What is the solution for y^{''}-2y^'-3y=cos(2t) ?

The solution for y^{''}-2y^'-3y=cos(2t) is y=c_{1}e^{3t}+c_{2}e^{-t}-4/65 sin(2t)-7/65 cos(2t)

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y^{''}-2y^'-3y=cos(2t)

y^{^{''}} - 2 y^{^{'}} - 3 y = cos (2 t)

y = c_{1} e^{3 t} + c_{2} e^{- t} - \frac{4}{65} sin (2 t) - \frac{7}{65} cos (2 t)

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = cos (2 t)

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{- t} - \frac{4}{65} sin (2 t) - \frac{7}{65} cos (2 t)

y = c_{1} e^{3 t} + c_{2} e^{- t} - \frac{4}{65} sin (2 t) - \frac{7}{65} cos (2 t)

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

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 e^{2 t}

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 0 + 6 sin (x)

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

y^{^{''}} + y = sin (x), y^{^{'}} (\frac{π}{2}) = \frac{3}{2}, y (0) = 0

What is the solution for y^{''}-2y^'-3y=cos(2t) ?
The solution for y^{''}-2y^'-3y=cos(2t) is y=c_{1}e^{3t}+c_{2}e^{-t}-4/65 sin(2t)-7/65 cos(2t)