What is the solution for y^{''}-y^'-6y=2sin(t) ?

The solution for y^{''}-y^'-6y=2sin(t) is y=c_{1}e^{3t}+c_{2}e^{-2t}-7/25 sin(t)+1/25 cos(t)

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y^{''}-y^'-6y=2sin(t)

y^{^{''}} - y^{^{'}} - 6 y = 2 sin (t)

y = c_{1} e^{3 t} + c_{2} e^{- 2 t} - \frac{7}{25} sin (t) + \frac{1}{25} cos (t)

Solution steps

y^{''} (t) - y^{'} (t) - 6 y = 2 sin (t)

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{- 2 t} - \frac{7}{25} sin (t) + \frac{1}{25} cos (t)

y = c_{1} e^{3 t} + c_{2} e^{- 2 t} - \frac{7}{25} sin (t) + \frac{1}{25} cos (t)

D (x) = 105.6 - 2.4 x

x^{^{''}} + 4 x^{^{'}} + 43 x = 10 cos (6 t)

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

y^{^{''}} + 4 y^{^{'}} + 5 y = 1 - 5 x + 8 cos (x) - 8 sin (x)

y^{^{''}} + 16 y = 4 e^{2 t} + 4 sin (2 t)

What is the solution for y^{''}-y^'-6y=2sin(t) ?
The solution for y^{''}-y^'-6y=2sin(t) is y=c_{1}e^{3t}+c_{2}e^{-2t}-7/25 sin(t)+1/25 cos(t)