What is the solution for y^{''}-2y^'-3y=8e^x-12 ?

The solution for y^{''}-2y^'-3y=8e^x-12 is y=c_{1}e^{3x}+c_{2}e^{-x}-2e^x+4

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y^{''}-2y^'-3y=8e^x-12

y^{^{''}} - 2 y^{^{'}} - 3 y = 8 e^{x} - 12

y = c_{1} e^{3 x} + c_{2} e^{- x} - 2 e^{x} + 4

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = 8 e^{x} - 12

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- x} - 2 e^{x} + 4

y = c_{1} e^{3 x} + c_{2} e^{- x} - 2 e^{x} + 4

y^{^{''}} + y = e^{x} + x

y^{^{''}} - y = x e^{x}

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 2 e^{- t} + 3

y^{^{''}} + 2 y^{^{'}} + 10 y = x^{2} e^{- x} cos (3 x)

y^{^{''}} + 4 y = 9 t, y (0) = 0, y^{^{'}} (0) = 7

What is the solution for y^{''}-2y^'-3y=8e^x-12 ?
The solution for y^{''}-2y^'-3y=8e^x-12 is y=c_{1}e^{3x}+c_{2}e^{-x}-2e^x+4