u^'x+\H(x)=(1-\H(x))/(1+\H(x))

Lösung

u^{^{'}} x + H (x) = \frac{1 - H ( x )}{1 + H ( x )}

u (x) = \int \frac{1 - 2 H ( x ) - H ( x ) ^{2}}{x ( 1 + H ( x ) )} d x

Schritte zur Lösung

u^{'} (x) x + H (x) = \frac{1 - H ( x )}{1 + H ( x )}

Isoliere

u^{'} (x) : u^{'} (x) = \frac{1 - 2 H ( x ) - H ( x ) ^{2}}{x ( 1 + H ( x ) )}

Wenn

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

dann

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

u (x) = \int \frac{1 - 2 H ( x ) - H ( x ) ^{2}}{x ( 1 + H ( x ) )} d x

u (x) = \int \frac{1 - 2 H ( x ) - H ( x ) ^{2}}{x ( 1 + H ( x ) )} d x

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

\frac{d y}{d x} y = \frac{2 x + 1}{x sin ( x )}

y^{^{'}} - b y = 2

x^{^{'}} = x^{2} - 4

y^{^{'}} = e^{2 x - 2 y}