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Find the unique solution of the second-order initial value problem: y'' + 6y = 0, y(0)...
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High School Teacher
First we solve differential equation `y''+6y=0`
This is equivalent to solving algebraic equation `z^2+6=0`
`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 is `y=C_1e^(-isqrt6x)+C_2e^(isqrt6x)`
And since `e^(ia)=cos a+isina` we have
`y=C_1(cos (-sqrt6x)+isin(-sqrt6x))6+C_2(cos sqrt6x+isin sqrt6x)`
Now we put new constants `A=C_1+C_2` and `B=(C_2-C_1)i` so we have
Now in order to calculate `A` and `B` we use initial values. First we find derivation `y'=-Asqrt6sinsqrt6x+Bsqrt6cossqrt6x`
From first equation we have `A=2.`
From second equation we have `Bsqrt6=-2=>B=-2/sqrt6`
Hence the final solution to our initial value problem is `y=2cos(sqrt6x)-2/sqrt6sin(sqrt6x)`
Posted by tiburtius on September 20, 2013 at 12:29 PM (Answer #1)
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