- #1
fresh
- 17
- 0
(x^2)(y^3) + x(1 + y^2)y' = 0
the integrating factor to make the above equation exact is (1)/(xy^3)
i have worked this equation out and have c = .5x^2 as the solution; however, the textbook says the solution is c = x^2 - y^(-2) + 2lnlyl
apparently they got this solution because h'(y) = y^(-3) + y(-1)
i found h'(y) to be equal to zero.
some sort of feedback would be greatly appreciated.
the integrating factor to make the above equation exact is (1)/(xy^3)
i have worked this equation out and have c = .5x^2 as the solution; however, the textbook says the solution is c = x^2 - y^(-2) + 2lnlyl
apparently they got this solution because h'(y) = y^(-3) + y(-1)
i found h'(y) to be equal to zero.
some sort of feedback would be greatly appreciated.