What is the solution for 16y^{''}-y=0,y(-4)=1,y^'(-4)=-1 ?

The solution for 16y^{''}-y=0,y(-4)=1,y^'(-4)=-1 is y=(e-(5e)/2)e^{t/4}+(5e^{(-t-4)/4})/(2)

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16y^{''}-y=0,y(-4)=1,y^'(-4)=-1

16 y^{^{''}} - y = 0, y (- 4) = 1, y^{^{'}} (- 4) = - 1

y = (e - \frac{5 e}{2}) e^{\frac{t}{4}} + \frac{5 e ^{\frac{- t - 4}{4}}}{2}

Solution steps

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

Solve linear ODE:

y = (e - \frac{5 e}{2}) e^{\frac{t}{4}} + \frac{5 e ^{\frac{- t - 4}{4}}}{2}

y = (e - \frac{5 e}{2}) e^{\frac{t}{4}} + \frac{5 e ^{\frac{- t - 4}{4}}}{2}

y^{^{''}} + 16 y^{^{'}} + 64 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} = 0, y (0) = 6 - 6 e^{4}, y (1) = 0

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

y^{^{''}} - \frac{3}{2} y^{^{'}} - y = 0

y^{^{''}} - 8 y^{^{'}} = 0, y (0) = 5 - 5 e^{8}, y (1) = 0

What is the solution for 16y^{''}-y=0,y(-4)=1,y^'(-4)=-1 ?
The solution for 16y^{''}-y=0,y(-4)=1,y^'(-4)=-1 is y=(e-(5e)/2)e^{t/4}+(5e^{(-t-4)/4})/(2)