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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - 13 y^{^{''}} + 12 y^{^{'}} = 0

y^{^{''}} - 5 y + 6 y = 0, y (0) = - \frac{3}{5}, y^{^{'}} (0) = - \frac{13}{5}

5 y^{^{''}} + 9 y^{^{'}} = 0

y^{^{''}} + 14 y^{^{'}} = 0, y (0) = - 7

y^{^{''''}} = 81 y

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