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

The solution for x^{''}+3x^'+2x=3cos(2t) is x=c_{1}e^{-t}+c_{2}e^{-2t}+9/20 sin(2t)-3/20 cos(2t)

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

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

x = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{9}{20} sin (2 t) - \frac{3}{20} cos (2 t)

Solution steps

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

Solve linear ODE:

x = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{9}{20} sin (2 t) - \frac{3}{20} cos (2 t)

x = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{9}{20} sin (2 t) - \frac{3}{20} cos (2 t)

2 x^{^{''}} + 5 x^{^{'}} + 4 x = 10 r (t)

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = e^{x} - x^{3} + 16

x^{^{''}} + 6 x^{^{'}} - 4 x = 8 e^{t}

x^{^{''}} + 4 x^{^{'}} + 43 x = 9 cos (6 t)

y^{^{''}} + 9 y = 2 cos (3 x) - 3 sin (3 x)

What is the solution for x^{''}+3x^'+2x=3cos(2t) ?
The solution for x^{''}+3x^'+2x=3cos(2t) is x=c_{1}e^{-t}+c_{2}e^{-2t}+9/20 sin(2t)-3/20 cos(2t)