y^{''}+2y^'+2y=3t^2

Lösung

y^{^{''}} + 2 y^{^{'}} + 2 y = 3 t^{2}

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{3 t ^{2}}{2} - 3 t + \frac{3}{2}

Schritte zur Lösung

y^{''} (t) + 2 y^{'} (t) + 2 y = 3 t^{2}

Solve linear ODE:

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{3 t ^{2}}{2} - 3 t + \frac{3}{2}

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{3 t ^{2}}{2} - 3 t + \frac{3}{2}

y^{^{''}} - y = a x e^{- x}, y (0) = \frac{1}{4}, y^{^{'}} (0) = 0

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

y^{^{''}} - 14 y^{^{'}} + 65 y = e^{7 x} sin (b x)

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

x^{^{''}} - 4 x^{^{'}} + 4 x = 2, x^{^{'}} (0) = 1