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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - 3 y^{^{''}} + 2 y^{^{'}} = 0

y^{^{''}} + 4 y^{^{'}} + 6.25 y = 0

2 y^{^{''}} + 3 y^{^{'}} + y = 0

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = 0, y^{^{'}} (0) = - 9

y^{^{''}} - 196 y = 0

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