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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

9 y^{^{''}} + 18 y^{^{'}} + 13 y = 0

q^{^{''}} + 20 q^{^{'}} + 100 q = 0

\frac{d ^{2} y}{d x ^{2}} - 366 \frac{d y}{d x} = 0

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

5 y^{^{'''}} + 45 y^{^{'}} - 1350 y = 0

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