y^{''}+4y=t^2+6e^t,y(0)=0,y^'(0)=5

Lösung

y^{^{''}} + 4 y = t^{2} + 6 e^{t}, y (0) = 0, y^{^{'}} (0) = 5

y = - \frac{43}{40} cos (2 t) + \frac{19}{10} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{6}{5} e^{t}

Schritte zur Lösung

y^{''} (t) + 4 y = t^{2} + 6 e^{t}

Solve linear ODE:

y = - \frac{43}{40} cos (2 t) + \frac{19}{10} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{6}{5} e^{t}

y = - \frac{43}{40} cos (2 t) + \frac{19}{10} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + \frac{6}{5} e^{t}

y^{^{''}} - 8 y^{^{'}} + 20 y = 100 x^{2} - 39 x e^{x}

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

y^{^{''}} + y = sin (3 x)

y^{^{'''}} - 6 y^{^{''}} + 12 y^{^{'}} - 8 y = 16 e^{- 2 t}

y^{^{''}} - 2 y^{^{'}} = cos (4 x)