What is the solution for y^{''}+2y^'=cos(t),y(0)=0 ?

The solution for y^{''}+2y^'=cos(t),y(0)=0 is y=-c_{2}+1/5+c_{2}e^{-2t}+2/5 sin(t)-1/5 cos(t)

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y^{''}+2y^'=cos(t),y(0)=0

y^{^{''}} + 2 y^{^{'}} = cos (t), y (0) = 0

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

Solution steps

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

Solve linear ODE:

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

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

What is the solution for y^{''}+2y^'=cos(t),y(0)=0 ?
The solution for y^{''}+2y^'=cos(t),y(0)=0 is y=-c_{2}+1/5+c_{2}e^{-2t}+2/5 sin(t)-1/5 cos(t)