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

Solution steps

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

Solve linear ODE:

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

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

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+e/2)e^{t/2}-(e^{(-t-2)/2})/(2)