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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} x}{d t ^{2}} + 16 x = 0

x^{^{''}} + 10 x^{^{'}} + 25 x = 0

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

y^{^{''''}} - 6 y^{^{'''}} + 9 y^{^{''}} = 0

y^{^{''}} - 9 y^{^{'}} - 52 y = 0

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