# Solving a challenging differential equation

How would one go about finding a closed form analytic solution to the following differential equation?

$$\frac{d^2y}{dx^2} +(x^4 +x^2+x+c)y(x) =0$$

where $c\in\mathbb{R}$

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One solution is $y=0$. –  Joel Reyes Noche Jan 12 at 15:29
Pretty sure the OP is looking for nontrivial solutions. –  Nameless Jan 13 at 0:06
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## 4 Answers

Maple's answer:

eq:=diff(y(x),x$2)+(x^4+x^3+x^2+c)*y(x)=0; dsolve(eq,y(x)) assuming real; - add comment You can have a closed form solution in terms of HeunT (the Heun Triconfluent function) $$y( x ) ={C_1}\,{{\rm e}^{\frac{1}{6}\,ix \left( 2\,{x}^{2}+3\right)}}{\it HeunT} \left(-\frac{{3}^{2/3}\sqrt[3]{2} \left(4\,c-1\right)}{8}, \frac{3\,i}{2}, -\frac{{2}^{2/3}\sqrt[3]{3}}{2}, \frac{i\sqrt [3]{2}\,{3}^{2/3}}{3}x \right) +{ C_2} {{\rm e}^{-\frac{1}{6}\,ix \left( 2\,{x}^{2}+3\right)}} {\it HeunT } \left( -\frac{{3}^{2/3}\sqrt[3]{2} \left(4\,c-1\right)}{8} ,-\frac{3\,i}{2}, -\frac{{2}^{2/3}\sqrt[3]{3}}{2}, -\frac{i\sqrt [3]{2}\,{3}^{2/3}}{3}x\right).$$ - How in the world would you know to use, think of using, or figure out that you need to use, Triconfluent Heun functions when given such an ode, if you don't mind me asking? It looks nothing like the ode's in your link, for starters. – bolbteppa Jan 12 at 8:53 add comment I am reasonably sure that this equation has no nice solution. If it did, it would be of the form$y(x) = \exp(\int u(x)),$where$u$is a rational function.If you want to know more, read this fairly nice summary of Kovacic's algorithm by Carolyn Smith. - add comment This is the way I would do it. Let$p(x)=x^4+x^2+x+c$. Put$z=y'$. Then the equation is equivalent to the system $$\begin{bmatrix}y\\ z\end{bmatrix}'=\begin{bmatrix}0&1\\-p(x)&0\end{bmatrix}\,\begin{bmatrix}y\\ z\end{bmatrix}.$$ Let us write$A(x)$for the$2\times 2$matrix above. For a given vector$\begin{bmatrix}c_1\\ c_2\end{bmatrix}$, a solution will be given by $$e^{B(x)}\,\begin{bmatrix}c_1\\ c_2\end{bmatrix},$$ where$B'(x)=A(x)$. That is$B(x)=\begin{bmatrix}0&x\\-q(x)&0\end{bmatrix}$, with$q(x)=x^5/5+x^3/3+x^2/2+cx$. If I didn't make a mistake, the exponential is $$e^{B(x)}=\begin{bmatrix}\cos\sqrt{xq(x)}&\sqrt\frac{x}{q(x)}\,\sin\sqrt{xq(x)}\\ -\sqrt\frac{q(x)}{x}\,\sin\sqrt{xq(x)}&\cos\sqrt{xq(x)}\end{bmatrix}.$$ As$y$will be the first coordinate of$e^{B(x)}\begin{bmatrix}c_1\\ c_2\end{bmatrix}\$, we get $$y(x)=c_1\,\cos\sqrt{\frac{x^6}5+\frac{x^4}3+\frac{x^3}2+cx^2}+c_2\,\frac{\sin\sqrt{\frac{x^6}5+\frac{x^4}3+\frac{x^3}2+cx^2}}{\sqrt{\frac{x^5}5+\frac{x^3}3+\frac{x^2}2+cx}}.$$

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Ahh.. I'm not sure what went wrong but I'm not getting it equal to zero when I plug it in to the ode. –  Millardo Peacecraft Jan 12 at 23:44
Yes, there is something wrong. But I cannot find it right away. –  Martin Argerami Jan 13 at 0:07
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