(dy)/(dx)=3x^4-6x^3+3x-8

Lösung

\frac{d y}{d x} = 3 x^{4} - 6 x^{3} + 3 x - 8

y = \frac{3 x ^{5}}{5} - \frac{3 x ^{4}}{2} + \frac{3 x ^{2}}{2} - 8 x + c_{1}

Schritte zur Lösung

\frac{d y}{d x} = 3 x^{4} - 6 x^{3} + 3 x - 8

Ersetze

\frac{d y}{d x}

mit

y^{'} (x)

y^{^{'}} (x) = 3 x^{4} - 6 x^{3} + 3 x - 8

Wenn

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

dann

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

y = \int 3 x^{4} - 6 x^{3} + 3 x - 8 d x

\int 3 x^{4} - 6 x^{3} + 3 x - 8 d x = \frac{3 x ^{5}}{5} - \frac{3 x ^{4}}{2} + \frac{3 x ^{2}}{2} - 8 x + c_{1}

y = \frac{3 x ^{5}}{5} - \frac{3 x ^{4}}{2} + \frac{3 x ^{2}}{2} - 8 x + c_{1}

t an g e n t f (x) = 5 x^{3} - 5 x^{2} + 2, (1, 2)

x \to - 1 - lim (3 x + 3 - 5)

y^{^{'}} = \frac{( y - y ^{2} )}{t}

d er i v a t i v e 5 5 x

in v erse l a pl a ce \frac{1}{s - 5}