# How to solve this non-linear differential equation?

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.

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.

• Just a comment, I think that you get x^2 u' = u^3 – Adrián Barquero Oct 21 '10 at 15:36
• @Adrián: indeed, that was a typo. – Mariano Suárez-Álvarez Oct 21 '10 at 15:37

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.

• May I observe that this is exactly what I wrote? – Mariano Suárez-Álvarez Oct 21 '10 at 15:52
• Indeed Mariano. I should have given my answer as a comment, but my rep wasn't high enough. – svenkatr Oct 21 '10 at 16:27

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}

• Ooops.. my LaTex code is not working. – Brian Oct 21 '10 at 16:24
• What are you trying to do with those \$ inside \text{}? – Agustí Roig Oct 21 '10 at 16:27
• How can I insert new line in LaTex code? – Brian Oct 21 '10 at 16:31
• Okay, it's working finally. – Brian Oct 21 '10 at 16:49
• I think it's right. Thanks guys. – Brian Oct 22 '10 at 14:44