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

The solution for y^{''}+2y^'-3y=x^2 is y=c_{1}e^x+c_{2}e^{-3x}-(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^{x} + c_{2} e^{- 3 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^{x} + c_{2} e^{- 3 x} - \frac{x ^{2}}{3} - \frac{4 x}{9} - \frac{14}{27}

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

y^{^{''}} + 4 y = sin (2 x) + 1

x^{^{''}} + 8 x^{^{'}} + 16 x = 6 sin (4 t)

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

x^{^{''}} + 16 x = 2 sin (t)

y^{^{''}} - 6 y^{^{'}} + 5 y = 9 e^{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^x+c_{2}e^{-3x}-(x^2)/3-(4x)/9-14/27