y^{''}+4y^'+12y=4x^2

Lösung

y^{^{''}} + 4 y^{^{'}} + 12 y = 4 x^{2}

y = e^{- 2 x} (c_{1} cos (2^{1 + \frac{1}{2}} x) + c_{2} sin (2^{1 + \frac{1}{2}} x)) + \frac{x ^{2}}{3} - \frac{2 x}{9} + \frac{1}{54}

Schritte zur Lösung

y^{''} (x) + 4 y^{'} (x) + 12 y = 4 x^{2}

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (2^{1 + \frac{1}{2}} x) + c_{2} sin (2^{1 + \frac{1}{2}} x)) + \frac{x ^{2}}{3} - \frac{2 x}{9} + \frac{1}{54}

y = e^{- 2 x} (c_{1} cos (2^{1 + \frac{1}{2}} x) + c_{2} sin (2^{1 + \frac{1}{2}} x)) + \frac{x ^{2}}{3} - \frac{2 x}{9} + \frac{1}{54}

y^{^{'''}} - 9 y^{^{''}} + 27 y^{^{'}} - 27 y = 9 e^{3 x}

y^{^{''}} + 6 y^{^{'}} + 5 y = t \cdot e^{- t}

y^{^{''}} - 6 y^{^{'}} - 16 y = - 24 t + 32 t^{2}

16 y^{^{''}} + y = 4 sec (\frac{π}{4})

y^{^{''}} + y^{^{'}} - 8 y = 48 t