Question

Find the differential equation representing the family of curves $v = \dfrac{A}{r} + B$, where A and B are arbitrary constants.

Answer 1

Hint:
The dependency of v on r is given with two constants A and B in the equation. So, we need to differentiate v with respect to r twice so that we can eliminate A and B to obtain the required differential equation.
Given:
The family of curves whose differential equation has to be obtained is represented by $v = \dfrac{A}{r} + B$ is given where A and B are arbitrary constants.

Stepwise solution:
Since, $v = \dfrac{A}{r} + B$ eq. (1)
Differentiate both sides of eq. (1) with respect to r, we have
$\dfrac{{dv}}{{dr}} = \, - \dfrac{A}{{{r^2}}}$
Let us rearrange the above equation to obtain A explicitly on one side, we get the following:
$ \Rightarrow \, - A = \,{r^2}\dfrac{{dv}}{{dr}}$ eq. (2)
Differentiate both sides of eq. (2) with respect to r, we have
$0 = {r^2}\dfrac{{{d^2}v}}{{d{v^2}}} + \dfrac{{dv}}{{dr}}.\,\dfrac{{d{r^2}}}{{dr}}$
\[ \Rightarrow \,0 = {r^2}\dfrac{{{d^2}v}}{{d{v^2}}} + 2r\dfrac{{dv}}{{dr}}\]
$ \Rightarrow\,r\dfrac{{{d^2}v}}{{d{r^2}}} + \,2\dfrac{{dv}}{{dr}} = 0$ eq. (3)
We have eliminated both the arbitrary constants A and B. So, the differential equation which represents the family of curves $v = \dfrac{A}{r} + B$ is given by $r\dfrac{{{d^2}v}}{{d{r^2}}} + \,2\dfrac{{dv}}{{dr}} = 0$.

Note:
The student must keep in mind that whenever one needs to find the differential equation of a family of curves, one has to differentiate the equation as many times as there are the numbers of arbitrary constants. The differential equation must not contain any arbitrary constant. Students often commit mistakes of eliminating only a few constants and make the differential equation which contains one or more constants. This is completely wrong.