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^2)/3+(4x)/9-14/27

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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 ^{2}}{3} + \frac{4 x}{9} - \frac{14}{27}

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 ^{2}}{3} + \frac{4 x}{9} - \frac{14}{27}

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{x ^{2}}{3} + \frac{4 x}{9} - \frac{14}{27}

x y^{^{'''}} - x y^{^{'}} = x^{4}

y^{^{''}} - y^{^{'}} = 5 e^{x}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} - 3 y = 2 e^{3 x}

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} - 32 y = (x + 1) e^{x}

2 y^{^{''}} + y^{^{'}} - y = e^{3 t}, y (0) = 2, y^{^{'}} (0) = 0

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^2)/3+(4x)/9-14/27