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

The solution for y^{''}+2y^'=x^2+1 is y=c_{1}+c_{2}e^{-2x}+(x^3)/6-(x^2)/4+(3x)/4

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

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

y = c_{1} + c_{2} e^{- 2 x} + \frac{x ^{3}}{6} - \frac{x ^{2}}{4} + \frac{3 x}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} + \frac{x ^{3}}{6} - \frac{x ^{2}}{4} + \frac{3 x}{4}

y = c_{1} + c_{2} e^{- 2 x} + \frac{x ^{3}}{6} - \frac{x ^{2}}{4} + \frac{3 x}{4}

y^{^{''}} - y^{^{'}} - 20 y = 27 e^{- 4 c}

(D - 2)^{2} y = 8 e^{2 x}

y^{^{''}} - 10 y^{^{'}} + 25 y = 102 t^{2} e^{5 t}

2 y^{^{''}} + 4 y^{^{'}} + 8 y = 8

y^{^{''}} - 10 y^{^{'}} + 25 y = 30 x^{2} + 3

What is the solution for y^{''}+2y^'=x^2+1 ?
The solution for y^{''}+2y^'=x^2+1 is y=c_{1}+c_{2}e^{-2x}+(x^3)/6-(x^2)/4+(3x)/4