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

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

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

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

y = (2 - c_{2}) e^{t} + c_{2} e^{- t}

Solution steps

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

Solve linear ODE:

y = (2 - c_{2}) e^{t} + c_{2} e^{- t}

y = (2 - c_{2}) e^{t} + c_{2} e^{- t}

x + y = - 3, x + 9 = y^{2}

\frac{d}{d x} (cos (x - \frac{π}{2}))

\frac{e ^{5 x}}{x ^{20}}

y = 3 x^{\frac{1}{2}} + x^{\frac{3}{2}}

- 6 x (sin (x) + cos (x))

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