What is the solution for y^{''}+2y^'+y=cos(5t) ?

The solution for y^{''}+2y^'+y=cos(5t) is y=c_{1}e^{-t}+c_{2}te^{-t}+5/338 sin(5t)-6/169 cos(5t)

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y^{''}+2y^'+y=cos(5t)

y^{^{''}} + 2 y^{^{'}} + y = cos (5 t)

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{5}{338} sin (5 t) - \frac{6}{169} cos (5 t)

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = cos (5 t)

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{5}{338} sin (5 t) - \frac{6}{169} cos (5 t)

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{5}{338} sin (5 t) - \frac{6}{169} cos (5 t)

What is the solution for y^{''}+2y^'+y=cos(5t) ?
The solution for y^{''}+2y^'+y=cos(5t) is y=c_{1}e^{-t}+c_{2}te^{-t}+5/338 sin(5t)-6/169 cos(5t)