Question

what is the integrating factor of the non-exact differential equation (3x^2+y^3+4)dx+(x^2-2xy)dy=0

Answer 1

$(3x^2+y^3+4)dx+(x^2-2xy)dy=0\\ M=3x^2+y^3+4, N=x^2-2xy\\ \frac{dM}{dy}=3y^2, \frac{dN}{dx}=2x-2y\\ \frac{dM}{dy} \ne \frac{dN}{dx}\\ \text{multiply through by}, x^my^n\\ x^my^n(3x^2+y^3+4)dx+ x^my^n(x^2-2xy)dy=0\\ (3x^{m+2}y^n+x^my^{n+3}+4x^my^n)dx+(x^{m+2}y^n-2x^{m+1}y^{n+1})dy=0\\ M=3x^{m+2}y^n+x^my^{n+3}+4x^my^n\\ N=x^{m+2}y^n-2x^{m+1}y^{n+1}\\ \frac{dM}{dy}=3nx^{m+2}y^{n-1}+(n+3)x^my^{n+2}+4nx^my^{n-1}\\ \frac{dN}{dx}=(m+2)x^{m+1}y^n-2(m+1)x^my^{n+1}\\ \frac{dM}{dy}=\frac{dN}{dx}\\ 3nx^{m+2}y^{n-1}+(n+3)x^my^{n+2}+4nx^my^{n-1}=(m+2)x^{m+1}y^n-2(m+1)x^my^{n+1}\\ 3nx^{m+2}y^{n-1}+(n+3)x^my^{n+2}+4nx^my^{n-1}-(m+2)x^{m+1}y^n+2(m+1)x^my^{n+1}=0\\ \text{set the coeefficients to be 0, to get the value of m and n}\\ \text{integrating factor is}\\, x^my^n \\ \text{for the value of m, n}\\ 3n=0, n=0\\ n+3=0, n=-3\\ 4n=0, n=0\\ -m-2=0, m=-2\\ 2m+2=0, m=-1\\ \text{the integrating factor becomes}\\ x^0y^{-2}$