What is the solution for y^{''}-2y^'-3y=12ex-9 ?

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

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y^{''}-2y^'-3y=12ex-9

y^{^{''}} - 2 y^{^{'}} - 3 y = 12 e x - 9

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

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = 12 e x - 9

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + 4 y = \frac{x}{e ^{2 x} ( 1 + x ^{2} )}

y^{^{''}} - y^{^{'}} + 64 y = 8 sin (8 t)

y^{^{''}} - 3 y^{^{'}} + 2 y = 2 x + 4

y^{^{''}} + 5 y^{^{'}} = x - 2 - 3 e^{2 x}

x^{^{''}} + 3 x^{^{'}} + 33 x = 8 cos (5 t)

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