside the ODE the generic solution $$u(r)$$ to verify the identity (I suppose the latter). And what does it mean to find the ODE that is solved by $$v(t) = v(log(r))$$, as if for that function there was an unique ODE having as a solution such function?

My first idea was to compute the first and second derivative of $$u(r) = ...$$ without doing the change of variable, and then substitute $$u(r), \frac{du}{dr}(r), \frac{d^2(u)}{dr}(r)$$ inside the ODE so I should get an identity but... does it make sense? Once I substituted them inside the ODE what should I do? Unfortunately I'm new to differential equations, One thing I read is that an ODE like the one above can be written as a system of differential equations of first order, but I can't figure out how to transform this one in particular, if only the term $$\frac{\lambda}{r^2}u$$ wasn't present I think it would have been easy. What would you suggest?

The exercise continues asking to find the exact solution with final values $$u(1)=1$$ and $$\frac{du}{dr}(1) = 1$$ but I think that once the general solutions are known, it is only a matter of substituting. For now if I try to substitute $$r=1$$ inside $$u(r) = r^{-\frac{N-2}{2}}v(log(r))$$ I can only conclude that $$log(r=1)=0$$ and that's it, because I don't know the function $$v(t)=v(log(r))$$.

Thanks for reading and for any type of help or suggestion.

• Tip: use the code \log to make it more distinct. $\log$ versus $log$. Commented May 30, 2020 at 16:51

The change of variable is a little tricky because you need to use the chain rule for second derivatives. Let $$U(t) = u(r(t))$$, with $$r(t) = e^t$$. Applying the chain rule twice, $$U'(t) = u'(r(t))r'(t) = u'(r(t))e^t$$ and $$U''(t) = u''(r(t)) (r'(t))^2 + u'(r(t)) r''(t) = u''(r(t))e^{2t} + u'(r(t))e^t.$$ The first equation is equivalent to $$\frac{du}{dr} = e^{-t} \frac{du}{dt}$$ and the second equation is equivalent to $$\frac{d^2 u}{dr^2} = e^{-2t}\left( \frac{d^2 u}{dt^2} - \frac{du}{dt}\right).$$ So the original ODE with independent variable $$r$$ is equivalent to the following linear second-order ODE with independent variable $$t$$: $$\frac{d^2 u}{dt^2} + (N-2)\frac{du}{dt} + \lambda u.$$