y^{''}+4y=t^2+4e^t,y(0)=0,y^'(0)=9

Lösung

y^{^{''}} + 4 y = t^{2} + 4 e^{t}, y (0) = 0, y^{^{'}} (0) = 9

y = - \frac{27}{40} cos (2 t) + \frac{41}{10} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{4}{5} e^{t}

Schritte zur Lösung

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

Solve linear ODE:

y = - \frac{27}{40} cos (2 t) + \frac{41}{10} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{4}{5} e^{t}

y = - \frac{27}{40} cos (2 t) + \frac{41}{10} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{4}{5} e^{t}

y^{^{''}} + 4 y = t^{2} + 4 e^{t}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 10 y^{^{'}} + 25 y = t^{- 3} e^{5 t}

y^{^{''}} + k^{2} y = k_{2} \cdot sin (k x)

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

y^{^{''}} - 8 y^{^{'}} + 16 y = 16 t^{6} e^{4 t}