(2xy+2y^2)dx+(x^2+4xy)dy=0

解

(2 x y + 2 y^{2}) d x + (x^{2} + 4 x y) d y = 0

y = \frac{- x ^{2} + x ^{4} + 8 c _{1} x}{4 x}, y = \frac{- x ^{2} - x ^{4} + 8 c _{1} x}{4 x}

解答ステップ

(2 x y + 2 y^{2}) d x + (x^{2} + 4 x y) d y = 0

完全 ODE を解く:

y = \frac{- x ^{2} + x ^{4} + 8 c _{1} x}{4 x}, y = \frac{- x ^{2} - x ^{4} + 8 c _{1} x}{4 x}

y = \frac{- x ^{2} + x ^{4} + 8 c _{1} x}{4 x}, y = \frac{- x ^{2} - x ^{4} + 8 c _{1} x}{4 x}