35v^'=2-0.2v,v(0)=-6

Lösung

35 v^{^{'}} = 2 - 0.2 v, v (0) = - 6

v = - 16 e^{- \frac{t}{175}} + 10

Schritte zur Lösung

35 v^{'} (t) = 2 - 0.2 v

Solve separable ODE:

v = - 16 e^{- \frac{t}{175}} + 10

v = - 16 e^{- \frac{t}{175}} + 10

\frac{d y}{d x} y = 3 x^{3 x}

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

5 x + y y^{^{'}} = 0, y (2) = 5

\frac{d y}{d x} = x^{2} (- 6 y + 9)

x^{3} \frac{d y}{d x} = sec (y)