What is the solution for y^{''}+8y^'-9y=0,y(0)=1,y^'(0)=0 ?

The solution for y^{''}+8y^'-9y=0,y(0)=1,y^'(0)=0 is y=(1-1/10)e^t+1/10 e^{-9t}

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y^{''}+8y^'-9y=0,y(0)=1,y^'(0)=0

y^{^{''}} + 8 y^{^{'}} - 9 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y = (1 - \frac{1}{10}) e^{t} + \frac{1}{10} e^{- 9 t}

Solution steps

y^{''} (t) + 8 y^{'} (t) - 9 y = 0

Solve linear ODE:

y = (1 - \frac{1}{10}) e^{t} + \frac{1}{10} e^{- 9 t}

y = (1 - \frac{1}{10}) e^{t} + \frac{1}{10} e^{- 9 t}

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = - 2, y^{^{'}} (0) = - 8

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

θ^{^{''}} (t) + 49 θ (t) = 0

y^{^{''}} + 12.26 y = 0

y^{^{''}} = \frac{- y ^{^{'}} - 5 y}{4}

What is the solution for y^{''}+8y^'-9y=0,y(0)=1,y^'(0)=0 ?
The solution for y^{''}+8y^'-9y=0,y(0)=1,y^'(0)=0 is y=(1-1/10)e^t+1/10 e^{-9t}