What is the solution for y^{''}-y^'-2y=8sin(t)-2cos(t) ?

The solution for y^{''}-y^'-2y=8sin(t)-2cos(t) is y=c_{1}e^{2t}+c_{2}e^{-t}-11/5 sin(t)+7/5 cos(t)

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y^{''}-y^'-2y=8sin(t)-2cos(t)

y^{^{''}} - y^{^{'}} - 2 y = 8 sin (t) - 2 cos (t)

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{11}{5} sin (t) + \frac{7}{5} cos (t)

Solution steps

y^{''} (t) - y^{'} (t) - 2 y = 8 sin (t) - 2 cos (t)

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{11}{5} sin (t) + \frac{7}{5} cos (t)

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{11}{5} sin (t) + \frac{7}{5} cos (t)

y^{^{''}} + 6 y^{^{'}} + 13 y = 13 x^{2}

2 y^{^{''}} - y^{^{'}} - y = t + 1, y (0) = 1, y^{^{'}} (0) = 0

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

x^{^{''}} - 6 x^{^{'}} + 9 x = 2 t e^{3 t}

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

What is the solution for y^{''}-y^'-2y=8sin(t)-2cos(t) ?
The solution for y^{''}-y^'-2y=8sin(t)-2cos(t) is y=c_{1}e^{2t}+c_{2}e^{-t}-11/5 sin(t)+7/5 cos(t)