What is the solution for y^{''}-4y^'-12y=2sin(t) ?

The solution for y^{''}-4y^'-12y=2sin(t) is y=c_{1}e^{6t}+c_{2}e^{-2t}-26/185 sin(t)+8/185 cos(t)

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y^{''}-4y^'-12y=2sin(t)

y^{^{''}} - 4 y^{^{'}} - 12 y = 2 sin (t)

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{26}{185} sin (t) + \frac{8}{185} cos (t)

Solution steps

y^{''} (t) - 4 y^{'} (t) - 12 y = 2 sin (t)

Solve linear ODE:

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{26}{185} sin (t) + \frac{8}{185} cos (t)

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{26}{185} sin (t) + \frac{8}{185} cos (t)

12 y^{^{''}} + 29 y^{^{'}} - 11 y = - 9

y^{^{''}} - 9 y = 2

y^{^{''}} - y = - 10 t + 4

y^{^{''}} - 10 y^{^{'}} + 25 y = e^{5} x ln (4 + x^{2})

\frac{d ^{2} y}{d x ^{2}} + y = cot (x)

What is the solution for y^{''}-4y^'-12y=2sin(t) ?
The solution for y^{''}-4y^'-12y=2sin(t) is y=c_{1}e^{6t}+c_{2}e^{-2t}-26/185 sin(t)+8/185 cos(t)