What is the solution for y^{''}+2y^'-3y=12e^x-6 ?

The solution for y^{''}+2y^'-3y=12e^x-6 is y=c_{1}e^x+c_{2}e^{-3x}+3xe^x+2

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y^{''}+2y^'-3y=12e^x-6

y^{^{''}} + 2 y^{^{'}} - 3 y = 12 e^{x} - 6

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

Solution steps

y^{''} (x) + 2 y^{'} (x) - 3 y = 12 e^{x} - 6

Solve linear ODE:

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

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

1 q^{^{''}} + 8 q^{^{'}} + \frac{1}{0.05} q = 80 t

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

y^{^{''}} + 4 y^{^{'}} + 4 y = (x + 5) e^{- 2 x}

y^{^{''}} + y^{^{'}} - 20 y = 5 + 2 \cdot x

y^{^{''}} - 5 y^{^{'}} + 6 y = e^{t}, y (0) = 1, y^{^{'}} (0) = 1

What is the solution for y^{''}+2y^'-3y=12e^x-6 ?
The solution for y^{''}+2y^'-3y=12e^x-6 is y=c_{1}e^x+c_{2}e^{-3x}+3xe^x+2