# Thread: how to solve this?

1. ## how to solve this?

Solve:
y''=yy'/x
And show that y=c3 is a solution of the problem too.
where y'=dy/dx

How to solve it?

2. ## Re: how to solve this?

The second question is easy: if y is a constant, then both y' and y'' are 0 so the equation becomes 0= 0.

3. ## Re: how to solve this?

Originally Posted by blocnt
Solve:
y''=yy'/x
And show that y=c3 is a solution of the problem too.
where y'=dy/dx

How to solve it?
Well I have a stab at it that might lead somewhere. But I can't finish it. Perhaps it will give someone an idea.

Do you have initial conditions on this?

Note that $\frac{1}{2} \frac{d}{dx}(y^2) = y y'$ and that $x y'' = \frac{d}{dx}(xy') - y'$.

Then
$y'' = \frac{y y'}{x}$

$x y'' = y y'$

$\int x y'' dx = \int y y' dx$

$\int \left ( \frac{d}{dx}(x y') - y' \right ) dx = \frac{1}{2}y^2$

$xy' - \int y' dx = \frac{1}{2}y^2$

$xy' - y = \frac{1}{2}y^2 + a$
where a is the accumulated integration constant.

My hope was for a Bernoulli equation, but we have that extra constant that seems to screw things up. I can't find a way to proceed from here.

-Dan

Edit: I was hoping that, since y = constant was a solution (leading to a =0 in the above equation) would help, but it leads to a Bernoulli equation where y = constant is not a solution to the Bernoulli, so this doesn't help.

4. ## Re: how to solve this?

I know a little late but the ODE is now separable.