What is the solution for y^{''}-2y^'-3y=4e^x-12 ?

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

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y^{''}-2y^'-3y=4e^x-12

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e^{x} - 12

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

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = 4 e^{x} - 12

Solve linear ODE:

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

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

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

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

y^{^{''}} - 6 y^{^{'}} + 5 y = - 9 e^{2 x}

y^{^{''}} + 2 y^{^{'}} + y = e^{- t} t^{- 1}

y^{^{''}} - 4 y^{^{'}} + 13 y = - 40 cos (x) + 13 x

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