2y^'+y+1=0

Lösung

2 y^{^{'}} + y + 1 = 0

y = e^{\frac{- t - c _{1}}{2}} - 1

Schritte zur Lösung

2 y^{'} (t) + y + 1 = 0

Solve separable ODE:

y = e^{\frac{- t - c _{1}}{2}} - 1

y = e^{\frac{- t - c _{1}}{2}} - 1

sin (x) y^{^{'}} = y cos (x) + 2 cos (x)

\frac{y ^{^{'}}}{y} = e^{2 x^{2} + 5}

3 x y^{^{'}} - y = 0

(1 + x^{2}) y^{^{'}} + y = - 1

y^{^{'}} = e^{3 x + 2 y}, y (2) = \frac{1}{3}