What is the solution for y^{''}-2y^'-3y=3x^2-5 ?

The solution for y^{''}-2y^'-3y=3x^2-5 is y=c_{1}e^{3x}+c_{2}e^{-x}-x^2+(4x)/3+1/9

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d y}{d x} = \frac{y ^{2}}{e ^{x}}

y = 1 4^{- \frac{8}{x}}

\frac{d}{d x} (20 + x)

\frac{\partial}{\partial x} ((x y - 3) e^{- x y})

y^{^{''}} - 4 y^{^{'}} + 3 y = 3 x^{2} - 5 x + 1

What is the solution for y^{''}-2y^'-3y=3x^2-5 ?
The solution for y^{''}-2y^'-3y=3x^2-5 is y=c_{1}e^{3x}+c_{2}e^{-x}-x^2+(4x)/3+1/9