What is the solution for y^{''}-y^'-6y=0,y(0)=2,y^'(0)=2 ?

The solution for y^{''}-y^'-6y=0,y(0)=2,y^'(0)=2 is y=(2-4/5)e^{3t}+4/5 e^{-2t}

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y^{''}-y^'-6y=0,y(0)=2,y^'(0)=2

y^{^{''}} - y^{^{'}} - 6 y = 0, y (0) = 2, y^{^{'}} (0) = 2

y = (2 - \frac{4}{5}) e^{3 t} + \frac{4}{5} e^{- 2 t}

Solution steps

y^{''} (t) - y^{'} (t) - 6 y = 0

Solve linear ODE:

y = (2 - \frac{4}{5}) e^{3 t} + \frac{4}{5} e^{- 2 t}

y = (2 - \frac{4}{5}) e^{3 t} + \frac{4}{5} e^{- 2 t}

\int_{1}^{3} (2 x^{2} - 2 x + 3) d x

\int sin (\frac{nπ x}{l}) d x

\int_{0}^{3 y - 5} x y^{2} d x

\int e^{- y^{2}} d y

t an g e n t y = \frac{2}{x}, (- 3, - \frac{2}{3})

What is the solution for y^{''}-y^'-6y=0,y(0)=2,y^'(0)=2 ?
The solution for y^{''}-y^'-6y=0,y(0)=2,y^'(0)=2 is y=(2-4/5)e^{3t}+4/5 e^{-2t}