What is the solution for y^{''}-y=0,y(0)= 1/5 ,y(1)= 1/5 cosh(1) ?

The solution for y^{''}-y=0,y(0)= 1/5 ,y(1)= 1/5 cosh(1) is y=(1/5-1/10)e^t+1/10 e^{-t}

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

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

y = (\frac{1}{5} - \frac{1}{10}) e^{t} + \frac{1}{10} e^{- t}

Solution steps

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

Solve linear ODE:

y = (\frac{1}{5} - \frac{1}{10}) e^{t} + \frac{1}{10} e^{- t}

y = (\frac{1}{5} - \frac{1}{10}) e^{t} + \frac{1}{10} e^{- t}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} - 3 y = 0

y^{^{''}} - 2 y^{^{'}} - 8 y = 0, y (0) = 1, y^{^{'}} (0) = - 3

y^{^{'''}} + 2 y^{^{''}} + 5 y^{^{'}} = 0

\frac{d ^{2} y}{d x ^{2}} + k y = 0

(D^{2} + 4)^{2} y = 0

What is the solution for y^{''}-y=0,y(0)= 1/5 ,y(1)= 1/5 cosh(1) ?
The solution for y^{''}-y=0,y(0)= 1/5 ,y(1)= 1/5 cosh(1) is y=(1/5-1/10)e^t+1/10 e^{-t}