(dy)/(dx)=-1/3 x^3

Lösung

\frac{d y}{d x} = - \frac{1}{3} x^{3}

y = - \frac{x ^{4}}{12} + c_{1}

Schritte zur Lösung

\frac{d y}{d x} = - \frac{1}{3} x^{3}

Ersetze

\frac{d y}{d x}

mit

y^{'} (x)

y^{^{'}} (x) = - \frac{1}{3} x^{3}

Wenn

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

dann

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

y = \int - \frac{1}{3} x^{3} d x

\int - \frac{1}{3} x^{3} d x = - \frac{x ^{4}}{12} + c_{1}

y = - \frac{x ^{4}}{12} + c_{1}

\frac{d ^{2}}{d x ^{2}} (x + \frac{1}{x})

\int \frac{cos ( 2 x )}{cos ( x )} d x

8 x^{2} y^{^{'}} = y^{^{'}} + 8 x e^{- y}

d er i v a t i v e f (x) = - 7 x - \frac{2}{x ^{7}}

\frac{\partial}{\partial x} (arctan (y))