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?
i mean: dv/v+Adt=Bdt/v dv/v+(A-B/v)dt=0
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