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

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August 25th 2010, 03:05 PM
HD09

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

I'm trying to solve the ODE

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

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

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

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

$z = Cy^4$

Now integrating wrt x to recover y,

$y = Cxy^4+K$

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

So what have I done wrong?
August 25th 2010, 04:05 PM
Ackbeet

Divide both sides by yy' and integrate.
August 26th 2010, 04:50 AM
Danny

Quote:

Originally Posted by HD09

I'm trying to solve the ODE

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

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

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

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

$z = Cy^4$

Now integrating wrt x to recover y,

$y = Cxy^4+K$

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

So what have I done wrong?

If $z = cy^4$ then $\dfrac{dy}{dx} = cy^4$ . Now separate $\dfrac{dy}{y^4} = cdx$ , integrate and solve for y.

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