derivative of x^2y^2+xy=0

Lösung

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

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

Schritte zur Lösung

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

Wenn

f^{^{'}} (x) = g (x)

dann

f (x) = \int g (x) d x

x^{2} y^{2} + x y = \int 0 d x

\int 0 d x = c_{1}

x^{2} y^{2} + x y = c_{1}

Isoliere

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

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

Vereinfache

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

y^{^{'}} = \frac{1 + y}{tan ( x )}

(x^{2} + 1) y^{^{'}} + 4 x y = 0

\frac{d y}{d x} = \frac{4 y}{x}, y = 2, x = 1

y^{^{'}} = sin (3 x) y^{2}

\frac{d y}{d x} = (x - 6) e^{- 2 y}, y (6) = ln (6)