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

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

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

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

y = \frac{e ^{t}}{2} - \frac{e ^{2}}{2} e^{- t}

Solution steps

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

Solve linear ODE:

y = \frac{e ^{t}}{2} - \frac{e ^{2}}{2} e^{- t}

y = \frac{e ^{t}}{2} - \frac{e ^{2}}{2} e^{- t}

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

\frac{d ^{2} y}{d x ^{2}} - 36 \frac{d y}{d x} = 0

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

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

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

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