Answer
(a)
The solution of this differential equation is $y=(-\sqrt{1+x^2}+2)^{-\tfrac{1}{2}}$ or $y=\dfrac{1}{\sqrt{2-\sqrt{1+x^2}}}$.
(c)
The interval in which the solution is defined is $(-\sqrt{3},\sqrt{3})$.
(b)
Work Step by Step
(a)
We have given $\dfrac{dy}{dx}=xy^3(1+x^2)^{\tfrac{-1}{2}}$
Now solve the differential equation as follows:
$\dfrac{dy}{dx}=xy^3(1+x^2)^{\tfrac{-1}{2}}$
$y^{-3}dy=\dfrac{x}{\sqrt{1+x^2}}\,dx$
Now integrate both sides as follows:
$\int y^{-3}dy=\int\dfrac{x}{\sqrt{1+x^2}}\,dx$
Let $1+x^2=u\implies x\,dx=\dfrac{du}{2}$
We get, $\int y^{-3}dy=\dfrac{1}{2}\int\dfrac{1}{\sqrt{u}}\,du$
$\implies -\dfrac{y^{-2}}{2}=\dfrac{1}{2}\sqrt{u}+C$
$\implies y^{-2}=-\sqrt{u}+C_0$
$\implies y=(-\sqrt{u}+C_0)^{-\tfrac{1}{2}}$
Now substitute back $u$.
$ y=(-\sqrt{1+x^2}+C_0)^{-\tfrac{1}{2}}$
Thus the explicit form is $y=(-\sqrt{1+x^2}+C_0)^{-\tfrac{1}{2}}$.
We have given the initial condition $y(0) = 1$.
Substitute $x=0$ and $y =1$
We get, $1=(-\sqrt{1}+C_0)^{-\tfrac{1}{2}}$
$\implies C_0-1=1$
$\implies C_0=2$
Hence, the solution of this differential equation is $y=(-\sqrt{1+x^2}+2)^{-\tfrac{1}{2}}$ or $y=\dfrac{1}{\sqrt{2-\sqrt{1+x^2}}}$.
(c)
We know the denominator can not be zero for a rational function.
Therefore, $\sqrt{2-\sqrt{1+x^2}}\ne 0$
$\implies 2-\sqrt{1+x^2}\ne 0$
$\implies 2 \ne \sqrt{1+x^2}$
$\implies 1+x^2\ne 4$
$\implies x^2\ne 3$
$\implies x\ne\sqrt{3} $ and $x\ne-\sqrt{3}$
Also, the value inside the square root can not be negative.
$2-\sqrt{1+x^2}\ge0$
$\implies 2\ge\sqrt{1+x^2}$
$\implies 4\ge1+x^2$
$\implies 3\ge x^2$
$\implies |\sqrt{3}|\ge x$
$\implies -\sqrt{3}\le x\le\sqrt{3}$
Also, $1+x^2> 0$ is always true for every real number.
Combining $-\sqrt{3}\le x\le\sqrt{3}$, $x\ne\sqrt{3} $ and $ x\ne-\sqrt{3}$
We get, $-\sqrt{3}< x
