What is the solution for y^{''}+2y^'+y=0,y(1)=6,y^'(0)=8 ?

The solution for y^{''}+2y^'+y=0,y(1)=6,y^'(0)=8 is y=(6e-(8+6e)/2)e^{-t}+((8+6e)e^{-t}t)/2

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y^{''}+2y^'+y=0,y(1)=6,y^'(0)=8

y^{^{''}} + 2 y^{^{'}} + y = 0, y (1) = 6, y^{^{'}} (0) = 8

y = (6 e - \frac{8 + 6 e}{2}) e^{- t} + \frac{( 8 + 6 e ) e ^{- t} t}{2}

Solution steps

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

Solve linear ODE:

y = (6 e - \frac{8 + 6 e}{2}) e^{- t} + \frac{( 8 + 6 e ) e ^{- t} t}{2}

y = (6 e - \frac{8 + 6 e}{2}) e^{- t} + \frac{( 8 + 6 e ) e ^{- t} t}{2}

What is the solution for y^{''}+2y^'+y=0,y(1)=6,y^'(0)=8 ?
The solution for y^{''}+2y^'+y=0,y(1)=6,y^'(0)=8 is y=(6e-(8+6e)/2)e^{-t}+((8+6e)e^{-t}t)/2