Solve:

y''=yy'/x

And show that y=c_{3}is a solution of the problem too.

where y'=dy/dx

How to solve it?

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- Oct 29th 2013, 01:20 PMblocnthow to solve this?
Solve:

y''=yy'/x

And show that y=c_{3}is a solution of the problem too.

where y'=dy/dx

How to solve it? - Oct 29th 2013, 01:34 PMHallsofIvyRe: 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.

- Oct 30th 2013, 08:32 AMtopsquarkRe: how to solve this?
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 $\displaystyle \frac{1}{2} \frac{d}{dx}(y^2) = y y'$ and that $\displaystyle x y'' = \frac{d}{dx}(xy') - y'$.

Then

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

$\displaystyle x y'' = y y'$

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

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

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

$\displaystyle xy' - y = \frac{1}{2}y^2 + a$

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. - Nov 27th 2013, 05:13 AMJesterRe: how to solve this?
I know a little late but the ODE is now separable.