# Solving Riccati differential equation

I was unable to solve the following differential equation. We've just started, so the level can't be so huge, however i didn't find an appropriate method for this one in particular.

$$\frac{dy}{dx} = x(y^2) + x^5$$

i found that,according to wolframalpha, this equation is called Riccati differential equation. There is a relatively simple way to solve it?

• the solution containes the Bessel function – Dr. Sonnhard Graubner Mar 15 '18 at 17:16
• It feels like $z=x^2$ so $dy/dz=(y^2+z^2)/2$ might help. – J.G. Mar 15 '18 at 17:24

Set $y=\frac{p}{q}$, $$\frac{p'}q-\frac{pq'}{q^2}=\frac{dy}{dx}=x\frac{p^2}{q^2}+x^5$$
which suggests $$q'=-xp$$ so that $$p'=x^5q$$ remains, which is now a system of 2 linear ODE, or in combination $$q''=-x(x^5q)-p\implies xq''-q'+x^7q=0$$ with the corresponding solution theory, power series expansion, numerical methods,...
Setting $q=xf(x^4/4)$ following the solution form of WolframAlpha gives $$q'=x^4f'(x^4/4)+f(x^4/4),\\ xq''=x^8f''(x^4/4)+5x^4f'(x^4/4)$$ so that insertion results in $$x^8f''(x^4/4)+5x^4f'(x^4/4)-x^4f'(x^4/4)-f(x^4/4)+x^8f(x^4/4)=0$$ and with $4t=x^4$ $$t^2f''(t)+tf'(t)+\Bigl(t^2-\frac1{16}\Bigr)f(t)=0$$ which is the equation of a Bessel function.