Non-Linear ODE

1. Jun 17, 2015

joshmccraney

Hi PF!

Can any of you help me reduce this ODE to find a solution?

$$y y''+2y'^2+xy'+\frac{1}{2}y = 0 \implies \\ (y y')'+y'^2+xy'+\frac{1}{2}y = 0 \implies\\ (yy')'+(xy)'+y'^2-\frac{1}{2}y=0$$

but here I am stopped. Am I even going the correct route? I know I can re-write this equation as terms primed, because a quadratic solves this ODE. Any help is awesome!

2. Jun 17, 2015

RUber

Are x and y both functions of t, so (xy)' = xy' + x'y? Or is y a function of x, so (xy)' = xy' +y ?

3. Jun 17, 2015

pasmith

I can see that if $y'' = 2a$ then every term of the ODE is a multiple of $x^2$, so certainly $y = ax^2$ is a solution (for two particular values of $a$).

You can of course just substitute $y = ax^2 + bx + c$ and see what constraints there are on $(a,b,c)$.

4. Jun 17, 2015

joshmccraney

Sorry, $y$ is the dependent variable and $x$ is the independent variable.

5. Jun 17, 2015

joshmccraney

And yes, I can verify that a quadratic solves the ODE, but I'm trying to see how to reduce this into an equation of primes. I'm curious and know it can be done, I just can't see how. Any ideas?

6. Jun 17, 2015

pasmith

Substituting $v = 2y' + x$ yields $\frac12 yv' + y'v = 0$. Thus either $y = 0$ or $v = 0$ or $$\frac12 \frac{d(\ln v)}{dx} + \frac{d(\ln y)}{dx} = 0.$$

(EDIT: Actually just multiplying $\frac12 yv' + y'v = 0$ by $2y$ suffices.)

Last edited: Jun 18, 2015
7. Nov 24, 2015

joshmccraney

This was awesome, pasmith! How would you go about solving this one $y y''+2y'^2+xy'+Ay = 0$ if $A$ could be any non-zero real number you wanted it to be?

8. Nov 25, 2015

pasmith

You may still be able to obtain quadratic solutions (if $y$ is a quadratic then so is $yy'' + 2y'^2 + xy' + Ay$) but setting $v = 2y' + x$ yields $$\frac{d}{dx}(y^2 v) = (1 - 2A)y^2$$ which only helps if $A = \frac12$.

Last edited: Nov 25, 2015
9. Nov 29, 2015

joshmccraney

Yea, it seems $A=1/2$ is the only way this works. It's very clever, though!

10. Nov 29, 2015

MidgetDwarf

Just out of curiosity is it all possible to solve this via power series? Or does the y*y" make it a no go?

11. Dec 1, 2015

pasmith

Try it. If $y = \sum_{n=0}^\infty a_nx^n$ then $yy''$ term gives you $$a_0a_{n+2}(n+2)(n+1) + a_1a_{n+1}(n+1)n + \dots + a_{k}a_{n+2-k}(n+ 2 - k)(n + 1 - k) + \dots + 2a_na_2$$ as the coefficient of $x^n$ and $y'^2$ gives you $$(n+1)a_1a_{n+1} + \dots + (k+1)(n - k + 1)a_{k+1}a_{n - k + 1} + \dots + (n+1)a_1a_{n+1}$$ so given $a_0$ and $a_1$ you can obtain $a_2$, then $a_3$, and so on.

12. Dec 1, 2015

MidgetDwarf

Thanks! Will try it out on hand and later run it trough mathematica.