# Solve $2\ddot{y}y - 3(\dot{y})^2 + 8x^2 = 0$

Solve differential equation

$$2\ddot{y}y - 3(\dot{y})^2 + 8x^2 = 0$$

I know that we have to use some smart substitution here, so that the equation becomes linear. The only thing I came up with is a smart guessed particular solution: $y = x^2$. If we plug this function in, we get:
$$2\cdot2\cdot x^2 - 3(2x)^2 +8x^2 = 4x^2 - 12x^2 + 8x^2= 0$$

I made a mistake. The coefficients where different in the exam: $$\begin{cases} 3\ddot{y}y + 3(\dot{y})^2 - 2x^2 = 0, \\ y(0) = 1, \\ \dot{y}(0) = 0. \end{cases}$$

Does it make the solution easier?

• It can be re-written as $5(y^\prime)^2-(y^2)^{\prime\prime}=8x^2$ but don't see how that helps. – John Wayland Bales Aug 14 '16 at 2:37
• Of course one can find the particular solutions $y=x^2$ and $y=-x^2$ . But the ODE isn't linear, so one cannot derive other solution from those particular ones. The most likely, the general solution of the ODE cannot be expressed with standard functions. If this question comes from a school problem or is a student training exercise, most probably there is a typo in it (for example $2y''y-2y'^2+8y^2=0$ is easily solvable). Or it isn't asked to solve the ODE, but to answer to the right question without need to solve the ODE. – JJacquelin Aug 14 '16 at 7:08
• @JJacquelin It was one of the questions in entrance exam (MA degree in CS). They will announce the results on Tuesday and tell us whether there was a typo. As for the equation, it actually contains terms $y''y$, $y'^2$ AND $x^2$ (not $y^2$). I have omitted initial conditions to simplify the problem. – Egor Okhterov Aug 14 '16 at 11:03
• Omitting initial conditions isn't a good idea. Specific conditions might be necessary to make the problem simpler that looking for the general solution of the ODE. The general solution might be not what it was asked for. – JJacquelin Aug 14 '16 at 12:55
• @JJacquelin Today I was able to look at the problems again. I've fixed the coefficients and added initial conditions. Is it now substantially different from the original problem I asked? – Egor Okhterov Aug 16 '16 at 19:21

Let $z=e^{ky}$. Then $z'=ky'e^{ky}$ and $$z''=ky''e^{ky}+k^2(y')^2 e^{ky}\text.$$

If we put $k=-\frac{3}{2}$ then $z''=-\frac{3}{2}y''e^{ky}+\frac{3}{2}^2(y')^2 e^{ky}$, or $\frac{2}{3}z''=-y''e^{ky}+\frac{3}{2}(y')^2 e^{ky}$, or $$-\frac{4}{3}z''=2y''e^{ky}-3(y')^2 e^{ky}\text{, or even}$$ $$-\frac{4z''}{3z}=2y''-3(y')^2\text.$$

Your original equation is now $$\frac{4z''}{3z}=8x^2\text,$$ which I hope will get you closer.

Indeed...

Given $z''=6x^2z$, expanding $z$ as a power series in $x$ yields $a_n=\frac{a_{n-2}}{(n+1)(n+2)}$ times a constant, which makes one think that $z$ must be something to do with $e^{x^2}$.

Put $$z=e^{ax^2+bx+c}$$ and differentiate twice, and you get something like $$z''=2az+4a^2x^2z+2abxz+b^2z\text,$$ which is enticing but has an embarrassing $xz$ term. Fortunately that is the only term in which $b$ isn't squared. So if you take $$z=e^{ax^2+bx+c}+e^{ax^2-bx+c}$$ and differentiate it twice, you will get rid of the unwanted $xz$ term and you will have constraints on $a$ and $b$ which make $z$ satisfy the differential equation.

General note: All this has been done on the back of one and a half envelopes, so do check it.

• mathworld.wolfram.com/ParabolicCylinderFunction.html There is no way I can see how this is a MA CS problem. – player100 Aug 16 '16 at 3:00
• Why? It only needs high school calculus (at least in the UK: the US may be different). – Martin Kochanski Aug 16 '16 at 11:47
• @MartinKochanski Please, have a look at the new equation. How do you think, is it easier to solve the new one? – Egor Okhterov Aug 16 '16 at 19:23
• @Pixar : Of course, it is much easier to solve the new one. – JJacquelin Aug 17 '16 at 1:43
• The only problem is that the substitution $z=e^{-\frac{3}{2}y}$ doesn't seem to work. When you take the equation $z'' = 6x^2z$ and let $z=e^{-\frac{3}{2}y}$, the resulting equation is $2y'' - 3(y')^2 + x^2 = 0$, so the $y''$ is not multiplied by $y$ as it is in the original equation. Good try. – Futurologist Aug 18 '16 at 4:07

$$\begin{cases} 3y''y + 3(y')^2 - 2x^2 = 0, \\ y(0) = 1, \\ y'(0) = 0. \end{cases}$$

$$y''y+y'^2=(y'y)'\quad\to\quad 3(y'y)'=2x^2$$ $$3y'y=\frac{2}{3}x^3+c_1$$ $y'(0)=0\quad\to\quad c_1=0$ $$y'y=\frac{2}{9}x^3$$ $$2y'y=(y^2)'=\frac{4}{9}x^3$$ $$y^2=\frac{1}{9}x^4+c_2$$ $y(0)=1\quad\to\quad c_2=1$ $$y=\sqrt{\frac{1}{9}x^4+1}$$

• I think that the first initial condition excludes the solution $y=-\sqrt{\frac{1}{ 9} x^4+1}$. Am I right? – Egor Okhterov Aug 17 '16 at 12:27
• You are right. Now corrected. Thank you. – JJacquelin Aug 17 '16 at 15:45

Hint:

Let $y=\dfrac{1}{u^2}$ ,

Then $y'=-\dfrac{2u'}{u^3}$

$y''=\dfrac{6(u')^2}{u^4}-\dfrac{2u''}{u^3}$

$\therefore\dfrac{2}{u^2}\left(\dfrac{6(u')^2}{u^4}-\dfrac{2u''}{u^3}\right)-3\left(-\dfrac{2u'}{u^3}\right)^2+8x^2=0$

$\dfrac{12(u')^2}{u^6}-\dfrac{4u''}{u^5}-\dfrac{12(u')^2}{u^6}=-8x^2$

$\dfrac{4u''}{u^5}=8x^2$

$u''=2x^2u^5$

This reduces to a special case of Emden-Fowler equation.

• Are you inferring that $x^2$ is the only solution that can be written down? – user99914 Aug 14 '16 at 17:18