Question

In each of problems 9 through 20: (a) find the solution of the given initial value problem in explicit form.

Answer 1

solution:

$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$$(\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}$