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

[Saladsamurai]

Homework Statement

I have the following ODE:

$$\frac{d(f^2g)}{dx} = \frac{b}{fg}\qquad(1)$$

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 $$f\frac{d(f^2g)}{dx} = \frac{b}{g}\qquad(2)$$

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

Any thoughts? Is this worth the time?

 

[fzero]

You can try to set

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

I seem to find $$m=f^2/4$$. This is just the standard use of an integrating factor.

 

[Saladsamurai]

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) =ax^1/2 so that (2) becomes:

$$af\frac{d(f^2x^{1/2})}{dx} = \frac{b}{ax^{1/2}}\qquad(3)$$

Carrying out the differentiation:

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

$$\Rightarrow\frac{f^3}{2x^{1/2}} + 2x^{1/2}f\frac{df}{dx} = \frac{b}{a^2x^{1/2}} \qquad(5)$$

$$\Rightarrow f^3 + 4xf\frac{df}{dx} - A = 0 \qquad(6)$$

where $A = \frac{2b}{a^2}$

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

EDIT: didn't see you there fzero Not sure which is easier Solving yours, or (6) .

 

[fzero]

I think my trick might not be too helpful, since you're still left with

$$\int e^m dx$$

to deal with.

 

[SammyS]

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.
...

$$\Rightarrow f^3 + 4xf\frac{df}{dx} - A = 0 \qquad(6)$$

where $A = \frac{2b}{a^2}$

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

That looks like it's separable.

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

$$\Rightarrow\ \frac{df}{\displaystyle \frac{A}{f}-f^2} = \frac{dx}{4x}$$

.