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=(1-4/5)e^{3t}+4/5 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{4}{5}) e^{3 t} + \frac{4}{5} e^{- 2 t}

Solution steps

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

Solve linear ODE:

y = (1 - \frac{4}{5}) e^{3 t} + \frac{4}{5} e^{- 2 t}

y = (1 - \frac{4}{5}) e^{3 t} + \frac{4}{5} e^{- 2 t}

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

\frac{d}{d x} (e^{x^{2} + 6 x})

\int (5 x + 1)^{4} d x

\frac{d ^{2}}{d x ^{2}} (\frac{1}{x ^{2} + 1})

f (x) = 9 - ∣ x - 4 ∣ at x

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=(1-4/5)e^{3t}+4/5 e^{-2t}