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

The solution for y^{''}-2y^'=x^2+2 is y=c_{1}e^{2x}+c_{2}-(x^3)/6-(x^2)/4-(5x)/4

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

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

y = c_{1} e^{2 x} + c_{2} - \frac{x ^{3}}{6} - \frac{x ^{2}}{4} - \frac{5 x}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} - \frac{x ^{3}}{6} - \frac{x ^{2}}{4} - \frac{5 x}{4}

y = c_{1} e^{2 x} + c_{2} - \frac{x ^{3}}{6} - \frac{x ^{2}}{4} - \frac{5 x}{4}

x^{^{''}} + 2 x^{^{'}} + 5 x = 8 cos (2 t) + 2 sin (2 t)

y^{^{''}} - y = e^{(2 x)}

y^{^{''}} + 16 y = 7 sin (x)

y^{^{'''}} - y^{^{''}} - 9 y^{^{'}} + 9 y = 7 - e^{x} + e^{3 x}

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

What is the solution for y^{''}-2y^'=x^2+2 ?
The solution for y^{''}-2y^'=x^2+2 is y=c_{1}e^{2x}+c_{2}-(x^3)/6-(x^2)/4-(5x)/4