Answer
(a)
The solution of this differential equation is $r=\dfrac{1}{-\ln{\theta}+0.5}$.
(c)
The interval in which the solution is defined is $R^\textbf{+}-\{e^{0.5}\}$.
(b)
Work Step by Step
(a)
We have given $\dfrac{dr}{d\theta}=\dfrac{r^2}{\theta}$.
Now solve the differential equation as follows:
$\dfrac{dr}{d\theta}=\dfrac{r^2}{\theta}$
$\implies\dfrac{dr}{r^2}=\dfrac{d\theta}{\theta}$
Now integrate both sides as follows:
$\int \dfrac{dr}{r^2}=\int\dfrac{d\theta}{\theta}$
We get, $-\dfrac{1}{r}=\ln{\theta}+C$
Or, $\dfrac{1}{r}=-\ln{\theta}+C_0$
Or, $r=\dfrac{1}{-\ln{\theta}+C_0}$
Thus the explicit form is $r=\dfrac{1}{-\ln{\theta}+C_0}$.
We have given the initial condition $r(1) = 2$.
Substitute $\theta=1$ and $r=2$
We get, $2=\dfrac{1}{-\ln{1}+C_0}$
$\implies C_0=\dfrac{1}{2}=0.5$
Hence, the solution of this differential equation is $r=\dfrac{1}{-\ln{\theta}+0.5}$.
(c)
We know the denominator can not be zero for a rational function.
Therefore, $-\ln{\theta}+0.5\ne 0$
$\implies \ln{\theta}\ne 0.5$
$\implies \theta\ne e^{0.5}$
Also, theta can not be negative.
Hence, the interval in which the solution is defined is $R^\textbf{+}-\{e^{0.5}\}$.
(b)
