What is the solution for y^{''}-y^'-6y=0,y(0)=1,y^'(0)=1 ?

The solution for y^{''}-y^'-6y=0,y(0)=1,y^'(0)=1 is y=(1-2/5)e^{3t}+2/5 e^{-2t}

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y^{''}-y^'-6y=0,y(0)=1,y^'(0)=1

y^{^{''}} - y^{^{'}} - 6 y = 0, y (0) = 1, y^{^{'}} (0) = 1

y = (1 - \frac{2}{5}) e^{3 t} + \frac{2}{5} e^{- 2 t}

Solution steps

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

Solve linear ODE:

y = (1 - \frac{2}{5}) e^{3 t} + \frac{2}{5} e^{- 2 t}

y = (1 - \frac{2}{5}) e^{3 t} + \frac{2}{5} e^{- 2 t}

\frac{d y}{d x} = 4 y + 3 e^{8 x} (x + 1)

\int \frac{r}{1 + r ^{2}} d r

\int \frac{e ^{x}}{1 + 49 e ^{2 x}} d x

\frac{\partial}{\partial y} (x sin (z))

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

What is the solution for y^{''}-y^'-6y=0,y(0)=1,y^'(0)=1 ?
The solution for y^{''}-y^'-6y=0,y(0)=1,y^'(0)=1 is y=(1-2/5)e^{3t}+2/5 e^{-2t}