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

  • Thread starter Helios
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In summary: It was the one in the first reply.In summary, the equation (1/x^2) d/dx[x^2(df/dx)] = -f^(3/2) is non-linear and does not have a general solution method. A series solution does not work. However, a simple solution of the form f(x) = A/x^4 may work, with A = 144. Further simplification may lead to a real solution for the equation.
  • #1
Helios
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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
 
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  • #2
A series solution doesn't work. The equation is nonlinear in "f".

Daniel.
 
  • #3
Helios said:
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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  • #4
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.
 
  • #5
Actually, I had that but wrote "\fra {2}{x}" instead of "\frac{2}{x}"!
 
  • #6
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.
 
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  • #7
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...
 
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  • #8
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
Yeah - sorry 'bout the sign. I wasn't using the equation in the initial post.
 

1. What does the equation (1/x^2) d/dx [x^2 (df/dx)] = -f^(3/2) represent?

The equation represents a second-order differential equation in which the dependent variable, f, is related to the independent variable, x, and its first and second derivatives.

2. How do you solve the equation (1/x^2) d/dx [x^2 (df/dx)] = -f^(3/2)?

To solve this equation, you can use the method of separation of variables. First, rearrange the equation to get df/dx on one side and f on the other side. Then, integrate both sides with respect to x and solve for f.

3. What is the general solution to the equation (1/x^2) d/dx [x^2 (df/dx)] = -f^(3/2)?

The general solution to this equation is f(x) = C/x, where C is a constant. This solution can be obtained by substituting u = f^(1/2) and solving the resulting first-order differential equation.

4. Can this equation be solved using any other methods?

Yes, this equation can also be solved using the method of power series. In this method, you assume that the solution can be represented by a power series and then find the coefficients by substituting the series into the equation.

5. What are the applications of the equation (1/x^2) d/dx [x^2 (df/dx)] = -f^(3/2)?

This equation has various applications in physics and engineering, such as in the study of heat transfer, fluid mechanics, and electrical circuits. It can also be used to model natural phenomena, such as population growth or radioactive decay.

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