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Find the unique solution of the second-order initial value problem: y'' + 6y = 0, y(0) = 2 y'(0) = -2

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Tibor Pejić eNotes educator | Certified Educator

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First we solve differential equation `y''+6y=0`

This is equivalent to solving algebraic equation `z^2+6=0`

`z^2=-6`

`z_(1,2)=pm sqrt6 i`

Now our solutions are `y_1=e^(z_1x)`  and `y_2=e^(z_2x)`

And solution of our equation is `y=C_1y_1+C_2y_2`  

where `C_1` and `C_2`   are complex constants.

Solution to our equation...

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