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

The solution for y^{''}-y^'-6y=cos(2t) is y=c_{1}e^{3t}+c_{2}e^{-2t}-1/52 sin(2t)-5/52 cos(2t)

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

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

y = c_{1} e^{3 t} + c_{2} e^{- 2 t} - \frac{1}{52} sin (2 t) - \frac{5}{52} cos (2 t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{- 2 t} - \frac{1}{52} sin (2 t) - \frac{5}{52} cos (2 t)

y = c_{1} e^{3 t} + c_{2} e^{- 2 t} - \frac{1}{52} sin (2 t) - \frac{5}{52} cos (2 t)

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 5 e^{4 x}

y^{^{''}} + 4 y = x^{2} + 1

\frac{d ^{2} y}{d x ^{2}} - 7 \frac{d y}{d x} + 6 y = e^{x}

y^{^{''}} + y = 12 cos^{2} (x)

What is the solution for y^{''}-y^'-6y=cos(2t) ?
The solution for y^{''}-y^'-6y=cos(2t) is y=c_{1}e^{3t}+c_{2}e^{-2t}-1/52 sin(2t)-5/52 cos(2t)