What is the solution for y^{''}+2y^'+y=0,y^'(0)=8,y(1)=6 ?

The solution for y^{''}+2y^'+y=0,y^'(0)=8,y(1)=6 is y=(-8+(6e+8)/2)e^{-t}+((6e+8)e^{-t}t)/2

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y^{''}+2y^'+y=0,y^'(0)=8,y(1)=6

y^{^{''}} + 2 y^{^{'}} + y = 0, y^{^{'}} (0) = 8, y (1) = 6

y = (- 8 + \frac{6 e + 8}{2}) e^{- t} + \frac{( 6 e + 8 ) e ^{- t} t}{2}

Solution steps

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

Solve linear ODE:

y = (- 8 + \frac{6 e + 8}{2}) e^{- t} + \frac{( 6 e + 8 ) e ^{- t} t}{2}

y = (- 8 + \frac{6 e + 8}{2}) e^{- t} + \frac{( 6 e + 8 ) e ^{- t} t}{2}

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 0, y (0) = 0

y^{^{''}} - 2 y^{^{'}} + 10 y = 0, y (0) = 0, y^{^{'}} (0) = 4

\frac{d ^{2} x}{d t ^{2}} + 2 \frac{d x}{d t} - 8 x = 0

8 x^{^{''}} + x = 0

1.5 x^{^{''}} + 12 x^{^{'}} + 24 x = 0

What is the solution for y^{''}+2y^'+y=0,y^'(0)=8,y(1)=6 ?
The solution for y^{''}+2y^'+y=0,y^'(0)=8,y(1)=6 is y=(-8+(6e+8)/2)e^{-t}+((6e+8)e^{-t}t)/2