Find the integrating factor in the form #x^n y^m# and then solve? #(2x^(-1) +x^(-2)y^(-1)-3y)dx+(3y^(-1)-4x+2x^(-1)y^(-2))dy=0#

1 Answer
Aug 12, 2018

Answer:

The integrating factor is: #x^2 y^3#

# f(bbx) = x^2 y^(3) - x^(3)y^4+ x y^2 = C#

Explanation:

# underbrace((2/x +1/(x^2y)-3y))_(f_x)dx+underbrace((3/y -4x+2/(xy^2)))_(f_y)dy=0#

But with the suggested I.F.

# underbrace((2x^(n-1)y^m + x^(n-2)y^(m-1)-3x^ny^(m+1)))_(f_x)dx+underbrace((3x^n y^(m-1) -4x^(n+1)y^m+2x^(n-1)y^(m-2))))_(f_y)dy=0#

#f_(xy) = 2mx^(n-1)y^(m-1) + (m-1)x^(n-2)y^(m-2)-3(m+1)x^ny^m #

#f_(yx) = 3nx^(n-1) y^(m-1) -4(n+1)x^ny^m+2(n-1)x^(n-2)y^(m-2)#

Comparing like exponents in terms:

#f_(xy) = f_(yx) implies {(2m = 3n),(m - 1 = 2 ( n - 1)),(3(m+1) = 4 ( n + 1)):} qquad implies {(n = 2),(m = 3):}#

The integrating factor is: #x^2 y^3#

-# {( f= int (2xy^3 + y^2-3x^2y^4 ) del x = x^2y^3 + xy^2- x^3y^4 + alpha(y)),(f= int (3x^2 y^(2) -4x^(3)y^3+2x y) del y = x^2 y^(3) - x^(3)y^4+ x y^2 + beta (x)):}#

#:. f(bbx) = x^2 y^(3) - x^(3)y^4+ x y^2 = C#