2
$\begingroup$

I'm having this question in my homework assignment in Linear Algebra and diffrential equation class, and trying to find the general solution for this second ODE.

$$y''y^3 = 1$$

Using substitution I said $p = y'$ and $p' = y'' \rightarrow \frac{dp}{dx}= \frac{dp}{dy} \times \frac{dy}{dx}$

Then we get $y^3 \frac{dp}{dx}$ now what's the next step should I integrate or differentiate the equation? What's the trick on these type of ODEs?

Thanks in advance!

$\endgroup$
  • $\begingroup$ The end result, following the hints given in some answers below, is $$y(x)=\pm\sqrt{\pm2x+c}.$$ $\endgroup$ – Did Feb 12 '15 at 9:30
  • $\begingroup$ @Did, I think you miss some $x^2$ term inside the square root. $\endgroup$ – mickep Feb 12 '15 at 9:38
  • $\begingroup$ @mickep Show me. $\endgroup$ – Did Feb 12 '15 at 9:39
  • $\begingroup$ @Did, I'll put it in an answer. $\endgroup$ – mickep Feb 12 '15 at 9:46
  • $\begingroup$ @Did, it is done... $\endgroup$ – mickep Feb 12 '15 at 9:59
5
$\begingroup$

Update This is now a complete solution.

Let $v(x)=y(x)^2$. Then $v'=2yy'$ and so (here we assume that $v\neq 0$) $$ v''=2(y')^2+2yy''=\frac{1}{2}\frac{(v')^2}{y^2}+\frac{2}{y^2}=\frac{1}{2}\frac{(v')^2}{v}+\frac{2}{v}, $$ or $$ vv''-\frac{1}{2}(v')^2-2=0 $$ This differential equation can be solved as follows. Differentiating, we find that $$ v'v''+vv'''-v'v''=0, $$ so $vv'''=0$. Hence $v'''=0$. But then $v$ must be a polynomial of degree $2$. Since we differentiated we cannot expect any polynomial of degree $2$ to work. We insert a second degee polynomial $v=a+bx+cx^2$ into the second order differential equation and look for conditions on $a$, $b$ and $c$ that gives us a solution. The condition becomes $$ 2ac-\frac{b^2}{2}-2=0. $$ Solving for $c$, we find that $$ v(x)=a+bx+\frac{4+b^2}{4a}x^2. $$ Then, one could go back to $y$ to get $$ y(x)=\pm\sqrt{a+bx+\frac{4+b^2}{4a}x^2}. $$

$\endgroup$
  • $\begingroup$ Mathematica gives $ y^2(x)= a + \frac{ (x+b)^2}{a} $ Can it be taken to the same form by adjustment of constants? $\endgroup$ – Narasimham Feb 14 '15 at 19:49
  • $\begingroup$ @Narasimham Indeed, if you in $d+(x+c)^2/d$ let $c=$2ab/(4+b^2)$ and $d=4a/(4+b^2)$ you end up with the expression I give above. $\endgroup$ – mickep Feb 14 '15 at 20:37
  • $\begingroup$ @mickup And can be cast in the form $ (a y/b)^2=(x−b)^2 $representing hyperbolas? I ask this (quickly,else we may have to move to chat) for two reasons. One, I did not see DE in this form for a conic before and second, that a variation of the DE could be generalized to represent all conics. – Narasimham 10 mins ago $\endgroup$ – Narasimham Feb 14 '15 at 21:09
4
$\begingroup$

Divide by y^3 then multiply by y' and integrate. That gets you to first order. I think the next integration is possible but a bit messy.

$\endgroup$
1
$\begingroup$

let $\frac{dy}{dx} = w.$ then you have a system of two differential equations $$\frac{dy}{dx} = w, \frac{dw}{dx} = \frac{1}{y^3}$$ from these two, you can get a differential equation $$\frac{dw}{dy} = \frac{1}{y^3}$$ which can be integrated to give you $$w = \frac{dy}{dx} = \frac{1}{2}(C -\frac{1}{y^2})$$ now you separate the variables to get $$\frac{y^2dy}{Cy^2 - 1} = \frac{dx}{2}$$

you can use partial fractions to find an integral for $y.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.