What is the solution for y^{''}+2y^'-15=x^2 ?

The solution for y^{''}+2y^'-15=x^2 is y=c_{1}+c_{2}e^{-2x}+(x^3)/6-(x^2)/4+(31x)/4

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y^{''}+2y^'-15=x^2

y^{^{''}} + 2 y^{^{'}} - 15 = x^{2}

y = c_{1} + c_{2} e^{- 2 x} + \frac{x ^{3}}{6} - \frac{x ^{2}}{4} + \frac{31 x}{4}

Solution steps

y^{''} (x) + 2 y^{'} (x) - 15 = x^{2}

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} + \frac{x ^{3}}{6} - \frac{x ^{2}}{4} + \frac{31 x}{4}

y = c_{1} + c_{2} e^{- 2 x} + \frac{x ^{3}}{6} - \frac{x ^{2}}{4} + \frac{31 x}{4}

What is the solution for y^{''}+2y^'-15=x^2 ?
The solution for y^{''}+2y^'-15=x^2 is y=c_{1}+c_{2}e^{-2x}+(x^3)/6-(x^2)/4+(31x)/4