y^{''}+y^'+4y=cos(3t)

解答

y^{^{''}} + y^{^{'}} + 4 y = cos (3 t)

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2})) + \frac{3}{34} sin (3 t) - \frac{5}{34} cos (3 t)

求解步骤

y^{''} (t) + y^{'} (t) + 4 y = cos (3 t)

求解线性方程:

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2})) + \frac{3}{34} sin (3 t) - \frac{5}{34} cos (3 t)

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2})) + \frac{3}{34} sin (3 t) - \frac{5}{34} cos (3 t)

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

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

y^{^{''}} + y - 2 y = 2 t - 40 cos (2 t)

y^{^{''}} - y = 2 x \cdot sin (x)

y^{^{''}} - 2 y^{^{'}} = cos (t), y (0) = 0, y^{^{'}} (0) = 1