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

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

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y^{''}-2y^'-3y=12ex-12

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

y = c_{1} e^{3 x} + c_{2} e^{- x} - 4 e x + \frac{8 e + 12}{3}

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} - 4 e x + \frac{8 e + 12}{3}

y = c_{1} e^{3 x} + c_{2} e^{- x} - 4 e x + \frac{8 e + 12}{3}

y^{^{''}} - y^{^{'}} - 2 y = 26 sin (x) + 40 x cos (x)

y^{^{''}} + 4 y^{^{'}} = x

y^{^{''}} + 2 y^{^{'}} - 3 y = (x + 1) e^{x}

\frac{d ^{2} y}{d x ^{2}} - 14 \frac{d y}{d x} + 49 y = (x + 1) e^{x}

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

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