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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d x}{d t} + 2 t x = 1

tan (4) x^{3}

x \to 0 lim (\frac{x}{x ^{2} + 7 x})

\int e^{x + 5} d x

\frac{d}{d x} (\frac{tan ( \frac{x}{2} ) + cot ( \frac{x}{2} )}{x})

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