solution:
[tex]find the solution of the inital value problem.\\
consider the differential equation y'=\frac{x(x^2+1)}{4y^3}\\
with initial condition y(0)=-\frac{1}{2}\\
y'=\frac{x(x^2+1)}{4y^3}\\
\frac{dy}{dx}=\frac{x(x^2+1)}{4y^3}\\
4y^3dy=x(x^2+1)dx\\
taken integral\\
4y^3dy=x(x^2+1)dx\\
\int 4y^3dy=\int x(x^2+1)dx\\
\frac{4y^4}{4}=\frac{x^4}{3}+\frac{x^2}{2}+c\\
put x=0,and y(0)=-\frac{-1}{\sqrt{2}}\\
y^4=\frac{x^4}{3}+\frac{x^2}{2}+c\\[/tex][tex](\frac{-1}{\sqrt{2}})^4=0+0+c\\
c=\frac{1}{4}\\
therefore, the solution is\\
y^4=\frac{x^4}{3}+\frac{x^2}{2}+\frac{1}{4}\\
y=(\frac{x^4}{3}+\frac{x^2}{2}+\frac{1}{4})^\frac{1}{4}[/tex]
