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A 2nd order DE

  • Thread starter Helios
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243
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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

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

Daniel.
 

HallsofIvy

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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
[tex]\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.
 
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dextercioby

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I think Halls meant

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

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

Daniel.
 

HallsofIvy

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Actually, I had that but wrote "\fra {2}{x}" instead of "\frac{2}{x}"!
 
There is a simple solution of the form [tex]f (x) = \frac{A}{x^4}[/tex], 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:
682
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you can substitute in:
[tex]g(x^n)=f(x)[/tex]
then you'll get:
[tex]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)[/tex]

let n be -1 to eliminate the first derivative term. Then maybe you can do some tricks and simplify...
 
Last edited:
243
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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.
 
Yeah - sorry 'bout the sign. I wasn't using the equation in the initial post.
 

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