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}

1

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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}