Consider the differential equation xy2 3 x2y1y 0 Find a s

Consider the differential equation xy2 + 3 + (x2y+1)y\' = 0 Find a solution to the initial value problem

Solution

xy^2+3+(x^2y+1)y\'=0 (xy^2+3) dx+(x^2y+1) dy=0 This is an exact equation We search for f(x,y) such that df/dx=xy^2+3 and df/dy=x^2y+1 df/dx=xy^2+3 f=x^2y^2/2+3x+C(y) df/dy=x^2y+1 f=x^2y^2/2+y+C(x) By comparing we get f=x^2y^2/2+y+3x The solution is the implicit equation x^2y^2/2+y+3x=C, where C is a constant