ax^'+x=0

Lösung

a x^{^{'}} + x = 0

x = e^{\frac{- t + c _{1}}{a}}

Schritte zur Lösung

a x^{'} (t) + x = 0

Solve separable ODE:

x = e^{\frac{- t + c _{1}}{a}}

x = e^{\frac{- t + c _{1}}{a}}

111 - 0.5 y = 55 y^{^{'}}

\frac{d y}{d x} = - 2 cos (x) sin (y)

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

x = \frac{y}{y ^{^{'}} + y}

20 (\frac{d v}{d t}) = - (20) (32) - 0.0025 v