1st order ODE, Reduction to Separable Form Help required

Hey guys,
I'm very new to this form of solving differential equations and I've come to a stump in a problem.

The problem is simply to find the general solution to $y'=(4x^2+y^2)/(xy)$

I have a guide that also shows the steps to the solution, but the area I am stuck on is identical to the guide, where i don't see how it got from one step to the next.

This is what I have so far;

Set $u=\frac{y}{x}, y=xu, y'=u+xu'$

$y'=\frac{4x^2}{xy}+\frac{y^2}{xy}$

$y'=\frac{4x}{y}+\frac{y}{x}$

$u+xu'=\frac{4}{u}+u$

Therefore;

$xu'=\frac{4}{u}$

Now, at this point I assumed it would be very easy to separate the variables to;

$uu'=\frac{4}{x}$

And solve by integration. However, the guide goes from

$xu'=\frac{4}{u}$

to

$2u du = \frac{8}{x} dx$
My question is... what step am i missing that u becomes 2u and 4/x becomes 8/x?

Thanks for your time

Re: 1st order ODE, Reduction to Separable Form Help required

Quote:

Originally Posted by Nuingaer

Hey guys,
I'm very new to this form of solving differential equations and I've come to a stump in a problem.

The problem is simply to find the general solution to $y'=(4x^2+y^2)/(xy)$

I have a guide that also shows the steps to the solution, but the area I am stuck on is identical to the guide, where i don't see how it got from one step to the next.

This is what I have so far;

Set $u=\frac{y}{x}, y=xu, y'=u+xu'$

$y'=\frac{4x^2}{xy}+\frac{y^2}{xy}$

$y'=\frac{4x}{y}+\frac{y}{x}$

$u+xu'=\frac{4}{u}+u$

Therefore;

$xu'=\frac{4}{u}$

Now, at this point I assumed it would be very easy to separate the variables to;

$uu'=\frac{4}{x}$

And solve by integration. However, the guide goes from

$xu'=\frac{4}{u}$

to

$2u du = \frac{8}{x} dx$
My question is... what step am i missing that u becomes 2u and 4/x becomes 8/x?

Thanks for your time

They multiplied both sides by 2. Why, I don't know. But it's valid...

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Re: 1st order ODE, Reduction to Separable Form Help required

A simpler method :