ay^'+y=c

Lösung

a y^{^{'}} + y = c

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

Schritte zur Lösung

a y^{'} (t) + y = c

Solve separable ODE:

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

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

\frac{d}{d θ} (f (θ)) f (θ) = sin (θ) + csc (θ)

4 ln (x) d x - x sec^{2} (y) d y = 0

\frac{d y}{d x} 2 x y = 1

\frac{d Q}{d x} = 15 - 3 \frac{Q}{400}

e^{t} y^{^{'}} = (1 + y^{2}) t