What is the solution for y^{''}-2y^'+2y=cos(x) ?

The solution for y^{''}-2y^'+2y=cos(x) is y=e^x(c_{1}cos(x)+c_{2}sin(x))-2/5 sin(x)+1/5 cos(x)

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 5 y^{^{'}} + 4 y = t e^{- t} H (t)

3 y^{^{''}} - 2 y^{^{'}} = t

(D^{4} - 2 D^{2} + 1) Y = cos (x)

D (D^{2} + 1) y = sin (x)

y^{^{'''}} - 2 y^{^{''}} - 5 y^{^{'}} + 6 y = 2 e^{x} (1 - 6 x)

What is the solution for y^{''}-2y^'+2y=cos(x) ?
The solution for y^{''}-2y^'+2y=cos(x) is y=e^x(c_{1}cos(x)+c_{2}sin(x))-2/5 sin(x)+1/5 cos(x)