What is the solution for 8y^{''}+32y^'=0,y(2)=1,y^'(-1)=1 ?

The solution for 8y^{''}+32y^'=0,y(2)=1,y^'(-1)=1 is y=1+1/(4e^{12)}-(e^{-4t-4})/4

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8y^{''}+32y^'=0,y(2)=1,y^'(-1)=1

8 y^{^{''}} + 32 y^{^{'}} = 0, y (2) = 1, y^{^{'}} (- 1) = 1

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

Solution steps

8 y^{''} (t) + 32 y^{'} (t) = 0

Solve linear ODE:

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

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

\frac{x ^{2}}{2 + x}

\frac{d x}{d y} = (4 x + 2 y + 1)

y = (8 x^{2} - 16 x + 16) e^{x}

\frac{d}{d x} (\frac{4 x}{x + 1})

\frac{\partial}{\partial y} (e^{\frac{y}{x}})

What is the solution for 8y^{''}+32y^'=0,y(2)=1,y^'(-1)=1 ?
The solution for 8y^{''}+32y^'=0,y(2)=1,y^'(-1)=1 is y=1+1/(4e^{12)}-(e^{-4t-4})/4