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

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

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y^{''}-y^'=(3x^2+6x+1)e^x

y^{^{''}} - y^{^{'}} = (3 x^{2} + 6 x + 1) e^{x}

y = c_{1} e^{x} + c_{2} + e^{x} x^{3} + e^{x} x - e^{x}

Solution steps

y^{''} (x) - y^{'} (x) = (3 x^{2} + 6 x + 1) e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} + e^{x} x^{3} + e^{x} x - e^{x}

y = c_{1} e^{x} + c_{2} + e^{x} x^{3} + e^{x} x - e^{x}

x \to + 1 lim (3 x^{3} - 4 x^{2} + x + 9)

y = e^{\frac{x}{2}}

\int 2 x y e^{x + y} d x

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