What is the solution for 2y^{''}+y^'-y=3e^{-x} ?

The solution for 2y^{''}+y^'-y=3e^{-x} is y=c_{1}e^{x/2}+c_{2}e^{-x}-e^{-x}x

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

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

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- x} - e^{- x} x

Solution steps

2 y^{''} (x) + y^{'} (x) - y = 3 e^{- x}

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- x} - e^{- x} x

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- x} - e^{- x} x

y^{^{''}} - y^{^{'}} = 9 t e^{2 t}

q^{^{''}} + 20 q^{^{'}} + 100 q = 24, q (0) = 0, q^{^{'}} (0) = 0

(D^{2} + 1) y = e^{i x}

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

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

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