Solving a second-order linear ODE: $\frac{d^2 y}{dx^2}+(x+1)\cdot \frac{dy}{dx}+5x^2\cdot y=0$

Question

Recently, a friend challenged me to find the general solution of the following differential equation:

$$\frac{d^2 y}{dx^2}+(x+1)\cdot \frac{dy}{dx}+5x^2\cdot y=0 \tag{1}$$

This is a second-order linear ordinary differential equation.

I have tried putting this ODE into the form of a Sturm-Liouville Equation by multiplying both sides by $e^{\int (x+1)~dx}$ to obtain: $$e^{\frac{x^2}{2}+x}\cdot\frac{d^2 y}{dx^2}+(x+1)\cdot e^{\frac{x^2}{2}+x}\cdot \frac{dy}{dx}+5x^2\cdot e^{\frac{x^2}{2}+x}\cdot y=0$$ By the reverse product rule: $$\frac{d}{dx}\left(e^{\frac{x^2}{2}+x}\cdot y'(x)\right)+5x^2\cdot e^{\frac{x^2}{2}+x}\cdot y=0 \tag{2}$$ Now, it is in Sturm-Liouville form, however I am unsure how to proceed from here.

Therefore, I have instead tried to do some substitution on the differential equation to eliminate the first order term to obtain this form: $$\frac{d^2 y}{dx^2}+q(x)\cdot y=0$$ Therefore, I have tried using the substitution: $$y=e^{-\frac{(1+x)^2}{4}}\cdot z$$ $$\ln y=\ln{z}-\frac{(1+x)^2}{4}$$ Differentiating implicitly both sides w.r.t $x$: $$\frac{y'}{y}=\frac{z'}{z}-\frac{1}{2}(x+1) \tag{3}$$ Differentiating again: $$\frac{y\cdot y''-(y')^2}{y^2}=\frac{z\cdot z''-(z')^2}{z^2}-\frac{1}{2}$$ Thus: $$\frac{y''}{y}-\left(\frac{y'}{y}\right)^2=\frac{z''}{z}-\left(\frac{z'}{z}\right)^2-\frac{1}{2}$$ Substituting $(3)$: $$\frac{y''}{y}=\left(\frac{z'}{z}-\frac{1}{2}(x+1)\right)^2+\frac{z''}{z}-\left(\frac{z'}{z}\right)^2-\frac{1}{2}$$ Expanding gives: $$\frac{y''}{y}=\left(\frac{z'}{z}\right)^2-\left(\frac{z'}{z}\right)(x+1)+\frac{1}{4}(x+1)^2+\left(\frac{z''}{z}\right)-\left(\frac{z'}{z}\right)^2-\frac{1}{2}$$ $$\frac{y''}{y}=-\left(\frac{z'}{z}\right)(x+1)+\frac{1}{4}(x+1)^2+\left(\frac{z''}{z}\right)-\frac{1}{2} \tag{4}$$ Going back to our original ODE $(1)$: $$y''+(x+1)y'+5x^2\cdot y=0$$ $$\frac{y''}{y}+(x+1)\frac{y'}{y}+5x^2=0$$ Substituting $(3)$ and $(4)$ gives: $$-\left(\frac{z'}{z}\right)(x+1)+\frac{1}{4}(x+1)^2+\left(\frac{z''}{z}\right)-\frac{1}{2}+(x+1)\left[\frac{z'}{z}-\frac{1}{2}(x+1)\right]+5x^2=0$$ Cancelling terms gives: $$\left(\frac{z''}{z}\right)-\frac{1}{4}(x+1)^2-\frac{1}{2}+5x^2=0$$ Which gives the ODE: $$z''+\left[5x^2-\frac{1}{2}-\frac{1}{4}(x+1)^2\right]z=0$$ When the $z$ term is expanded, it gives: $$z''+\frac{1}{4}(19x^2-2x-3)z=0 \tag{5}$$ I tried identifying this ODE as a known type, however I could not. Therefore, I am stuck at this point.

