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
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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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