(dy)/(dx)y=cos^{-1}(1/x)

Lösung

\frac{d y}{d x} y = cos^{- 1} (\frac{1}{x})

y = 2 x arccos (\frac{1}{x}) - ln (\frac{x ^{2} - 1}{x} + 1) + ln (\frac{x ^{2} - 1}{x} - 1) + c_{1}, y = - 2 x arccos (\frac{1}{x}) - ln (\frac{x ^{2} - 1}{x} + 1) + ln (\frac{x ^{2} - 1}{x} - 1) + c_{1}

Schritte zur Lösung

\frac{d y}{d x} y = arccos (\frac{1}{x})

Solve separable ODE:

y = 2 x arccos (\frac{1}{x}) - ln (\frac{x ^{2} - 1}{x} + 1) + ln (\frac{x ^{2} - 1}{x} - 1) + c_{1}, y = - 2 x arccos (\frac{1}{x}) - ln (\frac{x ^{2} - 1}{x} + 1) + ln (\frac{x ^{2} - 1}{x} - 1) + c_{1}

y = 2 x arccos (\frac{1}{x}) - ln (\frac{x ^{2} - 1}{x} + 1) + ln (\frac{x ^{2} - 1}{x} - 1) + c_{1}, y = - 2 x arccos (\frac{1}{x}) - ln (\frac{x ^{2} - 1}{x} + 1) + ln (\frac{x ^{2} - 1}{x} - 1) + c_{1}

s l o p e (3.4) (- 6.4)

x \to 0 - lim (ln (x^{8}))

d er i v a t i v e \frac{8}{x ^{2} + 4}

\int 9 x sec (x) tan (x) d x

\int 3 v (v^{2} + 2)^{2} d v