# Thread: Past papers with brief solutions :(

1. ## Past papers with brief solutions :(

Ive been going through exam papers and theres a question on differential equations which I cant get to grips with

m*dv/dt = mg-kv [g> k/mv]

dv/dt= g-kv/m

dt = g-kv/m dv

im not sure on how to integrate this can anyone help explain it?

2. ## Re: Past papers with brief solutions :(

This is a linear ODE, and you want to write it in standard form:

$\displaystyle \frac{dv}{dt}+\frac{k}{m}v=g$

Now, calculate the integrating factor $\displaystyle \mu(t)=e^{\int\frac{k}{m}\,dt}=e^{\frac{k}{m}t}$ and multiply the ODE by this:

$\displaystyle e^{\frac{k}{m}t}\frac{dv}{dt}+\frac{k}{m}e^{\frac{ k}{m}t}v=ge^{\frac{k}{m}t}$

Now, the left side is the differentiation of a product:

$\displaystyle \frac{d}{dt}\left(e^{\frac{k}{m}t}v \right)=ge^{\frac{k}{m}t}$

Can you proceed from here?

4. ## Re: Past papers with brief solutions :(

The integration for dt is simply "t".
The expression on the RHS is being integrated with respect to "v" so you should treat everything else as a constant and then proceed.
The expression ∫g-kv/m dv = g∫dv- k/m∫vdv
= gv- k/m(v^2/2)+c

Hope this helps.

5. ## Re: Past papers with brief solutions :(

hey thanks that helps alot guys I should be able to go from here thank you