2x^'+x=10

Lösung

2 x^{^{'}} + x = 10

x = - e^{\frac{- t - c _{1}}{2}} + 10

Schritte zur Lösung

2 x^{'} (t) + x = 10

Solve separable ODE:

x = - e^{\frac{- t - c _{1}}{2}} + 10

x = - e^{\frac{- t - c _{1}}{2}} + 10

sin (3 x) d x + 2 y \cdot cos^{3} (3 x) d y = 0

\frac{d y}{d x} y x + 1 = 4

y^{2} - \frac{y ^{^{'}}}{x} = 0

\frac{d y}{d x} x y + 2 ey = 2 e

\frac{d z}{d t} + e^{(t + z)} = 0