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

Lösung

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

y = (\frac{5 - 3 3}{4 3} + \frac{3}{2}) e^{(- 1 + 3) t} - \frac{5 - 3 3}{4 3} e^{(- 1 - 3) t} - \frac{t ^{2}}{2} - t - \frac{3}{2}

Schritte zur Lösung

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

Solve linear ODE:

y = (\frac{5 - 3 3}{4 3} + \frac{3}{2}) e^{(- 1 + 3) t} - \frac{5 - 3 3}{4 3} e^{(- 1 - 3) t} - \frac{t ^{2}}{2} - t - \frac{3}{2}

y = (\frac{5 - 3 3}{4 3} + \frac{3}{2}) e^{(- 1 + 3) t} - \frac{5 - 3 3}{4 3} e^{(- 1 - 3) t} - \frac{t ^{2}}{2} - t - \frac{3}{2}

y^{^{''}} + 3 \cdot y^{^{'}} + 2 y = sin (e^{x})

y^{^{''}} + 4 y^{^{'}} + 5 y = 5 x + e^{- x}

y^{^{'''}} + 12 y^{^{''}} - 256 y = e^{- 8 x}

8 y^{^{''}} + 3 y^{^{'}} = 12 x

y^{^{''}} + y^{^{'}} - 6 y = e^{2 t}