Thread: second order non linear ODE

1. second order non linear ODE

Does anyone can help me to solve this second order non linear ODE:

y'' + (2/x)(y') - (1/2y)(y')(y') = K,

y' = dy/dx

y'' = dy'/dx

y = y(x)

I've already guess y=Ax^2 satisfy this equation, but I want to solve it analitically..

I have transform the equation above with x=-(1/u) and f = ln H, that lead to a new equation:

f'' + (1/2)(f')(f') = -(K/u^4)exp{-f}

f = f(u)

f' = df/du

f'' = df'/du

but still cannot find the solution analitically.

Is y'' + (2/x)(y') - (1/2y)(y')(y') = K cannot be solved analitically?
It seem strange to me that an an ODE with a simple subtitution solution y=Ax^2 don't have any analytical solution..

Thanks before..

2. Originally Posted by ceramica
Does anyone can help me to solve this second order non linear ODE:

y'' + (2/x)(y') - (1/2y)(y')(y') = K,

y' = dy/dx

y'' = dy'/dx

y = y(x)

I've already guess y=Ax^2 satisfy this equation, but I want to solve it analitically..

I have transform the equation above with x=-(1/u) and f = ln H, that lead to a new equation:

f'' + (1/2)(f')(f') = -(K/u^4)exp{-f}

f = f(u)

f' = df/du

f'' = df'/du

but still cannot find the solution analitically.

Is y'' + (2/x)(y') - (1/2y)(y')(y') = K cannot be solved analitically?
It seem strange to me that an an ODE with a simple subtitution solution y=Ax^2 don't have any analytical solution..

Thanks before..
The ODE

$y'' + (2/x)(y') - (1/2y)(y')(y') = K,$

is difficult to read - and what is the last term $- \frac{1}{2y} y'^2$?

3. <br>

4. yes.. the ODE is:
$y''+ \frac{2}{x} y' - \frac{1}{2y} y'^2=K
$

just want to know whether this equation can be solved analitically or not.

5. Originally Posted by ceramica
yes.. the ODE is:
$y''+ \frac{2}{x} y' - \frac{1}{2y} y'^2=K$
just want to know whether this equation can be solved analitically or not.
Are solutions with $K = 0$ of any interest?

6. I've already solved it for $K = 0
$

It is $y = (1 + \frac{a}{bx})^2
$

I need to know whether this ODE can (or cannot) be solved analitically, so I can do numerical solution as soon as possible.

Thanks before.

7. Forget to tell, K is a constant, in general, is not zero.