Answer
(a)
The solution of this differential equation is $y=-\sqrt[4]{\dfrac{x^4}{4}+\dfrac{x^2}{2}+\dfrac{1}{4}}$.
(c)
The interval in which the solution is defined is all real numbers $R$.
(b)
Work Step by Step
(a)
We have given $\dfrac{dy}{dx}=\dfrac{x(x^2+1)}{4y^3}$
Now solve the differential equation as follows:
$\dfrac{dy}{dx}=\dfrac{x(x^2+1)}{4y^3}$
$4y^3\,dy=(x^3+x)dx$
$4y^3\,dy=x^3\,dx+xdx$
Now integrate both sides as follows:
$\int4y^3\,dy=\int x^3\,dx+\int xdx$
We get, $y^{4}=\dfrac{x^4}{4}+\dfrac{x^2}{2}+C$
Thus the explicit form is $y=\pm\sqrt[4]{\dfrac{x^4}{4}+\dfrac{x^2}{2}+C}$.
We have given the initial condition $y(0) = \dfrac{−1}{\sqrt{2}}$.
Thus only $y=-\sqrt[4]{\dfrac{x^4}{4}+\dfrac{x^2}{2}+C}$ is possible.
Substitute $x=0$ and $y = \dfrac{−1}{\sqrt2}$
We get, $\dfrac{-1}{\sqrt2}=-\sqrt[4]{\dfrac{(0)^4}{4}+\dfrac{(0)^2}{2}+C}$
$\dfrac{-1}{\sqrt2}=-\sqrt[4]{C}$
$\implies C=\dfrac{1}{4}$
Hence, the solution of this differential equation is $y=-\sqrt[4]{\dfrac{x^4}{4}+\dfrac{x^2}{2}+\dfrac{1}{4}}$.
(c)
We know that the value inside the root can not be negative.
Therefore, $\dfrac{x^4}{4}+\dfrac{x^2}{2}+\dfrac{1}{4}> 0$
Since even powers of any real number are always positive.
$\dfrac{x^4}{4}+\dfrac{x^2}{2}+\dfrac{1}{4}$ is always positive.
Hence, the interval in which the solution is defined is all real numbers $R$.
(b)
