# A 2nd order DE

#### 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

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#### dextercioby

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

Daniel.

#### HallsofIvy

Homework Helper
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
$$\frac{df^2}{dx^2}+ \frac{2}{x}\frac{df}{dx}= f^{\frac{3}{2}[/itex] As dextercioby said, it is non-linear. There are no general methods for solving non-linear differential equations- not even series solutions. Last edited by a moderator: #### dextercioby Science Advisor Homework Helper I think Halls meant [tex] \frac{d^{2}f}{dx^{2}}+\frac{2}{x}\frac{df}{dx}=f^{3/2}$$

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

Daniel.

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#### 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.

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#### tim_lou

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

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

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#### 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.