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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + 4 y = 5 sin (3 t)

y^{^{''}} + 3 y^{^{'}} + 6 y = x e^{2 x}

y^{^{''}} + 2 y^{^{'}} + 5 y = 1

y^{^{''}} - 4 y^{^{'}} + 4 y = x^{2} e^{2 x}

x^{^{''}} + 2 x^{^{'}} + x = e^{(- t)}

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