What is the solution for y^{''}+4y^'+3y=cos(t) ?

The solution for y^{''}+4y^'+3y=cos(t) is y=c_{1}e^{-t}+c_{2}e^{-3t}+1/5 sin(t)+1/10 cos(t)

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y^{''}+4y^'+3y=cos(t)

y^{^{''}} + 4 y^{^{'}} + 3 y = cos (t)

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + \frac{1}{5} sin (t) + \frac{1}{10} cos (t)

Solution steps

y^{''} (t) + 4 y^{'} (t) + 3 y = cos (t)

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + \frac{1}{5} sin (t) + \frac{1}{10} cos (t)

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + \frac{1}{5} sin (t) + \frac{1}{10} cos (t)

What is the solution for y^{''}+4y^'+3y=cos(t) ?
The solution for y^{''}+4y^'+3y=cos(t) is y=c_{1}e^{-t}+c_{2}e^{-3t}+1/5 sin(t)+1/10 cos(t)