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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''''}} + 9 y^{^{''}} = cos (3 t)

y^{^{''}} - 2 y^{^{'}} + 5 y = 4 e^{3 t}

y^{^{''}} + y = 1 - H (t - 3 π)

\frac{d ^{2} y}{d t ^{2}} + 4 y = 3 cos (2 t)

y^{^{'''}} + 6 y^{^{''}} - 32 y = e^{- 4 x}

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