(dy}{dx}=\frac{x^3-x^2+x)/x

Lösung

\frac{d y}{d x} = \frac{x ^{3} - x ^{2} + x}{x}

y = \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + x + c_{1}

Schritte zur Lösung

\frac{d y}{d x} = \frac{x ^{3} - x ^{2} + x}{x}

Ersetze

\frac{d y}{d x}

mit

y^{'} (x)

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

Wenn

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

dann

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

y = \int \frac{x ^{3} - x ^{2} + x}{x} d x

\int \frac{x ^{3} - x ^{2} + x}{x} d x = \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + x + c_{1}

y = \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + x + c_{1}

t an g e n t y = tan (x) + \frac{1}{3} tan^{3} (x)

x \to - 3 lim (x^{3} + 4 x^{2} - 7)

y^{^{''}} + 4 y = e^{- x} (7 - 4 x + 5 x^{2})

\frac{d y}{d x} = (y + 1) (y - 2), y (0) = 0

\frac{\partial}{\partial x} (6 ln (6 x^{3} + y^{2}))