u^'=\H(t)+2(\H(t))^2

Lösung

u^{^{'}} = H (t) + 2 (H (t))^{2}

u (t) = \int H (t) + 2 H (t)^{2} d t

Schritte zur Lösung

u^{'} (t) = H (t) + 2 (H (t))^{2}

Wenn

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

dann

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

u (t) = \int H (t) + 2 (H (t))^{2} d t

u (t) = \int H (t) + 2 (H (t))^{2} d t

Vereinfache

u (t) = \int H (t) + 2 H (t)^{2} d t

x \cdot y^{^{'}} = 2 \cdot y

y^{^{'}} = y, y (0) = 6

y^{^{'}} = 0.6 y (1 - \frac{y}{1120})

f (x) x + x f^{^{'}} (x) x + x^{2} f (x) x = 0

\frac{d y}{d x} = 20 x y