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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y = \frac{x ^{2} - 7 x + 2}{x - 2}

\int \frac{x}{( x ^{2} + 4 ) ^{\frac{3}{2}}} d x

f (x) = x^{2} + 2, [- 2, 1]

\frac{\partial}{\partial y} (2 y - 2 x)

\int (3 x^{11}) d x

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