What is the solution for y^{''}+2y^'+y=0,y^'(0)=8,y(1)=8 ?

The solution for y^{''}+2y^'+y=0,y^'(0)=8,y(1)=8 is y=(-8+(8e+8)/2)e^{-t}+((8e+8)e^{-t}t)/2

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

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

y = (- 8 + \frac{8 e + 8}{2}) e^{- t} + \frac{( 8 e + 8 ) e ^{- t} t}{2}

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = 0

Solve linear ODE:

y = (- 8 + \frac{8 e + 8}{2}) e^{- t} + \frac{( 8 e + 8 ) e ^{- t} t}{2}

y = (- 8 + \frac{8 e + 8}{2}) e^{- t} + \frac{( 8 e + 8 ) e ^{- t} t}{2}

What is the solution for y^{''}+2y^'+y=0,y^'(0)=8,y(1)=8 ?
The solution for y^{''}+2y^'+y=0,y^'(0)=8,y(1)=8 is y=(-8+(8e+8)/2)e^{-t}+((8e+8)e^{-t}t)/2