y^'=(2x+1)/(x^2-3x+2),y(3)=0

Lösung

y^{^{'}} = \frac{2 x + 1}{x ^{2} - 3 x + 2}, y (3) = 0

y = - 3 ln (x - 1) + ln (8 (x - 2)^{5})

Schritte zur Lösung

y^{'} (x) = \frac{2 x + 1}{x ^{2} - 3 x + 2}

Wenn

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

dann

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

y = \int \frac{2 x + 1}{x ^{2} - 3 x + 2} d x

\int \frac{2 x + 1}{x ^{2} - 3 x + 2} d x = - 3 ln (x - 1) + 5 ln (x - 2) + c_{1}

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

Wende die Anfangsbedingung an:

y = - 3 ln (x - 1) + ln (8 (x - 2)^{5})

y = - 3 ln (x - 1) + ln (8 (x - 2)^{5})

x \frac{d y}{d x} = 1 + 3 y + y^{2}

\frac{d}{d x} (x cos (x) + 2 tan (x))

t an g e n t - x^{2} + 7 x + 2, at x = 3

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

x \to 0 + lim (\frac{x}{x})