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

The solution for y^{''}-y^'-2y=0,y(0)=2,y^'(0)=2 is y=(2-2/3)e^{2t}+2/3 e^{-t}

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

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

y = (2 - \frac{2}{3}) e^{2 t} + \frac{2}{3} e^{- t}

Solution steps

y^{''} (t) - y^{'} (t) - 2 y = 0

Solve linear ODE:

y = (2 - \frac{2}{3}) e^{2 t} + \frac{2}{3} e^{- t}

y = (2 - \frac{2}{3}) e^{2 t} + \frac{2}{3} e^{- t}

\frac{\partial}{\partial x} (1 + 3 x)

x \cdot d y = (x + y) \cdot d x

\frac{( x - ln ( 1 + x ) )}{x ^{2}}

x \to - 5 lim (\frac{x ^{2} - 6}{5 - x})

y = x + \frac{3}{x}, (3, 4)

What is the solution for y^{''}-y^'-2y=0,y(0)=2,y^'(0)=2 ?
The solution for y^{''}-y^'-2y=0,y(0)=2,y^'(0)=2 is y=(2-2/3)e^{2t}+2/3 e^{-t}