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=2e^{-t}+(2e-2)te^{-t}

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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 e^{- t} + (2 e - 2) t e^{- t}

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d t ^{2}} + 7 \frac{d y}{d t} + 12 y = 0

\frac{d ^{2} y}{d t ^{2}} - 18 \frac{d y}{d t} + 145 y = 0

y^{^{'''}} - 3 y^{^{''}} - 15 y^{^{'}} + 25 y = 0

y^{^{''}} + 14 y^{^{'}} + 48 y = 0, y (0) = 0, y^{^{'}} (0) = - 6

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = 5

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=2e^{-t}+(2e-2)te^{-t}