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

解

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

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

解答ステップ

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

置換を使用して解く:

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

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