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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 5 y = e^{3 x} sin (2 x)

x^{^{''}} + x = tan (t)

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

y^{^{''}} + 9 y = 8 sin (t), y (0) = 0, y^{^{'}} (0) = 4

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

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