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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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