Fluid flows steadily under a constant pressure gradient along a straight tube of circular cross-section of radius $a$. The velocity $v$ of a particle of the fluid is parallel to the axis of the tube and depends only on the distance from the axis. The equation satisfied by $v$ is
$$\frac 1r \frac{\mathrm d}{\mathrm dr} \big( r \frac{\mathrm dv}{\mathrm dr}\big) = -k$$
Find the general solution for $v$.
Here's what I did:
$$\frac 1r \frac{\mathrm d}{\mathrm dr} \big( r \frac{\mathrm dv}{\mathrm dr}\big) = -k$$
$$\int \frac 1r \frac{\mathrm d}{\mathrm dr} \big( r \frac{\mathrm dv}{\mathrm dr}\big) \, \mathrm dr = \int -k \, \mathrm dr$$
Integrate by parts, with
$u = \frac 1r$ and $\mathrm du = -\frac1{r^2} \mathrm dr$
$v = r \frac{\mathrm dv}{\mathrm dr}$ and $\mathrm dv = \frac{\mathrm d}{\mathrm dr} \big( r \frac{\mathrm dv}{\mathrm dr}\big) \, \mathrm dr$
$$\frac 1r \cdot \big( r \frac{\mathrm dv}{\mathrm dr}\big) - \int \big( r \frac{\mathrm dv}{\mathrm dr}\big) \cdot \frac{-1}{r^2} \, \mathrm dr = -kr + c$$
$$\frac{\mathrm dv}{\mathrm dr} - \int - \frac 1r \, \mathrm dv = -kr + c$$
$$\frac{\mathrm dv}{\mathrm dr} + \frac vr = -kr + c$$
$$\frac{\mathrm dv}{\mathrm dr} = -kr - \frac vr + c$$
$$v = \int (-kr - \frac vr + c) \, \mathrm dr $$
$$v = - \frac 12 kr^2 - v \ln r + c_1r + c_2$$
$$v + v \ln r = - \frac 12 kr^2 + c_1r + c_2$$
$$v(1 + \ln r) = - \frac 12 kr^2 + c_1r + c_2$$
$$v = - \frac{kr^2 + c_1r + c_2}{1 + \ln r}$$
But apparently this is wrong. The solution was $v(t) = -\frac 14 kr^2 + A \ln r + B$, but I do not understand how this was acquired. Where did I go wrong?
You first have $$\frac 1r \frac{\mathrm d}{\mathrm dr} \big( r \frac{\mathrm dv}{\mathrm dr}\big) = -k$$
Multiplying both sides by r,
$$ \frac{\mathrm d}{\mathrm dr} \big( r \frac{\mathrm dv}{\mathrm dr}\big) = -kr$$
Integrating both sides,
This gives $$\mathrm {d} \big( r \frac{\mathrm dv}{\mathrm dr}\big) = -kr \mathrm dr$$
Integrating, $$\int \mathrm {d} \big( r \frac{\mathrm dv}{\mathrm dr}\big) = \int-kr \mathrm dr \implies$$
$$ r \frac{\mathrm dv}{\mathrm dr} = -\frac{1}{2} kr^2 +A$$
Dividing by r, $$ \frac{\mathrm dv}{\mathrm dr} = -\frac{1}{2} kr +\frac{A}{r}$$
So, $$ \mathrm dv = -\frac{1}{2} kr +\frac{A}{r} {\mathrm dr}$$ and, $$ \int \mathrm dv = \int -\frac{1}{2} kr +\frac{A}{r} {\mathrm dr} \implies$$
$$v(t) = -\frac{1}{4} kr^2 +{A}\ln{r} + B$$ which is the desired answer. So, as you can see, integration by parts was not necessary, though a good insight.