(2x^3-x^2y)+(2x^2y-x^3)y^'=0

Lösung

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

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

Schritte zur Lösung

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

Solve using substitution:

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

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

\frac{d}{d x} (arctan (\frac{x + y}{1 - x y}))

\frac{d}{d x} (4 x - 4)

\frac{\partial}{\partial y} (3 x^{2} (5 x + 7 y))

d er i v a t i v e f (x) = x^{3} - 6 x^{2} + 9 x

x \to \frac{π}{4} lim (tan (2 x))