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

The solution for y^{''}-2y^'-3y=8ex-9 is y=c_{1}e^{3x}+c_{2}e^{-x}-(8ex)/3+(16e+27)/9

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y^{''}-2y^'-3y=8ex-9

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

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{8 e x}{3} + \frac{16 e + 27}{9}

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} - \frac{8 e x}{3} + \frac{16 e + 27}{9}

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{8 e x}{3} + \frac{16 e + 27}{9}

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

y^{^{''}} + 5 y^{^{'}} + 4 y = 24

d^{2} (d^{2} + 4) y = 96 x^{2}

\frac{d ^{2} y}{d t ^{2}} + 12 \frac{d y}{d t} + 32 = 32 H (t)

y^{^{''''}} + 2 y^{^{''}} + y = 9 cos (2 x)

What is the solution for y^{''}-2y^'-3y=8ex-9 ?
The solution for y^{''}-2y^'-3y=8ex-9 is y=c_{1}e^{3x}+c_{2}e^{-x}-(8ex)/3+(16e+27)/9