Note that I am trying to avoid a series solution for this differential equation. I am aware that the result will be in terms of non-elementary functions. Wolfram|Alpha suggests that the solution will be in terms of the Hermite polynomial $H_n(z)$ defined as: $$H_n(z)=\frac{n!}{2\pi i} \oint e^{-t^2+2tz}\cdot t^{-n-1}~dt$$ And the Kummer confluent hypergeometric function $_1F_1(a;b;x)$ defined as: $$_1F_1(a;b;x)=1+\frac{a}{b}x+\frac{a(a+1)}{b(b+1)}\frac{x^2}{2!}+\cdots=\sum_{k=0}^{\infty} \frac{(a)_k}{(b)_k}\frac{x^k}{k!}$$ Where $(a)_k$ and $(b)_k$ are Pochhammer Symbols.

In conclusion, I would appreciate some guidance on how to continue solving this ODE analytically. I was thinking that equation $(5)$ seems simpler to solve from what we have, however if $(1)$ seems easier, please feel free to continue from the original ODE.

Thanks in advance.

Edit:

I figured that I could simplify $(5)$ further by completing the square: $$\frac{d^2 z}{dx^2}+\left[\frac{19}{4}\left(x-\frac{1}{19}\right)^2-\frac{29}{38}\right]z=0$$ And then applying the substitution $u=x-\frac{1}{19}$ and $du=dx$. Evaluating $\frac{d^2 z}{dx^2}$: $$\frac{dz}{dx}=\frac{dz}{du}\cdot \frac{du}{dx}=\frac{dz}{du}$$ Thus, differentiating w.r.t $x$ gives: $$\frac{d^2 z}{dx^2}=\frac{d}{dx}\left(\frac{dz}{du}\right)=\frac{d}{du}\left(\frac{dz}{du}\right)\frac{du}{dx}=\frac{d^2 z}{du^2}$$ Therefore, we reduce it to the form: $$\frac{d^2 z}{du^2}+\left[\frac{19}{4}u^2-\frac{29}{38}\right]z=0 \tag{6}$$

@projectilemotion - No solution from your friend? then I would go on to say that he is no friend of yours posing such a question without a solution ;). — Chinny84, Mar 16 '17 at 03:56
Are you sure that your friend is ... a friend ? This problem is just a monster. — Claude Leibovici, Mar 16 '17 at 14:04
@ClaudeLeibovici Yes, I am sure. I did not have much else to do, so I asked my friend to give me a random differential equation to solve. In fact, the first problem he gave me was even more difficult, so I asked him to make it easier: $$\frac{d^2 y}{dx^2}+(x^{(8+2x)^2}+4x^2)\cdot \frac{dy}{dx}+(x^{2x+2}+16)\cdot y=x^2+1$$ — projectilemotion, Mar 16 '17 at 14:26
If you let $y=e^{-(1+x)^2/4}f$, the equation in $f$ is $f''+qf=0$, where $q$ is a quadratic polynomial. Not sure that helps, but it doesn't hurt to have the equation in potential form. — Disintegrating By Parts, Mar 16 '17 at 23:30
@projectilemotion : It looks like you did. After that you may be able to reduce to the harmonic oscillator in some way, but the solutions will involve these solutions along the imaginary axis because of the sign of the quadratic term. — Disintegrating By Parts, Mar 17 '17 at 21:16
@TrialAndError Yes, indeed. I managed to put the differential equation in the form of a Weber Differential Equation and solve it in terms of parabolic cylinder functions as shown on my answer. — projectilemotion, Mar 17 '17 at 22:43

projectilemotion · Accepted Answer · 2017-04-27T17:43:07.373

I figured it out. Equation $(6)$ can be put into the form of a Weber Differential Equation defined as: $$\frac{d^2 y}{dx^2}+\left(\nu + \frac{1}{2}-\frac{x^2}{4}\right)y=0 \tag{7}$$ With general solution: $$y(x)=c_1D_{\nu}(x)+c_2 D_{-\nu-1}(ix)$$ Where $D_{\nu}(x)$ represents the parabolic cylinder function.

