# A 2nd order DE

1. Jan 5, 2007

### 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

2. Jan 5, 2007

### dextercioby

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

Daniel.

3. Jan 5, 2007

### HallsofIvy

Staff Emeritus
This is the same as
$$\frac{df^2}{dx^2}+ \frac{2}{x}\frac{df}{dx}= f^{\frac{3}{2}[/itex] As dextercioby said, it is non-linear. There are no general methods for solving non-linear differential equations- not even series solutions. Last edited: Jan 5, 2007 4. Jan 5, 2007 ### dextercioby I think Halls meant [tex] \frac{d^{2}f}{dx^{2}}+\frac{2}{x}\frac{df}{dx}=f^{3/2}$$

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

Daniel.

5. Jan 5, 2007

### HallsofIvy

Staff Emeritus

6. Jan 6, 2007

### 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.

Last edited: Jan 6, 2007
7. Jan 6, 2007

### 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...

Last edited: Jan 6, 2007
8. Jan 10, 2007

### 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.

9. Jan 12, 2007

### Matthew Rodman

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