y^{''}-8y^'+12y=2x(e^{2x}-e^{-2x})

Lösung

y^{^{''}} - 8 y^{^{'}} + 12 y = 2 x (e^{2 x} - e^{- 2 x})

y = c_{1} e^{6 x} + c_{2} e^{2 x} - \frac{e ^{2 x} x ^{2}}{4} - \frac{e ^{2 x} x}{8} - \frac{e ^{- 2 x} x}{16} - \frac{3}{128} e^{- 2 x}

Schritte zur Lösung

y^{''} (x) - 8 y^{'} (x) + 12 y = 2 x (e^{2 x} - e^{- 2 x})

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} e^{2 x} - \frac{e ^{2 x} x ^{2}}{4} - \frac{e ^{2 x} x}{8} - \frac{e ^{- 2 x} x}{16} - \frac{3}{128} e^{- 2 x}

y = c_{1} e^{6 x} + c_{2} e^{2 x} - \frac{e ^{2 x} x ^{2}}{4} - \frac{e ^{2 x} x}{8} - \frac{e ^{- 2 x} x}{16} - \frac{3}{128} e^{- 2 x}

4 q^{^{''}} + 8 q^{^{'}} + 80 q = 88 t

3 y^{^{''}} - 8 y^{^{'}} - 3 y = 5

y^{^{''}} + 4 y^{^{'}} - 5 = 2 t^{2} - 3 t + 6

y^{^{''}} + 25 y = 6 x, y (0) = 1, y^{^{'}} (0) = 5

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