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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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