What is the solution for 64y^{''}-y=0,y(-8)=1,y^'(-8)=-1 ?

The solution for 64y^{''}-y=0,y(-8)=1,y^'(-8)=-1 is y=(e-(9e)/2)e^{t/8}+(9e^{(-t-8)/8})/(2)

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64y^{''}-y=0,y(-8)=1,y^'(-8)=-1

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

y = (e - \frac{9 e}{2}) e^{\frac{t}{8}} + \frac{9 e ^{\frac{- t - 8}{8}}}{2}

Solution steps

64 y^{''} (t) - y = 0

Solve linear ODE:

y = (e - \frac{9 e}{2}) e^{\frac{t}{8}} + \frac{9 e ^{\frac{- t - 8}{8}}}{2}

y = (e - \frac{9 e}{2}) e^{\frac{t}{8}} + \frac{9 e ^{\frac{- t - 8}{8}}}{2}

What is the solution for 64y^{''}-y=0,y(-8)=1,y^'(-8)=-1 ?
The solution for 64y^{''}-y=0,y(-8)=1,y^'(-8)=-1 is y=(e-(9e)/2)e^{t/8}+(9e^{(-t-8)/8})/(2)