What is the solution for y^{''}+5y^'+4y=e^{-x},y(0)=1,y^'(0)=0 ?

The solution for y^{''}+5y^'+4y=e^{-x},y(0)=1,y^'(0)=0 is y=(1+2/9)e^{-x}-2/9 e^{-4x}+(e^{-x}x)/3

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y^{''}+5y^'+4y=e^{-x},y(0)=1,y^'(0)=0

y^{^{''}} + 5 y^{^{'}} + 4 y = e^{- x}, y (0) = 1, y^{^{'}} (0) = 0

y = (1 + \frac{2}{9}) e^{- x} - \frac{2}{9} e^{- 4 x} + \frac{e ^{- x} x}{3}

Solution steps

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

Solve linear ODE:

y = (1 + \frac{2}{9}) e^{- x} - \frac{2}{9} e^{- 4 x} + \frac{e ^{- x} x}{3}

y = (1 + \frac{2}{9}) e^{- x} - \frac{2}{9} e^{- 4 x} + \frac{e ^{- x} x}{3}

y^{^{''}} - 8 y + 7 = 0, y (0) = 1, y^{^{'}} (0) = 2

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

y^{^{''}} - 7 y^{^{'}} + 12 y = t^{2}

y^{^{''}} - 2 y^{^{'}} + y = 2000 x^{2} - 150 e^{x}

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

What is the solution for y^{''}+5y^'+4y=e^{-x},y(0)=1,y^'(0)=0 ?
The solution for y^{''}+5y^'+4y=e^{-x},y(0)=1,y^'(0)=0 is y=(1+2/9)e^{-x}-2/9 e^{-4x}+(e^{-x}x)/3