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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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