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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 x

y^{^{''}} - 3 y^{^{'}} + 2 y = 2 x

x^{^{''}} - 4 x^{^{'}} + 5 x = cos (t)

y^{^{''}} + 3 y^{^{'}} + 2 y = 4 e^{2 x}, y (0) = - 3, y^{^{'}} (0) = 5

y^{^{'''}} + 4 y^{^{'}} = t e^{- 2 t} + 3 cos (2 t)

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