# Thread: 2nd order non-linear ODE, no 'x' term

1. ## 2nd order non-linear ODE, no 'x' term

I'm trying to solve the ODE

$\displaystyle yy"=4(y')^2$

I'm "close" to the answer provided, but I can't see where I'm going wrong. This is what I've done:

let $\displaystyle z=y'$, so $\displaystyle y" = \frac{dz}{dx} = \frac{dy}{dx}\frac{dz}{dy}=z\frac{dz}{dy}$

Then the ODE becomes $\displaystyle yz\frac{dz}{dy}=4z^2$, which is separable and gives me

$\displaystyle z = Cy^4$

Now integrating wrt x to recover y,

$\displaystyle y = Cxy^4+K$

However... the answer provided is $\displaystyle y = (Cx+K)^{-1/3}$, and I don't think manipulating what I have will get me here

So what have I done wrong?

2. Divide both sides by yy' and integrate.

3. Originally Posted by HD09
I'm trying to solve the ODE

$\displaystyle yy"=4(y')^2$

I'm "close" to the answer provided, but I can't see where I'm going wrong. This is what I've done:

let $\displaystyle z=y'$, so $\displaystyle y" = \frac{dz}{dx} = \frac{dy}{dx}\frac{dz}{dy}=z\frac{dz}{dy}$

Then the ODE becomes $\displaystyle yz\frac{dz}{dy}=4z^2$, which is separable and gives me

$\displaystyle z = Cy^4$

Now integrating wrt x to recover y,

$\displaystyle y = Cxy^4+K$

However... the answer provided is $\displaystyle y = (Cx+K)^{-1/3}$, and I don't think manipulating what I have will get me here

So what have I done wrong?
If $\displaystyle z = cy^4$ then $\displaystyle \dfrac{dy}{dx} = cy^4$. Now separate $\displaystyle \dfrac{dy}{y^4} = cdx$, integrate and solve for y.