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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

2 y^{^{''}} - 16 y^{^{'}} = 0

y^{^{''}} + 10 y^{^{'}} + 16 y = 0, y (0) = 1, y^{^{'}} (0) = - 12

y^{^{''}} + 9 π^{2} y = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (- 1) = 2, y^{^{'}} (- 1) = 5

y^{^{''}} + 8 y^{^{'}} + 41 y = 0, y^{^{'}} (0) = 2, y (0) = 1

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