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

The solution for y^{''}-2y^'+y=e-2x is y=c_{1}e^x+c_{2}xe^x-2x+e-4

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

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

y = c_{1} e^{x} + c_{2} x e^{x} - 2 x + e - 4

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} - 2 x + e - 4

y = c_{1} e^{x} + c_{2} x e^{x} - 2 x + e - 4

y^{^{''}} - 3 y^{^{'}} + 5 y = 2 e^{t}

y^{^{''}} + 4 y^{^{'}} + 6 y = 1^{- t}

y^{^{''}} + y = t - \frac{π}{2}, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 1

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 2

What is the solution for y^{''}-2y^'+y=e-2x ?
The solution for y^{''}-2y^'+y=e-2x is y=c_{1}e^x+c_{2}xe^x-2x+e-4