# A simple differential equation problem.

1. Nov 24, 2006

### MathematicalPhysicist

is there a solution to this euqation: (i got to in a problem in physics so the notation is appropiate):
$$(m_1+m_2)\frac{dv}{dt}=m_1g-m_2gsin(a)-f_s-\beta v-V\rho g$$
is there a solution or perhaps i got it wrong in equations?

2. Nov 24, 2006

### arildno

Well, the notation might be appropriate, but your definition of the symbols is not! :grumpy:
Assuming that everything is constant apart from v,
just rewrite your diff. eq as:
$$\frac{dv}{dt}+Av=B$$
where A and B are constants.
You may use an integrating factor to solve this problem.
This will also work for non-constant A and B (depending solely on t, and not on v!!), but you might not get an explicit formula in terms of elementary functions then.

Last edited: Nov 24, 2006
3. Nov 24, 2006

### MathematicalPhysicist

what's wrong with my interpratation of the symbols?

4. Nov 24, 2006

### arildno

It is lacking. How should I know what you mean by your symbols when you don't define them?

5. Nov 24, 2006

### MathematicalPhysicist

well, the m's are masses, the angle a doesnt change with time (perhaps here lies your critic), V stands for volume and rho stand for water density.
that's it.

6. Nov 24, 2006

### MathematicalPhysicist

perhaps, v=Acos(b)+Bsin(b) is a possible solution here?

7. Nov 24, 2006

### arildno

Hmm..and I guess that g is the acceleration due to gravity and beta is the coefficient of air resistance. But you didn't sayu that.
Furthermore f_s is some sort of applied force I don't know what is.

8. Nov 24, 2006

### MathematicalPhysicist

what integrating factor to use here?
dv/v(A+B/v)+dt=0 will that suffice here?

9. Nov 24, 2006

### CPL.Luke

this is a pretty simple integrating factor

if you mmove the resistance term (Bv) to the left hand side and divide by (m1+m2) than you have an quation of the form

dv/dt +Pv = Q

where Q is some function of t, and so is P

in a problem like this the integrating factor is e^(integral of P dt)

multiplying by that and simplifying will get you an equation of the form

d(ve^(integral P dt))/dt =Qe^(integral P dt)

integrate, and then you'll have a simple algebra problem

also notice that the solution only has 1 parameter, all first order differential equation have only one parameter in the solution.

10. Nov 30, 2006

### Pseudo Statistic

Why use an integrating factor? The ODE's clearly separable...

11. Dec 1, 2006

### arildno

Why use separation of variables?
You can clearly use an integrating factor..

It's just a matter of preference what method you use.

12. Dec 1, 2006

### CPL.Luke

hmm it doesn't appear to be seperable.

if you note its of the form dv/dt +Bv=C

as somebody else mentioned, that is not a seperable equation

13. Dec 1, 2006

### arildno

So, you disagree that the diff.eq may be rewritten as:
$$\frac{1}{C-Bv}\frac{dv}{dt}=1$$ ??

14. Dec 1, 2006

### CPL.Luke

oh my mistake, didn't see that manipulation