What is the solution for y^{''}-2y^'-3y=3x+1 ?

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

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y^{''}-2y^'-3y=3x+1

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 x + 1

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

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = 3 x + 1

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} = 12 x^{2}

y^{^{''}} - 6 y^{^{'}} + 9 y = 5 x^{- 2 x}

y^{^{''}} - 3 y^{^{'}} - 5 = e^{2 x} - 3 cos (3 x)

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

y^{^{''}} + 10 y^{^{'}} + 16 y = 949 sin (3 t)

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