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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} x}{d t ^{2}} + a \cdot \frac{d x}{d t} = 0

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + 4 y = 0

y^{^{''}} + 6 y^{^{'}} + 10 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 8

(d^{2} - 2) y = 0

\frac{d ^{6} y}{d x ^{6}} - 2 \frac{d ^{3} y}{d x ^{3}} + y = 0

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