# The general solution to the differential equation

I am having problems in classifying the differential equation $y''=y(x^2)$ in categories like homogeneous, exact, bernoulli, separable and non-exact so I could have the general solution.

Or would someone help me find the solution

• Almost never do such equations have closed-form solutions. If you can even prove a solution exists. – Chappers Jun 17 '15 at 16:22
• @Chappers,Do solved you mean it can not be solved? – DOCTOR NGILAZI BANDA JOSHUA Jun 17 '15 at 16:27
• is it $y(x^2)$ or $y(x)^2$? – Dr. Sonnhard Graubner Jun 17 '15 at 16:40
• @Dr. Sonnhard Graubner ,Square of x is multiplying y – DOCTOR NGILAZI BANDA JOSHUA Jun 17 '15 at 16:42
• Oh, so you mean $y''=x^2y$? – Chappers Jun 17 '15 at 16:42

the solution containes the Bessel-function $$y \left( x \right) =C_1 \,\sqrt {x} \, {{I}_{1/4}\left(1/2\,{x}^ {2}\right)}+ C_2 \,\sqrt {x} \, {{K}_{1/4}\left(1/2\,{x}^{2} \right)}$$

The General solution to this differential equation would be an infinite series:

$$y(x) = \sum^\infty_{n=0} a_nx^n$$

In this case, the coefficents are pretty odd. Since $a_2 = 0, a_3 = 0$, and the recurrence is every 4 terms. In this case we have:

$$a_4 = a_0/12, a_5 = a_1/20, a_8 = a_4/56, a_9 = a_5/72$$

To find the exact relationship requires some art. I would go with:

$$a_{4k} = \frac{a_0}{(4^k)k!(4k-1)(4k-5)...etc)} , a_{4k+1} = \frac{a_1}{4^kk!(4k+1)(4k-3)...etc}$$

The etc. can be rewritten as factorials if needed.