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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} + 2 y = \frac{e ^{3 x}}{1 + e ^{x}}

\frac{d ^{2} y}{d t ^{2}} - 4 y = t cos (2 t)

y^{^{''}} + 25 y = 6 sin (x)

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2 x} (cos (x) - 8 sin (x))

4 y^{^{''}} - 8 y^{^{'}} + 8 y = e^{x} sec (x)

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