u^'=2x\H(x)

Lösung

u^{^{'}} = 2 x H (x)

u (x) = \int 2 x H (x) d x

Schritte zur Lösung

u^{'} (x) = 2 x H (x)

Wenn

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

dann

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

u (x) = \int 2 x H (x) d x

u (x) = \int 2 x H (x) d x

y^{^{'}} = \frac{7 cos ( 7 x )}{3 + 2 y}, y (0) = - 1

y^{^{'}} = y \cdot ln (y)

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

d y = (\frac{y}{x} - cos (e) c^{2} \frac{y}{x}) d x

2 x e^{x^{2} + y^{2}} + 2 y e^{x^{2} + y^{2}} \frac{d y}{d x} = 0