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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y = \frac{2}{cos ( x )}

y^{^{''}} - 6 y^{^{'}} + 5 y = - 9 e^{2 x}

y^{^{''}} + 2 y^{^{'}} + y = e^{- t} t^{- 1}

y^{^{''}} - 4 y^{^{'}} + 13 y = - 40 cos (x) + 13 x

y^{^{''}} + 25 y = cos (5 x) csc^{2} (5 x)

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