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?
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?
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?
Do you know how to use an Integrating factor - Wikipedia, the free encyclopedia?
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.