(dy)/(dx)=((2x+5))/((x^2+2x)),y(1)=1

Lösung

\frac{d y}{d x} = \frac{( 2 x + 5 )}{( x ^{2} + 2 x )}, y (1) = 1

y = \frac{5}{2} ln (x) - \frac{1}{2} ln (x + 2) + 1 + \frac{ln ( 3 )}{2}

Schritte zur Lösung

\frac{d y}{d x} = \frac{( 2 x + 5 )}{( x ^{2} + 2 x )}

Ersetze

\frac{d y}{d x}

mit

y^{'} (x)

y^{^{'}} (x) = \frac{( 2 x + 5 )}{( x ^{2} + 2 x )}

Wenn

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

dann

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

y = \int \frac{2 x + 5}{x ^{2} + 2 x} d x

\int \frac{2 x + 5}{x ^{2} + 2 x} d x = \frac{5}{2} ln (x) - \frac{1}{2} ln (x + 2) + c_{1}

y = \frac{5}{2} ln (x) - \frac{1}{2} ln (x + 2) + c_{1}

Wende die Anfangsbedingung an:

y = \frac{5}{2} ln (x) - \frac{1}{2} ln (x + 2) + 1 + \frac{ln ( 3 )}{2}

y = \frac{5}{2} ln (x) - \frac{1}{2} ln (x + 2) + 1 + \frac{ln ( 3 )}{2}

\int 2 x^{\frac{3}{2}} d x

a re a \frac{1}{x}, \frac{1}{x ^{2}}, 1, 2

\frac{d}{d x} (\frac{x ^{3} - x}{1 + x ^{2}})

s l o p e in t erce pt (0, 1), (8, 6)

d er i v a t i v e \frac{3 x - 9}{x}