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

The solution for y^{''}+y^'-2y=cos(x) is y=c_{1}e^x+c_{2}e^{-2x}+1/10 sin(x)-3/10 cos(x)

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

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

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{1}{10} sin (x) - \frac{3}{10} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{1}{10} sin (x) - \frac{3}{10} cos (x)

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{1}{10} sin (x) - \frac{3}{10} cos (x)

y^{^{''}} + 5 y^{^{'}} + 6 y = x^{2}

y^{^{''}} + 9 y = e^{t}

y^{^{''}} + y^{^{'}} - 6 y = x

y^{^{''}} - 7 y^{^{'}} + a = 0, y (0) = 1, y^{^{'}} (0) = 10

y^{^{''}} + 4 y = - 6, y (\frac{π}{8}) = - \frac{1}{2}, y^{^{'}} (\frac{π}{8}) = 2

What is the solution for y^{''}+y^'-2y=cos(x) ?
The solution for y^{''}+y^'-2y=cos(x) is y=c_{1}e^x+c_{2}e^{-2x}+1/10 sin(x)-3/10 cos(x)