y^{''}-4y^'=te^{-2t},y(0)=0

Lösung

y^{^{''}} - 4 y^{^{'}} = t e^{- 2 t}, y (0) = 0

y = (- c_{2} - \frac{1}{18}) e^{4 t} + c_{2} + \frac{e ^{- 2 t} t}{12} + \frac{1}{18} e^{- 2 t}

Schritte zur Lösung

y^{''} (t) - 4 y^{'} (t) = t e^{- 2 t}

Solve linear ODE:

y = (- c_{2} - \frac{1}{18}) e^{4 t} + c_{2} + \frac{e ^{- 2 t} t}{12} + \frac{1}{18} e^{- 2 t}

y = (- c_{2} - \frac{1}{18}) e^{4 t} + c_{2} + \frac{e ^{- 2 t} t}{12} + \frac{1}{18} e^{- 2 t}

\frac{d ^{2} y}{d x ^{2}} + 8 \frac{d y}{d x} + 16 y = e^{4 x}

y^{^{''}} + 16 y^{^{'}} + 10000 y = 400 sin (200 x)

y^{^{''}} + 4 y^{^{'}} = e^{3 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = \frac{- 2 e ^{2 t}}{t ^{2} + 1}

y^{^{''}} + 7 y^{^{'}} + 10 y = 100 sin (10 t)