Answer
(a)
The solution of this differential equation is $y=-\sqrt{2\ln{(1+x^2)}+4}$.
(c)
The interval in which the solution is defined is $[\sqrt{e^{-2}-1},\infty]$.
(b)
Work Step by Step
(a)
We have given $\dfrac{dy}{dx}=\dfrac{2x}{y+x^2y}$
Now solve the differential equation as follows:
$\dfrac{dy}{dx}=\dfrac{2x}{y(1+x^2)}$
$y\,dy=\dfrac{2x}{(1+x^2)}\,dx$
Now integrate both sides as follows:
$\int y\,dy=\int\dfrac{2x}{(1+x^2)}\,dx$
Now let $1+x^2=t \implies 2x\,dx=dt$
We get, $\int y\,dy=\int\dfrac{1}{t}\,dt$
$\dfrac{y^{2}}{2}=\ln{t}+C$
Now substitute back $t$.
We get, $\dfrac{y^{2}}{2}=\ln{(1+x^2)}+C$
Or, $y=\pm\sqrt{2\ln{(1+x^2)}+C_0}$
Thus the explicit form can be $y=\sqrt{2\ln{(1+x^2)}+C_0}$ or $y=-\sqrt{2\ln{(1+x^2)}+C_0}$.
We have given the initial condition $y(0) = -2$.
Thus, only $y=-\sqrt{2\ln{(1+x^2)}+C_0}$ is possible.
Substitute $x=0$ and $y =-2$
We get, $-2=-\sqrt{2\ln{(1)}+C_0}$
$\implies 2=\sqrt{C_0}$
$\implies C_0=4$
Hence, the solution of this differential equation is $y=-\sqrt{2\ln{(1+x^2)}+4}$.
(c)
We know that the value inside the square root can not be negative.
Therefore, $2\ln{(1+x^2)}+4\ge 0$
$\implies 2\ln{(1+x^2)}\ge -4$
$\implies \ln{(1+x^2)} \ge -2$
$\implies 1+x^2\ge e^{-2}$
$\implies x\ge \sqrt{e^{-2}-1}$
Hence, the interval in which the solution is defined is $[\sqrt{e^{-2}-1},\infty]$.
(b)
