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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int 2 (x - 1) d x

y^{^{''}} + 2 y^{^{'}} + 36 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\int_{1}^{3} (x - 1) d x

\frac{d}{d x} (2 tan (x) sec^{2} (x))

ln (tan (x))

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