What is the solution for -2y^{''}+8y^'=0,y(-1)=2,y^'(2)=2 ?

The solution for -2y^{''}+8y^'=0,y(-1)=2,y^'(2)=2 is y=2-1/(2e^{12)}+(e^{4t-8})/2

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-2y^{''}+8y^'=0,y(-1)=2,y^'(2)=2

- 2 y^{^{''}} + 8 y^{^{'}} = 0, y (- 1) = 2, y^{^{'}} (2) = 2

y = 2 - \frac{1}{2 e ^{12}} + \frac{e ^{4 t - 8}}{2}

Solution steps

- 2 y^{''} (t) + 8 y^{'} (t) = 0

Solve linear ODE:

y = 2 - \frac{1}{2 e ^{12}} + \frac{e ^{4 t - 8}}{2}

y = 2 - \frac{1}{2 e ^{12}} + \frac{e ^{4 t - 8}}{2}

y^{^{''}} + 7 y^{^{'}} + 12 y = 0, y (0) = 1, y^{^{'}} (0) = 9

y^{^{''}} = - y^{^{'}} - y

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + y^{^{'}} = 0, y (\frac{π}{4}) = 0, y^{^{'}} (\frac{π}{4}) = - 3

y^{^{''}} + (\frac{v}{a}) y^{^{'}} = 0

What is the solution for -2y^{''}+8y^'=0,y(-1)=2,y^'(2)=2 ?
The solution for -2y^{''}+8y^'=0,y(-1)=2,y^'(2)=2 is y=2-1/(2e^{12)}+(e^{4t-8})/2