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

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

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

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

y = \frac{1}{2} e^{t} - \frac{1}{2} e^{- t}

Solution steps

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

Solve linear ODE:

y = \frac{1}{2} e^{t} - \frac{1}{2} e^{- t}

y = \frac{1}{2} e^{t} - \frac{1}{2} e^{- t}

\frac{d}{d x} (8 cos^{4} (x))

x \to - 1 lim (f (x))

1. 1^{x} + 9^{x} d x

\frac{40}{x ^{9}}

x \to - 1 lim (\frac{x ^{3}}{x ^{2} + 1})

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