What is the solution for y^{''}-2y^'-3y=8e^x-9 ?

The solution for y^{''}-2y^'-3y=8e^x-9 is y=c_{1}e^{3x}+c_{2}e^{-x}-2e^x+3

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y^{''}-2y^'-3y=8e^x-9

y^{^{''}} - 2 y^{^{'}} - 3 y = 8 e^{x} - 9

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

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = 8 e^{x} - 9

Solve linear ODE:

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

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

5 y^{^{''}} - 10 y^{^{'}} + 10 y = e^{x} sec (x)

y^{^{''}} - 16 y = 32 + 45 e^{x}

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

y^{^{''}} + y = x sin (x)

y^{^{'''}} + y^{^{''}} + 4 y^{^{'}} + 4 y = 100 x sin (3 x)

What is the solution for y^{''}-2y^'-3y=8e^x-9 ?
The solution for y^{''}-2y^'-3y=8e^x-9 is y=c_{1}e^{3x}+c_{2}e^{-x}-2e^x+3