What is the solution for y^{''}-2y^'-3y=4ex-12 ?

The solution for y^{''}-2y^'-3y=4ex-12 is y=c_{1}e^{3x}+c_{2}e^{-x}-(4ex)/3+(8e+36)/9

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y^{''}-2y^'-3y=4ex-12

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e x - 12

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{4 e x}{3} + \frac{8 e + 36}{9}

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = 4 e x - 12

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{4 e x}{3} + \frac{8 e + 36}{9}

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{4 e x}{3} + \frac{8 e + 36}{9}

4 y^{^{''}} + 4 y^{^{'}} + y = e^{\frac{- x}{2}}

y^{^{''}} + 4 y^{^{'}} + 5 y = - 15 x + e^{- x}

y^{^{''}} + 6 y^{^{'}} + 9 y = 5 e^{- 3 t}

4 y^{^{''}} - 4 y^{^{'}} + y = 24 e^{\frac{t}{2}}

y^{^{''''}} + 36 y^{^{''}} = sin (6 t) + t e^{7 t} + 4

What is the solution for y^{''}-2y^'-3y=4ex-12 ?
The solution for y^{''}-2y^'-3y=4ex-12 is y=c_{1}e^{3x}+c_{2}e^{-x}-(4ex)/3+(8e+36)/9