Find an integrating factor making ydx2x3ydy0 exact and use i

Find an integrating factor making ydx+(2x+3y)dy=0 exact, and use it to find implicit solutions to the differential equation.Then find an explicit solution to the initial value problem given by this equation and the initial condition y(1)=-1

Solution

Let us multiply by y. We get

y^2dx+(2xy+3y^2)dy=0

(y^2)_y=2y

(2xy)_x=2y

subscript denotes partial derivative w.r.t. indicated variable.

Let, F be the solution

Hence, dF=y^2dx+(2xy+3y^2)dy=0

dF=F_xdx+F_ydy

So, F_x=y^2,F_y=2xy+3y^2

F_x=y^2

Integrating this w.r.t. x treating y as a parameter gives

F=xy^2+g(y)=C , C is constant of integration

F_y=2xy+g\'(y)=2xy+3y^2

Hence, g\'=3y^2

g=y^3

F=xy^2+y^3=C

y(1)=-1

So, C=1-1=0

C=0

So solution is

xy^2+y^3=0