y^{''}+2y^'-y=t^{-1}*e^t

Lösung

y^{^{''}} + 2 y^{^{'}} - y = t^{- 1} \cdot e^{t}

y = c_{1} e^{(- 1 + 2) t} + c_{2} e^{(- 1 - 2) t} + e^{- t + 2 t} \cdot \int \frac{2 e ^{- 2 t + 2 t}}{4 t} d t + e^{- t - 2 t} \cdot \int - \frac{2 e ^{2 t + 2 t}}{4 t} d t

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} e^{(- 1 + 2) t} + c_{2} e^{(- 1 - 2) t} + e^{- t + 2 t} \cdot \int \frac{2 e ^{- 2 t + 2 t}}{4 t} d t + e^{- t - 2 t} \cdot \int - \frac{2 e ^{2 t + 2 t}}{4 t} d t

y = c_{1} e^{(- 1 + 2) t} + c_{2} e^{(- 1 - 2) t} + e^{- t + 2 t} \cdot \int \frac{2 e ^{- 2 t + 2 t}}{4 t} d t + e^{- t - 2 t} \cdot \int - \frac{2 e ^{2 t + 2 t}}{4 t} d t

y^{^{''}} - 2 y^{^{'}} + 5 y = 2 x + e^{- x}

x^{^{''}} - x^{^{'}} - 12 x = e^{4 t}, x (0) = 1, x^{^{'}} (0) = 0

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

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

y^{^{''}} - 4 y^{^{'}} + 4 y = - 3 + 2 x + 2 x^{2}