What is the solution for y^{''}+4y^'-5y=0,y(0)=0,y^'(0)=1 ?

The solution for y^{''}+4y^'-5y=0,y(0)=0,y^'(0)=1 is y= 1/6 e^t-1/6 e^{-5t}

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y^{''}+4y^'-5y=0,y(0)=0,y^'(0)=1

y^{^{''}} + 4 y^{^{'}} - 5 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y = \frac{1}{6} e^{t} - \frac{1}{6} e^{- 5 t}

Solution steps

y^{''} (t) + 4 y^{'} (t) - 5 y = 0

Solve linear ODE:

y = \frac{1}{6} e^{t} - \frac{1}{6} e^{- 5 t}

y = \frac{1}{6} e^{t} - \frac{1}{6} e^{- 5 t}

y^{^{'''}} - y^{^{''}} + 2 y^{^{'}} - 2 y = 0

77 y^{^{''''}} + 106 y^{^{''}} + 9 y = 0

- 2 y^{^{''}} + y = 0

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = 3, y^{^{'}} (0) = 2

y^{^{''}} + 16 y = 0, y (0) = 1, y^{^{'}} (0) = 0

What is the solution for y^{''}+4y^'-5y=0,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+4y^'-5y=0,y(0)=0,y^'(0)=1 is y= 1/6 e^t-1/6 e^{-5t}