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

The solution for y^{''}-2y^'+2y=cos(2x) is y=e^x(c_{1}cos(x)+c_{2}sin(x))-1/5 sin(2x)-1/10 cos(2x)

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} + 3 y^{^{'}} + 2 y = 26 x + 47

y^{^{''}} + y^{^{'}} = x, y (0) = 8, y^{^{'}} (0) = 0

y^{^{''}} - y^{^{'}} - 6 y = 23 sin (3 x)

y^{^{''}} + 4 y = 3 cos (2 t), y (0) = 1, y^{^{'}} (0) = 0

What is the solution for y^{''}-2y^'+2y=cos(2x) ?
The solution for y^{''}-2y^'+2y=cos(2x) is y=e^x(c_{1}cos(x)+c_{2}sin(x))-1/5 sin(2x)-1/10 cos(2x)