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

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

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y^{''}-2y^'+1=xe^x

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} - 6 y = 6 e^{2 x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e^{x} - 9 + 24 x + 4

y^{^{''''}} - 2 y^{^{''}} + 1 = 0

x^{^{''}} + b \cdot x^{^{'}} + a = 0

x^{^{''}} - 6 x - 16 x = sin (2 t)

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