(12x^2-20y^2)dx+(-20xy+36x^{-1}y^3)dy=0

解

(12 x^{2} - 20 y^{2}) d x + (- 20 x y + 36 x^{- 1} y^{3}) d y = 0

y = - \frac{- 5 x ^{2} + - 2 x ^{4} - c _{1}}{9 x ^{2}} x, y = - - \frac{- 5 x ^{2} + - 2 x ^{4} - c _{1}}{9 x ^{2}} x, y = \frac{5 x ^{2} + - 2 x ^{4} - c _{1}}{3}, y = - \frac{5 x ^{2} + - 2 x ^{4} - c _{1}}{3}

解答ステップ

(12 x^{2} - 20 y^{2}) d x + (- 20 x y + 36 x^{- 1} y^{3}) d y = 0

置換を使用して解く:

y = - \frac{- 5 x ^{2} + - 2 x ^{4} - c _{1}}{9 x ^{2}} x, y = - - \frac{- 5 x ^{2} + - 2 x ^{4} - c _{1}}{9 x ^{2}} x, y = \frac{5 x ^{2} + - 2 x ^{4} - c _{1}}{3}, y = - \frac{5 x ^{2} + - 2 x ^{4} - c _{1}}{3}

y = - \frac{- 5 x ^{2} + - 2 x ^{4} - c _{1}}{9 x ^{2}} x, y = - - \frac{- 5 x ^{2} + - 2 x ^{4} - c _{1}}{9 x ^{2}} x, y = \frac{5 x ^{2} + - 2 x ^{4} - c _{1}}{3}, y = - \frac{5 x ^{2} + - 2 x ^{4} - c _{1}}{3}