Does the equation have a power series solution?

Question

Does the equation

$$(x^2 + 25)y'' + xy' + x^3y= 0$$

have a power series solution $y = \sum^\infty_{n=0}c_nx^n$?

If yes, determine the minimal radius of convergence the solution is guaranteed to have.

You do not have to solve the equation.

My attempt

$$\sum^\infty_{n=2}c_nn(n-1)x^n + 25\sum^\infty_{n=2}c_nn(n-1)x^{n-2}+\sum^\infty_{n=1}c_nnx^n +\sum^\infty_{n=0}c_nx^{n+3} = 0$$

$$\sum^\infty_{n=0}c_nn(n-1)x^n + 25\sum^\infty_{n=0}c_{n+2}(n+2)(n+1)x^{n}+\sum^\infty_{n=0}c_nnx^n +\sum^\infty_{n=3}c_{n-3}x^{n} = 0$$

$$2c_2x^2 + \sum^\infty_{n=3}c_nn(n-1)x^n + 2c_0 + 6c_1x + 12c_2x^2 + 25\sum^\infty_{n=3}c_{n+2}(n+2)(n+1)x^{n}+ c_1x + 2c_2x^2+\sum^\infty_{n=3}c_nnx^n +\sum^\infty_{n=3}c_{n-3}x^{n} = 0$$

We set

$$c_0 = 0, c_1 = 0, c_2 = 0$$

$$\sum^\infty_{n=3}[c_n(n-1) + 25c_{n+2}(n+2)(n+1) + c_nn + c_{n-3}] x^n = 0$$

$$c_n(n-1) + 25c_{n+2}(n+2)(n+1) + c_nn + c_{n-3} = 0$$

$$25c_{n+2}(n+2)(n+1) = -c_nn - c_{n-3} -c_n(n-1)$$

$$c_{n+2} = \frac{-c_nn - c_{n-3} -c_n(n-1)}{25(n+2)(n+1)}$$

I know that the power series does not have a solution in the form of $y = \sum^\infty_{n=0}c_nx^n$ but I'm quite stuck on how to expain it. My intuition says it has something to do with the fact that when $n = 1$, $c_{n-3} = c_{-2}$ and that is undefined. I am not sure of this however.

Any tips?

Answer 1

You may want to study your method of taking a series solution. It is not correct. Regardless, we can write the equation in the form: $$y'' + \frac{x}{x^2 + 25}y' + \frac{x^3}{x^2 + 25}y = 0$$ Both of those coefficients are analytic everywhere, so a series solution exists everywhere.

Answer 2

You made some errors.

$$2c_2x^2 + \sum^\infty_{n=3}c_nn(n-1)x^n + \color{red}{25(}\color{}2c_\color{red}2\color{} + 6c_\color{red}3\color{}x + 12c_\color{red}4\color{}x^2\color{red}{)}\color{} + 25\sum^\infty_{n=3}c_{n+2}(n+2)(n+1)x^{n}+ c_1x + 2c_2x^2+\sum^\infty_{n=3}c_nnx^n +\sum^\infty_{n=3}c_{n-3}x^{n} = 0$$

and

$$\sum^\infty_{n=3}[c_n\color{red}n\color{}(n-1) + 25c_{n+2}(n+2)(n+1) + c_nn + c_{n-3}] x^n = 0$$

Should work out from there. $c_0, c_1$ are free.