(x+1)y^'=x+6

Lösung

(x + 1) y^{^{'}} = x + 6

y = 5 ln (x + 1) + x + c_{1}

Schritte zur Lösung

(x + 1) y^{'} (x) = x + 6

Isoliere

y^{'} (x) : y^{'} (x) = \frac{x + 6}{x + 1}

Wenn

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

dann

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

y = \int \frac{x + 6}{x + 1} d x

\int \frac{x + 6}{x + 1} d x = 5 ln (x + 1) + x + c_{1}

y = 5 ln (x + 1) + x + c_{1}

\frac{\partial}{\partial x} (3 x^{4} - sin (5 x) + 4 x)

x \to - 2 lim (x + 2)

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

x \to 0 lim (2 x + 1 - 1)

\frac{d}{d x} ((ln (3 x))^{5})