Homework Question: Can this ODE be solved using separation of variables?

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Homework Help Overview

The discussion revolves around an ordinary differential equation (ODE) of the form \(\frac{d(f^2g)}{dx} = \frac{b}{fg}\), where \(f\) is an unknown function and \(g\) is a known function. Participants are exploring whether this ODE can be solved using separation of variables or other techniques.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to manipulate the ODE to find a form suitable for integration, questioning the feasibility of expressing one side as a derivative. Some participants suggest using integrating factors and explore the implications of substituting a specific function for \(g(x)\). Others express doubt about the effectiveness of these approaches, noting the complexity of the resulting equations.

Discussion Status

The discussion is active, with participants sharing different approaches and expressing uncertainty about the best method to tackle the problem. Some guidance on potential techniques has been offered, but there is no explicit consensus on a solution or method yet.

Contextual Notes

Participants are grappling with the nonlinear nature of the equations that arise from their manipulations, and there is a recognition of the challenges posed by the specific form of \(g(x)\) chosen for substitution.

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Homework Statement



I have the following ODE:

[tex] \frac{d(f^2g)}{dx} = \frac{b}{fg}\qquad(1)[/tex]

Where b is a known constant, f is an unknown function of x that I am seeking, and g is a known function of x.

Now, my next step was to actually plug in my known function of g(x), carry out the differentiation and seek a solution to that ODE. But I am getting to the point with this DE's that I like to look for 'tricks' that allow me to integrate directly by writing one side as a derivative. I was wondering if that could be done here? If I write (1) as [tex] f\frac{d(f^2g)}{dx} = \frac{b}{g}\qquad(2)[/tex]

I was thinking that the left side could be written in the form dP/dx.

Any thoughts? Is this worth the time?
 
Last edited:
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You can try to set

[tex]d \left( \frac{1}{2} e^m f^3 g^2 \right) =e^m f g ~d(f^2g).[/tex]

I seem to find [tex]m=f^2/4[/tex]. This is just the standard use of an integrating factor.
 
Meh, screw the trick ... I plugged in for g(x) and it's a nightmare, so I doubt there is a trick. Let's see if we can just solve the ODE.

g(x) =ax1/2 so that (2) becomes:

[tex] <br /> af\frac{d(f^2x^{1/2})}{dx} = \frac{b}{ax^{1/2}}\qquad(3)<br /> [/tex]

Carrying out the differentiation:

[tex] <br /> f\left [ f^2(1/2)x^{-1/2} + x^{1/2}(2)f\frac{df}{dx}\right ] = \frac{b}{a^2x^{1/2}}\qquad(4)<br /> [/tex]

[tex] <br /> \Rightarrow\frac{f^3}{2x^{1/2}} + 2x^{1/2}f\frac{df}{dx} = \frac{b}{a^2x^{1/2}}<br /> \qquad(5)<br /> [/tex]

[tex] <br /> \Rightarrow f^3 + 4xf\frac{df}{dx} - A = 0<br /> \qquad(6)<br /> [/tex]

where [itex]A = \frac{2b}{a^2}[/itex]

I'm not really sure how to solve (6) as it is very nonlinear ... any thoughts?

EDIT: didn't see you there fzero :smile: Not sure which is easier :redface: Solving yours, or (6) .
 
I think my trick might not be too helpful, since you're still left with

[tex]\int e^m dx[/tex]

to deal with.
 
Saladsamurai said:
Meh, screw the trick ... I plugged in for g(x) and it's a nightmare, so I doubt there is a trick. Let's see if we can just solve the ODE.
...

[tex]\Rightarrow f^3 + 4xf\frac{df}{dx} - A = 0<br /> \qquad(6)[/tex]

where [itex]A = \frac{2b}{a^2}[/itex]

I'm not really sure how to solve (6) as it is very nonlinear ... any thoughts?
...

That looks like it's separable.

[tex](6)\Rightarrow f^2 + 4x\frac{df}{dx} - \frac{A}{f} = 0[/tex]

[tex]\Rightarrow\ \frac{df}{\displaystyle \frac{A}{f}-f^2} = \frac{dx}{4x}[/tex]

.
 

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