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

The solution for y^{''}+y^'-2y=2+5cos(2x) is y=c_{1}e^x+c_{2}e^{-2x}+(sin(2x)-3cos(2x))/4-1

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

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

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{sin ( 2 x ) - 3 cos ( 2 x )}{4} - 1

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{sin ( 2 x ) - 3 cos ( 2 x )}{4} - 1

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{sin ( 2 x ) - 3 cos ( 2 x )}{4} - 1

y^{^{'''}} + 4 y^{^{'}} = tan (x)

y^{^{''}} - 10 y^{^{'}} + 25 y = 25 x + 10

y^{^{''}} - 7 y^{^{'}} + 12 = 2 e^{4 t}, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 14 y^{^{'}} + 49 y = 42 x + 5

d (x) = 5 (x - 9)^{2} - 32

What is the solution for y^{''}+y^'-2y=2+5cos(2x) ?
The solution for y^{''}+y^'-2y=2+5cos(2x) is y=c_{1}e^x+c_{2}e^{-2x}+(sin(2x)-3cos(2x))/4-1