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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} - 6 y = 2 t^{2} + 2 t - 5 e^{3 t}

d (x) = 2 x - 4

\frac{d ^{2} y}{d x ^{2}} - 4 y = 4 sin (2 x) - 3 e^{2} x

12 y^{^{''}} - 5 y^{^{'}} - 3 y = e^{x}

y^{^{''}} + 4 y = sin (6 x)

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