a simple differential equation problem.

[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?
thanks in advance.

[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.

[MathematicalPhysicist]

what's wrong with my interpratation of the symbols?

[arildno]

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

[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.

[MathematicalPhysicist]

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

[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.

[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?

[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.

[Pseudo Statistic]

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

[arildno]

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

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

[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

[arildno]

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

[CPL.Luke]

oh my mistake, didn't see that manipulation