(x+1)dy=(x+6)dx

Lösung

(x + 1) d y = (x + 6) d x

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

Schritte zur Lösung

(x + 1) d y = (x + 6) d x

Lass

y

die abhängige Variable sein. Teile durch

d x

(x + 1) \frac{d y}{d x} = x + 6

Ersetze

\frac{d y}{d x}

mit

y^{'} (x)

(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} (x^{2} + x)

y^{^{''}} - 2 y^{^{'}} + y = 0, y (0) = 2, y^{^{'}} (0) = 6

a re a y = x^{2}, y = x^{4}

\int \frac{x + 4}{x ^{2} + 2 x + 5} d x

\frac{d}{d x} (\frac{- x ^{2} - 7}{( x ^{2} - 7 ) ^{2}})