What is the solution for y^{''}+8y^'+16y=6sin(4t) ?

The solution for y^{''}+8y^'+16y=6sin(4t) is y=c_{1}e^{-4t}+c_{2}te^{-4t}-3/16 cos(4t)

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y^{''}+8y^'+16y=6sin(4t)

y^{^{''}} + 8 y^{^{'}} + 16 y = 6 sin (4 t)

y = c_{1} e^{- 4 t} + c_{2} t e^{- 4 t} - \frac{3}{16} cos (4 t)

Solution steps

y^{''} (t) + 8 y^{'} (t) + 16 y = 6 sin (4 t)

Solve linear ODE:

y = c_{1} e^{- 4 t} + c_{2} t e^{- 4 t} - \frac{3}{16} cos (4 t)

y = c_{1} e^{- 4 t} + c_{2} t e^{- 4 t} - \frac{3}{16} cos (4 t)

y^{^{''}} - 2 y - 3 y^{^{'}} = 3 t^{2} - 5

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

y^{^{''}} + 2 y^{^{'}} - 35 y = x^{2}

y^{^{''}} + 16 y = cos (3 t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - y^{^{'}} - 6 y = - 14 t + 12 t^{2}

What is the solution for y^{''}+8y^'+16y=6sin(4t) ?
The solution for y^{''}+8y^'+16y=6sin(4t) is y=c_{1}e^{-4t}+c_{2}te^{-4t}-3/16 cos(4t)