Analitic solution to ${df(x)\over{dx}} + {1\over2} f^2 (x) = ax^2 +bx + c$

Question

What is the analytic solution to the following differential equation? $${df(x)\over{dx}} + {1\over2} f^2 (x) = ax^2 +bx + c$$

I know that this is a nonlinear nonhomogeneous first-order ODE. However, I do not know how to solve it.

Answer 1

From Wikipedia's page, I just learned that it is possible to convert Riccatti equation to a second order linear DE.

Let $u$ be a solution of the following the second order linear DE: $$u''-\frac{1}{2}(ax^2+bx+c)u=0.\tag{*}$$ Then consider $y_p=\frac{2u'}{u}$. By differentiating we have $$y_p'=\frac{2u''u-2(u')^2}{u^2}=\frac{(ax^2+bx+c)u^2-2(\frac{1}{2}uy_p)^2}{u^2}=(ax^2+bx+c)-\frac{1}{2}y_p^2.$$ Hence, $y_p$ satisfies the Riccatti Equation $y'+\frac{1}{2}y^2=ax^2+bx+c.$

The general solution can be found by the usual procedure for solving Riccatti equation knowing a particular solution.

So all remains is to solve $(*)$ :https://www.wolframalpha.com/input?i=y%27%27-%281%2F2%29%28ax%5E2%2Bbx%2Bc%29y%3D0 I don't know the parabolic cylinder function. I would try to solve $(*)$ by usual series method. I think.