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

The solution for y^{''}-y^'-2y=cos(2t) is y=c_{1}e^{2t}+c_{2}e^{-t}-1/20 sin(2t)-3/20 cos(2t)

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

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

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{1}{20} sin (2 t) - \frac{3}{20} cos (2 t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{1}{20} sin (2 t) - \frac{3}{20} cos (2 t)

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{1}{20} sin (2 t) - \frac{3}{20} cos (2 t)

y^{^{''}} + 4 y = cos (9 x)

y^{^{''}} + 4 y = csc^{2} (2 x)

y^{^{''}} - 2 y^{^{'}} + y = 4 cos (t)

y^{^{''}} + y^{^{'}} - 6 y = - 5 e^{2 x}

y^{^{''}} + 4 y^{^{'}} + 3 y = 4 e^{- 3 x} + 18 x + 15

What is the solution for y^{''}-y^'-2y=cos(2t) ?
The solution for y^{''}-y^'-2y=cos(2t) is y=c_{1}e^{2t}+c_{2}e^{-t}-1/20 sin(2t)-3/20 cos(2t)