What is the solution for y^{''}+8y^'-9y=0,y(1)=2,y^'(1)=0 ?

The solution for y^{''}+8y^'-9y=0,y(1)=2,y^'(1)=0 is y=e^{t-10}(2e^9-(e^9)/5)+(e^9)/5 e^{-9t}

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y^{''}+8y^'-9y=0,y(1)=2,y^'(1)=0

y^{^{''}} + 8 y^{^{'}} - 9 y = 0, y (1) = 2, y^{^{'}} (1) = 0

y = e^{t - 10} (2 e^{9} - \frac{e ^{9}}{5}) + \frac{e ^{9}}{5} e^{- 9 t}

Solution steps

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

Solve linear ODE:

y = e^{t - 10} (2 e^{9} - \frac{e ^{9}}{5}) + \frac{e ^{9}}{5} e^{- 9 t}

y = e^{t - 10} (2 e^{9} - \frac{e ^{9}}{5}) + \frac{e ^{9}}{5} e^{- 9 t}

\frac{d y}{d x} = \frac{e ^{2 x}}{4 y ^{3}}

\frac{d}{d x} (ln (cos^{2} (x) + 5 x^{3}))

y^{^{''}} = - k^{2} y

\frac{d}{d x} (a x^{3} + b x^{2} + c x + d)

f (x) = \frac{2 x ^{3}}{3} - 12 x

What is the solution for y^{''}+8y^'-9y=0,y(1)=2,y^'(1)=0 ?
The solution for y^{''}+8y^'-9y=0,y(1)=2,y^'(1)=0 is y=e^{t-10}(2e^9-(e^9)/5)+(e^9)/5 e^{-9t}