31y^{''}+800y=100cos(4t),y(0)=0,y^'(0)=0

解答

31 y^{^{''}} + 800 y = 100 cos (4 t), y (0) = 0, y^{^{'}} (0) = 0

y = \frac{25}{76} cos (4 t) - \frac{25}{76} cos (\frac{20 2 t}{31})

求解步骤

31 y^{''} (t) + 800 y = 100 cos (4 t)

求解线性方程:

y = \frac{25}{76} cos (4 t) - \frac{25}{76} cos (\frac{20 2 t}{31})

y = \frac{25}{76} cos (4 t) - \frac{25}{76} cos (\frac{20 2 t}{31})

x^{^{''''}} + 2 x^{^{''}} + x = 4 t e^{t}

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

2 y^{^{''}} - 7 y^{^{'}} + 3 y = 4 x^{2}

y^{^{''}} + 3 y^{^{'}} + 2 y = (1 + e^{2 x})^{- 1}

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