1/2 y^{''}+1/2 y^'+6y=10cos(3t)

解答

\frac{1}{2} y^{^{''}} + \frac{1}{2} y^{^{'}} + 6 y = 10 cos (3 t)

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{47 t}{2}) + c_{2} sin (\frac{47 t}{2})) + \frac{10}{3} sin (3 t) + \frac{10}{3} cos (3 t)

求解步骤

\frac{1}{2} y^{''} (t) + \frac{1}{2} y^{'} (t) + 6 y = 10 cos (3 t)

求解线性方程:

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{47 t}{2}) + c_{2} sin (\frac{47 t}{2})) + \frac{10}{3} sin (3 t) + \frac{10}{3} cos (3 t)

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{47 t}{2}) + c_{2} sin (\frac{47 t}{2})) + \frac{10}{3} sin (3 t) + \frac{10}{3} cos (3 t)

y^{^{'''}} - 6 y^{^{''}} - y^{^{'}} + 6 y = e^{5 x}

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

y^{^{''}} - 4 \cdot y = e^{- α \cdot x}

y^{^{'''}} + 3 y^{^{''}} + 2 y^{^{'}} = (x^{2} + 4 x + 8) e^{2} x

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x - 5 e x