(dy)/(dx)= 1/(sqrt(1-x)),y(0)=1

Lösung

\frac{d y}{d x} = \frac{1}{1 - x}, y (0) = 1

y = - 2 1 - x + 3

Schritte zur Lösung

\frac{d y}{d x} = \frac{1}{1 - x}

Ersetze

\frac{d y}{d x}

mit

y^{'} (x)

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

Wenn

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

dann

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

y = \int \frac{1}{1 - x} d x

\int \frac{1}{1 - x} d x = - 2 1 - x + c_{1}

y = - 2 1 - x + c_{1}

Wende die Anfangsbedingung an:

y = - 2 1 - x + 3

y = - 2 1 - x + 3

d er i v a t i v e f (x) = \frac{x ^{2} + 2 x - 4}{( x + 1 ) ^{2}}

t an g e n t x^{3} - x^{2} + 1

\int x \cdot ln (\frac{1 - x}{1 + x}) d x

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

\int_{1}^{2} \frac{x}{2} - \frac{2}{x} d x