u^'=e^{6\H(t)+9t}

Lösung

u^{^{'}} = e^{6 H (t) + 9 t}

u (t) = \int e^{6 H (t) + 9 t} d t

Schritte zur Lösung

u^{'} (t) = e^{6 H (t) + 9 t}

Wenn

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

dann

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

u (t) = \int e^{6 H (t) + 9 t} d t

u (t) = \int e^{6 H (t) + 9 t} d t

4 y^{^{'}} + 2 y = 120, y (0) = 1

\frac{d y}{d t} = \frac{6 t ^{2}}{5 y}

\frac{\frac{d y}{d x}}{y} = 5

\frac{d y}{d x} = \frac{- x}{y}, y (4) = - 3

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