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

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

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y^{''}+2y^'-y=0,y(0)=0,y^'(0)=2sqrt(2)

y^{^{''}} + 2 y^{^{'}} - y = 0, y (0) = 0, y^{^{'}} (0) = 22

y = e^{(- 1 + 2) t} - e^{(- 1 - 2) t}

Solution steps

y^{''} (t) + 2 y^{'} (t) - y = 0

Solve linear ODE:

y = e^{(- 1 + 2) t} - e^{(- 1 - 2) t}

y = e^{(- 1 + 2) t} - e^{(- 1 - 2) t}

x \to \frac{1}{2} lim (3 x - 5)

\int \frac{x ^{2} - 2 x + 5}{9 - x ^{2}} d x

y = - x^{2}, (8, - 64)

e^{x}, e^{- 4 x}, ln (4)

\int - x^{2} + x + 6 d x

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