What is the solution for y^{''}+2y^'+y=e^{-x},y(0)=0,y^'(0)=1 ?

The solution for y^{''}+2y^'+y=e^{-x},y(0)=0,y^'(0)=1 is y=(e^{-x}x^2)/2+e^{-x}x

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y^{''}+2y^'+y=e^{-x},y(0)=0,y^'(0)=1

y^{^{''}} + 2 y^{^{'}} + y = e^{- x}, y (0) = 0, y^{^{'}} (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 6 y^{^{'}} + 9 y = 2 e^{(- 3 x)} \frac{1}{x ^{2} + 1}

y^{^{''}} + 4 y^{^{'}} - 5 y = (- 4 x^{2} - 8 x - 3) e^{3 x}

y^{^{''}} - 4 y^{^{'}} + 5 = 4 sin (x) - \frac{25}{8} x

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

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{t} sin (t)

What is the solution for y^{''}+2y^'+y=e^{-x},y(0)=0,y^'(0)=1 ?
The solution for y^{''}+2y^'+y=e^{-x},y(0)=0,y^'(0)=1 is y=(e^{-x}x^2)/2+e^{-x}x