Solve $t y''(t) +2 y'(t) - ty(t)=0$

Question

I have been trying to solve the differential equation $t y''(t) +2 y'(t) - ty(t)=0$ through the Frobenius method. WolframAlpha says the solution should be $ y(t) = c_1 \frac{e^{-t}}{t} + c_2\frac{e^t}{t}$. I got that $r=0 \lor r=-1$. For $r=0$, we get that $a_{n+1}=\frac{a_{n-1}}{(n+1)(n+2)}$, so $a_{2k}=\frac{1}{(2k+1)!}$, but this will not get me the correct solution. Could anyone try to solve it and tell me where I went wrong?

Answer 1

There is no mistake in your calculations. As you derived the solution is actually given by $$ y(t)=a_0\sum_{k=0}^\infty \frac{t^{2k}}{(2k+1)!}\,. $$ Note that since $$ \sinh t=\frac{e^{t}-e^{-t}}{2}=\sum_{k=0}^\infty\frac{t^{2k+1}}{(2k+1)!}\,, $$ hence you found the totally legitimate solution $$ y(t)=a_0\frac{\sinh t}{t}\,. $$

To reconcile it with the solution given by Wolfram Alfa just note that the linear combination of $e^t/t$ and $e^{-t}/t$ gives you exactly your solution.

Since the second value of $r=-1$ (you did find it also correctly), and hence your $r$'s are divided by integer, Frobenius method does allow you to find a second linearly independent solution to your problem. It can be done, however, using any of the standard methods of reducing the order of your ODE (or using Abel's formula).

Answer 2

Sorry for not typing. This is a solution without Frobenius method!