# First order nonlinear differential equation Help needed.

1. Sep 7, 2011

### plushval

Hi All,

I have been trying to solve following nonlinear differential equation, but I couldn't solve it.

0 = a*[f(t)]^{z/(z-1)} + (-t+C)*f(t) + b*[df(t)/dt]

where a, b and C are constants and 0< z<1.

Karen

2. Sep 7, 2011

3. Sep 7, 2011

### plushval

Thanks a lot for the suggestion, but unfortunately I could not make integrating factor technique work since integrating factors turns out to be a function of both "t" and "f". Any other suggestions?

Kind regards

4. Sep 7, 2011

### jackmell

I got a suggestion: When you get stuck, fall back and work on a simpler one first, then build it back up to the original problem:

$$f'+f^{1/2}+(c-t)f=0$$

now, what happens if I let $u=f^{1/2}$

you can convert that to a DE in u right?. You know, f=u^2, f'=2uu'. You can do that? Ok, put all that back in, solve it for u, then of course f=u^2. Ok, how about:

$$f'+f^{3/4}+(c-t)f=0$$

same dif. Let u=f^{1/4}. Now when you figure out what u is, then raise it to the fourth power to get f.

$$f'+f^{n/m}+(c-t)f=0$$

How about when n/m is not rational?

Last edited: Sep 7, 2011
5. Sep 7, 2011

### plushval

I will definitely try to solve it in the way that you suggested jackmell. Thanks a lot!

As a quick note, z/(z-1) is negative; hence it may cause some problems.

6. Sep 7, 2011

### plushval

As I said, z/(z-1) taking negative values is very problematic, therefore unfortunately, I could not solve the ODE using the approach you suggested, jackmall. I appreciate your time, though.

Any other suggestions would be highly appreciated.

Kind regards

7. Sep 8, 2011

### jackmell

If:

$$f'+\frac{1}{f^{1/2}}+(c-t)f=0$$

then what happens if we let:

$$u=f^{3/2}$$

If:

$$f'+\frac{1}{f^{1/4}}+(c-t)f=0$$

then what happens if we let:

$$u=f^{5/4}$$