Substituting $\gamma=\sqrt[4]{-19}\cdot u$ and $d\gamma =\sqrt[4]{-19}~du$ puts it in the form of $(7)$. Let's evaluate $\frac{d^2 z}{du^2}$: $$\frac{dz}{du}=\frac{dz}{d\gamma}\cdot \frac{d\gamma}{du}=\frac{dz}{d\gamma}\cdot \sqrt[4]{-19}$$ $$\frac{d^2 z}{du^2}=\frac{d}{du}\left(\frac{dz}{d\gamma}\cdot \sqrt[4]{-19}\right)=\frac{d}{d\gamma}\left(\frac{dz}{d\gamma}\cdot \sqrt[4]{-19}\right)\cdot \frac{d\gamma}{du}=\frac{d^2 z}{d\gamma^2}\sqrt{19}\cdot i$$ Substituting into $(6)$ gives: $$i \frac{d^2 z}{d\gamma^2}\sqrt{19}+\left[\frac{19}{4}\cdot \frac{\gamma^2}{\sqrt{19}i}-\frac{29}{38}\right]z=0$$ Dividing both sides by $i\sqrt{19}$ gives: $$\frac{d^2z}{d\gamma^2}+\left[-\frac{\gamma^2}{4}+\frac{29i}{38\sqrt{19}}\right]z=0 \tag{8}$$ Thus, it is now in the form of $(7)$. Hence, we let $\nu=\frac{29i}{38\sqrt{19}}-\frac{1}{2}$ and obtain the general solution for $z$ as a function of $\gamma$: $$\large z(\gamma)=c_1\cdot D_{\frac{29i}{38\sqrt{19}}-\frac{1}{2}}(\gamma)+c_2\cdot D_{-\frac{1}{2}-\frac{29i}{38\sqrt{19}}}(i\gamma)$$ Thus, if we substitute back for $u$, we obtain: $$\large z(u)=c_1\cdot D_{\frac{29i}{38\sqrt{19}}-\frac{1}{2}}(\sqrt[4]{-19}\cdot u)+c_2\cdot D_{-\frac{1}{2}-\frac{29i}{38\sqrt{19}}}(i\sqrt[4]{-19}\cdot u)$$ Substituting back for $x$, we obtain: $$z(x)=c_1\cdot D_{\frac{29i}{38\sqrt{19}}-\frac{1}{2}}\left(\sqrt[4]{-19}\cdot \left(x-\frac{1}{19}\right)\right)+c_2\cdot D_{-\frac{1}{2}-\frac{29i}{38\sqrt{19}}}\left(i\sqrt[4]{-19}\cdot \left(x-\frac{1}{19}\right)\right)$$ Finally, substituting back for $y$, we obtain: $$y\cdot e^{\frac{(1+x)^2}{4}}=c_1\cdot D_{\frac{29i}{38\sqrt{19}}-\frac{1}{2}}\left(\sqrt[4]{-19}\cdot \left(x-\frac{1}{19}\right)\right)+c_2\cdot D_{-\frac{1}{2}-\frac{29i}{38\sqrt{19}}}\left(i\sqrt[4]{-19}\cdot \left(x-\frac{1}{19}\right)\right)$$ Which gives the explicit solution: $$\small \bbox[5px,border:2px solid #C0A000]{y(x)=c_1\cdot D_{\frac{29i}{38\sqrt{19}}-\frac{1}{2}}\left(\sqrt[4]{-19}\cdot \left(x-\frac{1}{19}\right)\right)\cdot e^{-\frac{(1+x)^2}{4}}+c_2\cdot D_{-\frac{1}{2}-\frac{29i}{38\sqrt{19}}}\left(i\sqrt[4]{-19}\cdot \left(x-\frac{1}{19}\right)\right)\cdot e^{-\frac{(1+x)^2}{4}}} \tag{9}$$ This is equivalent to the solution given by Wolfram|Alpha, but I find this solution to be a neater one.

Solving a second-order linear ODE: $\frac{d^2 y}{dx^2}+(x+1)\cdot \frac{dy}{dx}+5x^2\cdot y=0$

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