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I'm trying to solve this non-linear differential equation using substitution $\dfrac{y}{x}$ to $t$. However, I can't solve this equation.

$$ \text{xy$\prime $} = \left(\frac{y}{x}\right)^3+y $$

How to solve this equation and what's the general solution?

Thanks.

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3 Answers 3

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if you change the dependent variable to $u=y/x$, as you said, the equation becomes $x^2u'=u^3$, which can be immediately be solved by separating variables.

Show us what you tried and we may help you with it.

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  • $\begingroup$ Just a comment, I think that you get x^2 u' = u^3 $\endgroup$ Oct 21, 2010 at 15:36
  • $\begingroup$ @Adrián: indeed, that was a typo. $\endgroup$ Oct 21, 2010 at 15:37
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The substitution $u = \frac{y}{x}$ works. What you have is

$y = ux$

i.e., $ y' = x u' + u$

or $x y' = x^2 u' + y$

Substituting this into your equation, we get

$x^2 u' = u^3$

Which can be solved using standard methods.

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    $\begingroup$ May I observe that this is exactly what I wrote? $\endgroup$ Oct 21, 2010 at 15:52
  • $\begingroup$ Indeed Mariano. I should have given my answer as a comment, but my rep wasn't high enough. $\endgroup$
    – svenkatr
    Oct 21, 2010 at 16:27
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Here is my solution. Is this right?

\begin{align} \text{xy$\prime $} = \left(\frac{y}{x}\right)^3+y \\ \\ \frac{y}{x}=u, \\ y = \text{ux}, \\ \text{y$\prime $} = \text{u$\prime $x} + u \\ \\ x^2\text{u$\prime $} = u^3 \\ \int \frac{1}{u^3} \, du =\int \frac{1}{x^2} \, dx \\ -\frac{1}{2u^2} = -\frac{1}{x} + c \\ u^{2 }= \frac{1}{\frac{2}{x}+c} \\ y^{2 }= \frac{x^2}{\frac{2}{x}+c} \\ y = x\left(\frac{2}{x}+c\right)^{-\frac{1}{2}} \end{align}

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  • $\begingroup$ Ooops.. my LaTex code is not working. $\endgroup$
    – Brian
    Oct 21, 2010 at 16:24
  • $\begingroup$ What are you trying to do with those $ inside \text{}? $\endgroup$ Oct 21, 2010 at 16:27
  • $\begingroup$ How can I insert new line in LaTex code? $\endgroup$
    – Brian
    Oct 21, 2010 at 16:31
  • $\begingroup$ Okay, it's working finally. $\endgroup$
    – Brian
    Oct 21, 2010 at 16:49
  • $\begingroup$ I think it's right. Thanks guys. $\endgroup$
    – Brian
    Oct 22, 2010 at 14:44

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