What is the solution for 4y^{''}-y=0,y(-2)=1,y^'(-2)=-1 ?

The solution for 4y^{''}-y=0,y(-2)=1,y^'(-2)=-1 is y=(e-(3e)/2)e^{t/2}+(3e^{(-t-2)/2})/(2)

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4y^{''}-y=0,y(-2)=1,y^'(-2)=-1

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

y = (e - \frac{3 e}{2}) e^{\frac{t}{2}} + \frac{3 e ^{\frac{- t - 2}{2}}}{2}

Solution steps

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

Solve linear ODE:

y = (e - \frac{3 e}{2}) e^{\frac{t}{2}} + \frac{3 e ^{\frac{- t - 2}{2}}}{2}

y = (e - \frac{3 e}{2}) e^{\frac{t}{2}} + \frac{3 e ^{\frac{- t - 2}{2}}}{2}

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 5, y^{^{'}} (0) = - 1

2 y^{^{''}} - 3 y^{^{'}} + y = 0

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

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

y^{^{''}} - 4 y^{^{'}} + 9 y = 0

What is the solution for 4y^{''}-y=0,y(-2)=1,y^'(-2)=-1 ?
The solution for 4y^{''}-y=0,y(-2)=1,y^'(-2)=-1 is y=(e-(3e)/2)e^{t/2}+(3e^{(-t-2)/2})/(2)