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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

x^{^{''}} + 4 x^{^{'}} + 5 x = e^{- 2} t sin (t)

d, d (t) = - 16 t^{2} + 32 t + 40

x^{^{''}} + 81 x = sin (7 t)

y^{^{''}} + 2 y^{^{'}} = t e^{- t}

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