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Question asks to find the general solution of the differential equation.

$$\frac{1}{r}\left(\frac{d}{dr}\left(r\frac{dw}{dr}\right)\right)-\frac{\lambda^2}{r^2}w=0.$$

The answer given is $w(r)=c_1r^{\lambda}+c_2r^{-\lambda}$, but I am not sure how the book got that.

I know I can't use characteristic equation because it has a function as a coefficient instead of a constant. I am sure I must use change of variables method.

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By the Product Rule, $\dfrac{d}{dr}\left(r\dfrac{dw}{dr}\right) = r\dfrac{d^2w}{dr^2}+\dfrac{dw}{dr}$.

Thus, the differential equation becomes $\dfrac{d^2w}{dr^2}+\dfrac{1}{r}\dfrac{dw}{dr}-\dfrac{\lambda^2}{r^2}w = 0$.

Now, try something in the form $w = r^{\alpha}$. Then, $\dfrac{dw}{dr} = \alpha r^{\alpha-1}$ and $\dfrac{d^2w}{dr^2} = \alpha(\alpha-1) r^{\alpha-2}$.

Plug these in to get $\alpha(\alpha-1) r^{\alpha-2} + \alpha r^{\alpha-2} - \lambda^2r^{\alpha - 2} = (\alpha^2-\lambda^2)r^{\alpha-2} = 0 \leadsto \alpha = \pm \lambda$.

Thus, two linearly independent solutions are $w = r^{\lambda}$ and $w = r^{-\lambda}$.

By linearity, the general solution is $w = C_1r^{\lambda}+C_2r^{-\lambda}$.

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  • $\begingroup$ Jimmy, how did you get $$\dfrac{d}{dr}\left(r\dfrac{dw}{dr}\right) = r\dfrac{d^2w}{dr^2}+\dfrac{dw}{dr}?$$ Is that just the product rule? $\endgroup$ – Robben Jul 25 '14 at 5:28
  • $\begingroup$ That's just the chain rule. $\dfrac{d}{dr}\left[r\dfrac{dw}{dr}\right] = r\dfrac{d}{dr}\left[\dfrac{dw}{dr}\right] + \dfrac{d}{dr}\left[r\right]\dfrac{dw}{dr} = r\dfrac{d^2w}{dr^2}+\dfrac{dw}{dr}$. $\endgroup$ – JimmyK4542 Jul 25 '14 at 5:30
  • $\begingroup$ Thanks! Last question. How did you know to use the form $w=r^{\alpha}$? $\endgroup$ – Robben Jul 25 '14 at 5:31
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    $\begingroup$ One reason to try that is that each term of his equation does the same thing to a function like $r^\alpha$: it lowers the degree by two. When that kind of situation happens, it's natural to try powers of $r$. $\endgroup$ – Semiclassical Jul 25 '14 at 5:37
  • $\begingroup$ EDIT: Yes, it is the product rule. I called it the wrong thing. As for using $r^{\alpha}$, see this: en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation $\endgroup$ – JimmyK4542 Jul 25 '14 at 5:37
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First rewrite the equation by multiplying both sides by $r^2$:

$$\frac{1}{r}\left(\frac{d}{dr}\left(r\frac{dw}{dr}\right)\right)-\frac{\lambda^2}{r^2}w=0\\ \implies r\left(\frac{d}{dr}\left(r\frac{dw}{dr}\right)\right)-\lambda^2w=0.$$

Make the substiution $r=e^x$ and define $w(r)=w(e^x)=:y(x)$. Then, $x=\log{r}$, $\frac{dx}{dr}=\frac{1}{r}$, and

$$r\frac{d}{dr}w(r)=r\frac{dx}{dr}\frac{d}{dx}y(x)=r\cdot\frac{1}{r}\frac{d}{dx}y(x)=\frac{d}{dx}y(x),$$

and

$$r\frac{d}{dr}\left(r\frac{d}{dr}w(r)\right)=\frac{d^2}{dx^2}y(x).$$

Hence, the equation becomes the 2nd order linear ODE with constant coefficients:

$$y''-\lambda^2y=0.$$

This familiar equation of course has the general solution,

$$y(x)=c_1e^{\lambda x}+c_2e^{-\lambda x}.$$

Substituting back $x=\log{r},$

$$w(r)=c_1e^{\lambda\log{r}}+c_2e^{\lambda\log{r}}=c_1r^{\lambda}+c_2r^{-\lambda}.$$

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