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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + 2 y = 2 e^{t}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 81 y = cosh (9 x)

y^{^{''}} + 4 y = x e^{x} + x sin (2 x)

y^{^{''}} + 49 y = csc (7 x)

y^{^{''}} + y^{^{'}} + 2 = 0

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