(4xy+3y^2-x)dx+x(x+2y)dy=0

解

(4 x y + 3 y^{2} - x) d x + x (x + 2 y) d y = 0

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

解答ステップ

(4 x y + 3 y^{2} - x) d x + x (x + 2 y) d y = 0

完全 ODE を解く:

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

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