# General Solution of the ODE $\eta' - 2 \zeta \eta = K\left(\zeta^2+1\right)^2$

I am trying to find the function $\Phi(\zeta)$ which is determined by the ordinary differential equation $$\Phi'' - 2 \zeta \Phi' = K\left(\zeta^2+1\right)^2$$ where $K$ is a constant from my model, i will like to retain.

I am interested in finding the general solution of this ode. I was thinking if we assume $\Phi' = \eta$ then our ode will become.

$$\eta' - 2 \zeta \eta = K\left(\zeta^2+1\right)^2$$ which if i am not mistaken is still a non-linear ode.

How can we proceed here ? I am also open to other methods of solving this ODE.

• Observe that $e^{-\zeta^2} \eta$ satisfies $$\left( e^{-\zeta^2} \eta\right)' = e^{-\zeta^2} \left( \eta' - 2 \zeta \eta\right)$$ this allows you to explicitly write down an integral formula for $\eta$. – Willie Wong Jun 14 '16 at 3:46
• You can get some way by using an integrating factor with this. – David Jun 14 '16 at 3:48
• A bit subjective, but I don't like your choice of variables. – Zach466920 Jun 14 '16 at 4:24
• Note that neither of these ODEs are non-linear (as you've written in your post). The ODEs would only be non-linear if the unknown functions $\Phi$ or $\eta$ appeared non-linearly. These ODEs are both linear, but are non-constant coefficient and also inhomogeneous. – okrzysik Jun 14 '16 at 5:10
• @okrzysik, Thank you very much for highlighting that, I took my ODE class, quite a number of years ago, so got confused. I will revise my post. – Comic Book Guy Jun 14 '16 at 5:15

Changing notations, let us consider the differential equation $$y'-2x y=K(x^2+1)^2$$ Using the solution of the homogeneous equation, just as Willie Wong suggested, let $y=z\ e^{x^2}$ which reduces the equation to $$e^{x^2}\, z'=K(x^2+1)^2$$ So, we are left with the problem of computing integrals $$J_n=\int x^{2n}\,e^{-x^2}\,dx$$ $(n=0,1,2).$ We find, using integration by parts, $$J_0=\frac{\sqrt{\pi }}{2} \, \text{erf}(x)$$ $$J_1=\frac{\sqrt{\pi }}{4} \, \text{erf}(x)-\frac{1}{2} e^{-x^2} x$$ $$J_2=\frac{3\sqrt{\pi }}{8} \, \text{erf}(x)-e^{-x^2} \left(\frac{x^3}{2}+\frac{3 x}{4}\right)$$ Combining all the above $$z=K \left(\frac{11\sqrt{\pi }}{8} \,\text{erf}(x)-e^{-x^2} \left(\frac{x^3}{2}+\frac{7 x}{4}\right)\right)+C$$ then $y$.
For the next step $\int y\,dx$, it seems to be much more complex and I suppose that part of the result would include some hypergeometric functions.
• @ComicBookGuy. You are welcome ! Patience was just required for $J_1$ and $J_2$. – Claude Leibovici Jun 14 '16 at 5:25