What is the solution for y^{''}-2y^'-3y=4ex-9 ?

The solution for y^{''}-2y^'-3y=4ex-9 is y=c_{1}e^{3x}+c_{2}e^{-x}-(4ex)/3+(8e+27)/9

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y^{''}-2y^'-3y=4ex-9

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e x - 9

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{4 e x}{3} + \frac{8 e + 27}{9}

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = 4 e x - 9

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{4 e x}{3} + \frac{8 e + 27}{9}

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{4 e x}{3} + \frac{8 e + 27}{9}

y^{^{''}} - 8 y^{^{'}} + 16 y = 24 x + 2

\frac{d ^{2} y}{d x ^{2}} + y = - 16 cos (3 x)

y^{^{''}} - 4 y^{^{'}} + 3 y = \frac{1}{1 + e ^{- x}}

x^{^{''}} + 8 x^{^{'}} + 16 x = 2 sin (4 t)

y^{^{''}} - 3 y^{^{'}} + 2 y = 7 e^{3 x}

What is the solution for y^{''}-2y^'-3y=4ex-9 ?
The solution for y^{''}-2y^'-3y=4ex-9 is y=c_{1}e^{3x}+c_{2}e^{-x}-(4ex)/3+(8e+27)/9