11y^{''}+12y^'+96y=5cos(4t)

解答

11 y^{^{''}} + 12 y^{^{'}} + 96 y = 5 cos (4 t)

y = e^{- \frac{6 t}{11}} (c_{1} cos (\frac{2 255 t}{11}) + c_{2} sin (\frac{2 255 t}{11})) + \frac{15}{544} sin (4 t) - \frac{25}{544} cos (4 t)

求解步骤

11 y^{''} (t) + 12 y^{'} (t) + 96 y = 5 cos (4 t)

求解线性方程:

y = e^{- \frac{6 t}{11}} (c_{1} cos (\frac{2 255 t}{11}) + c_{2} sin (\frac{2 255 t}{11})) + \frac{15}{544} sin (4 t) - \frac{25}{544} cos (4 t)

y = e^{- \frac{6 t}{11}} (c_{1} cos (\frac{2 255 t}{11}) + c_{2} sin (\frac{2 255 t}{11})) + \frac{15}{544} sin (4 t) - \frac{25}{544} cos (4 t)

\frac{\partial}{\partial y} (x^{2} sin (4 y))

\frac{\partial}{\partial z} (ln (x^{2} + y^{2}) - z)

n = 0 \sum \infty 2 n + 1

\int 39 sec^{4} (x) d x

y^{^{'}} = (t + y)^{2} - 1, y (2) = 5