Solving Non-Linear ODE: Tips and Guidance from PF Community

[member 428835]

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!

 

[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 ?

 

[pasmith]

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!

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).

 

[member 428835]

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

 

[member 428835]

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?

 

[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.)

 

[member 428835]

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?

 

[pasmith]

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?

You may still be able to obtain quadratic solutions (if y is a quadratic then so is yy&#039;&#039; + 2y&#039;^2 + xy&#039; + Ay) but setting v = 2y&#039; + x yields <br /> \frac{d}{dx}(y^2 v) = (1 - 2A)y^2 which only helps if A = \frac12.

 

[member 428835]

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

 

[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?

 

[pasmith]

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

Try it. If y = \sum_{n=0}^\infty a_nx^n then yy&#039;&#039; 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&#039;^2 gives you <br /> (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.

 

[MidgetDwarf]

Try it. If y = \sum_{n=0}^\infty a_nx^n then yy&#039;&#039; 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&#039;^2 gives you <br /> (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.

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