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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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