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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y = 10 e^{5 x}

y^{^{''}} - 2 y^{^{'}} + y = sin (x) + e^{x}

y^{^{'''}} - y^{^{''}} + y^{^{'}} - y = sec (t)

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

y^{^{''}} + 2 y^{^{'}} + 5 y = 1, y^{^{'}} (0) = 0, y (0) = 0

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