What is the solution for y^{''}-5y^'+6y=cos(2x) ?

The solution for y^{''}-5y^'+6y=cos(2x) is y=c_{1}e^{3x}+c_{2}e^{2x}-5/52 sin(2x)+1/52 cos(2x)

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

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

y = c_{1} e^{3 x} + c_{2} e^{2 x} - \frac{5}{52} sin (2 x) + \frac{1}{52} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{2 x} - \frac{5}{52} sin (2 x) + \frac{1}{52} cos (2 x)

y = c_{1} e^{3 x} + c_{2} e^{2 x} - \frac{5}{52} sin (2 x) + \frac{1}{52} cos (2 x)

y^{^{''}} - 2 y^{^{'}} + y = e^{t}, y (0) = 0, y (0) = 5

y^{^{''}} + 16 y = 3 sec (4 x)

y^{^{''}} + 4 y = 6 cos (x)

y^{^{''}} + 9 y^{^{'}} = e^{t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 x - sin (2 x)

What is the solution for y^{''}-5y^'+6y=cos(2x) ?
The solution for y^{''}-5y^'+6y=cos(2x) is y=c_{1}e^{3x}+c_{2}e^{2x}-5/52 sin(2x)+1/52 cos(2x)