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!

  • $\begingroup$ 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

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 $$ y(x)=\pm\sqrt{a+bx+\frac{4+b^2}{4a}x^2}. $$

  • $\begingroup$ 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

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.


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 $y.$


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