Solving (1/x^2) d/dx [x^2 (df/dx)] = -f^(3/2)

[Helios]

help me solve
( 1 / x^2 ) d / dx [ x^2 ( df / dx ) ] = - f^( 3 / 2 )
I think it needs a series sol'n but it's tough

 

[dextercioby]

A series solution doesn't work. The equation is nonlinear in "f".

Daniel.

 

[HallsofIvy]

help me solve
( 1 / x^2 ) d / dx [ x^2 ( df / dx ) ] = - f^( 3 / 2 )
I think it needs a series sol'n but it's tough

This is the same as
\frac{df^2}{dx^2}+ \frac{2}{x}\frac{df}{dx}= f^{\frac{3}{2}[/itex]<br /> As dextercioby said, it is non-linear. There are no general methods for solving non-linear differential equations- not even series solutions.

 

[dextercioby]

I think Halls meant

\frac{d^{2}f}{dx^{2}}+\frac{2}{x}\frac{df}{dx}=f^{3/2}

but the exact form is less relevant. The nonlinearity is.

Daniel.

 

[HallsofIvy]

Actually, I had that but wrote "\fra {2}{x}" instead of "\frac{2}{x}"!

 

[Matthew Rodman]

There is a simple solution of the form f (x) = \frac{A}{x^4}, if that's any help. (I haven't worked out what the value of A is yet).

edit: by my first estimate, A = 144.

 

[tim_lou]

you can substitute in:
g(x^n)=f(x)
then you'll get:
nx^{n-1}g&#039;&#039;(x^n)+(n^2-n)x^{n-2}g&#039;(x^n)+2nx^{n-2}g&#039;(x^n)=g^{\frac{3}{2}}(x^n)

let n be -1 to eliminate the first derivative term. Then maybe you can do some tricks and simplify...

 

[Helios]

Thanks Matt, but..

Thanks but a minus sign is being omitted here! Look at the original. It seems that 144 i / r^4 is a sol'n but it's imaginary. But it might be a clue. My intuition tells me that the real sol'n will have a nice form.

 

[Matthew Rodman]

Yeah - sorry 'bout the sign. I wasn't using the equation in the initial post.