What is the solution for y^{''}+2y^'+y=8e^x,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+2y^'+y=8e^x,y(0)=0,y^'(0)=0 is y=-2e^{-x}-4e^{-x}x+2e^x

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y^{''}+2y^'+y=8e^x,y(0)=0,y^'(0)=0

y^{^{''}} + 2 y^{^{'}} + y = 8 e^{x}, y (0) = 0, y^{^{'}} (0) = 0

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

Solution steps

y^{''} (x) + 2 y^{'} (x) + y = 8 e^{x}

Solve linear ODE:

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

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

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

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

y^{^{''}} - 10 y^{^{'}} + 25 y = 3 e^{5 t}

y^{^{''}} + y = H (t - 1)

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

What is the solution for y^{''}+2y^'+y=8e^x,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+2y^'+y=8e^x,y(0)=0,y^'(0)=0 is y=-2e^{-x}-4e^{-x}x+2e^x