Question

.

Answer 1

I don't know how to solve differential equation but I know this.
\[
v=\frac{ds}{dt}
\]

Answer 2

I can tell you that \(e^{-kt}\) is the solution to the homogeneous part. I don't know how to solve the inhomogeneous though.

Answer 3

It is separable right

Answer 4

\[\dfrac{ds}{dt} = v_0 -ks\]

\[\dfrac{ds}{v_0 - ks} = dt\]

\[\int\dfrac{ds}{v_0 - ks} = \int dt\]

Answer 5

I think you only add the constant of integration to one side. Like this:
\[
\int\frac{1}{v_0-ks}\,ds=\int dt\\
-\frac{1}{k}\ln(v_0-ks)+c=t
\]

Answer 6

I guess so. Just remember to use the initial condition to solve for the s(t) needed then take the second derivative..

Answer 7

Ah you add the constant to one side only because:
\[
\int\frac{1}{v_0-ks}\,ds=\int dt\\
-\frac{1}{k}\ln(v_0-ks)+c=t+k_1\\
-\frac{1}{k}\ln(v_0-ks)=t+k\quad\text{with }k=k_1-c\\
\]