u^'=e^{9x-\H(x)}

Lösung

u^{^{'}} = e^{9 x - H (x)}

u (x) = \int e^{9 x - H (x)} d x

Schritte zur Lösung

u^{'} (x) = e^{9 x - H (x)}

Wenn

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

dann

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

u (x) = \int e^{9 x - H (x)} d x

u (x) = \int e^{9 x - H (x)} d x

y^{^{'}} = y (5 x^{2} + 3)

x + 1 y^{^{'}} = y + 1

(t^{2} - 16 t + 48) (\frac{d y}{d t}) = y

\frac{d z}{d t} = 2 t e^{6 z}

\frac{d}{d x} ((y) \cdot 10) + 5 y = 5