y^'=e^{6x}-4x

Lösung

y^{^{'}} = e^{6 x} - 4 x

y = \frac{1}{6} e^{6 x} - 2 x^{2} + c_{1}

Schritte zur Lösung

y^{'} (x) = e^{6 x} - 4 x

Wenn

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

dann

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

y = \int e^{6 x} - 4 x d x

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

y = \frac{1}{6} e^{6 x} - 2 x^{2} + c_{1}

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

\frac{\partial}{\partial x} (x^{2} - 4 x y + y^{3} + 4 y)

d er i v a t i v e \frac{- 200 ln ( \frac{f}{40} )}{( f + 20 )}

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

\int_{1}^{\infty} \frac{1}{n ^{2}}