How do I find an integrating factor and solve (2x^2y + x) dy + (xy^2 + y) dx = 0?

Answer 1

Let M= xy² +y and N= 2x²y +x test for exactness ∂M/∂y = 2xy+1 and ∂N/∂x =4xy+1. Since the two partials are not equal the equation is not exact. Since N = xf(xy) and M= yf(xy) use the Integration factor I =1/(xM-yN) which gives -1/x²y^{2. Now rearrange the equation and divide by I -(xdy + ydx)/x^2y^2 -1(2x^2ydy + xy^2dx)/x^2y^2 =0 this now gives -d(-1/xy) = 2dy/y +dx/x or d(1/xy) = 2dy/y +dx/x. Doing the integration 1/xy= lny^2 +lnx + c. Now 1/xy = lnxy^2 + c.}