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

The solution for x^{''}+3x^'+2x=cos(t) is x=c_{1}e^{-t}+c_{2}e^{-2t}+3/10 sin(t)+1/10 cos(t)

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

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

x = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{3}{10} sin (t) + \frac{1}{10} cos (t)

Solution steps

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

Solve linear ODE:

x = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{3}{10} sin (t) + \frac{1}{10} cos (t)

x = c_{1} e^{- t} + c_{2} e^{- 2 t} + \frac{3}{10} sin (t) + \frac{1}{10} cos (t)

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

y^{^{''}} - 8 y^{^{'}} + 12 y = 2 x e^{2 x} - 2 x e^{- 2 x}

x^{^{''}} + x = t + 1

y^{^{''}} + 6 y^{^{'}} + 9 y = e^{3 x} ln (1 + x^{2})

y^{^{''}} + 36 y = 36 sec (6 t)

What is the solution for x^{''}+3x^'+2x=cos(t) ?
The solution for x^{''}+3x^'+2x=cos(t) is x=c_{1}e^{-t}+c_{2}e^{-2t}+3/10 sin(t)+1/10 cos(t)