What is the solution for y^{''}+2y^'+y=0,y(0)=1,y(1)=3 ?

The solution for y^{''}+2y^'+y=0,y(0)=1,y(1)=3 is y=e^{-t}+(3e-1)e^{-t}t

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y^{''}+2y^'+y=0,y(0)=1,y(1)=3

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = 1, y (1) = 3

y = e^{- t} + (3 e - 1) e^{- t} t

Solution steps

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

Solve linear ODE:

y = e^{- t} + (3 e - 1) e^{- t} t

y = e^{- t} + (3 e - 1) e^{- t} t

y^{^{''''}} - 5 y^{^{'''}} + 5 y^{^{''}} + 5 y^{^{'}} - 6 y = 0

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

y^{^{''}} = - y^{^{'}}

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

y^{^{''}} + 9 y = 0, y (π) = 0, y (0) = 0

What is the solution for y^{''}+2y^'+y=0,y(0)=1,y(1)=3 ?
The solution for y^{''}+2y^'+y=0,y(0)=1,y(1)=3 is y=e^{-t}+(3e-1)e^{-t}t