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}

\frac{d ^{2} x}{d t ^{2}} + 2 a \frac{d x}{d t} + b^{2} x = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y = e^{3 x}

y^{^{'''}} - y^{^{''}} + 100 y^{^{'}} - 100 y = 0

y^{^{''}} + 9 y^{^{'}} + 14 y = 0, y (0) = 0

(\frac{d ^{2} y}{d x ^{2}}) + \frac{d y}{d x} - 6 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+e/2)e^{t/2}-(e^{(-t-2)/2})/(2)