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

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### 1 Answer

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 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)`

`y=(C_1+C_2)cossqrt6x+(C_2-C_1)isinsqrt6x`

Now we put new constants `A=C_1+C_2` and `B=(C_2-C_1)i` so we have

`y=Acossqrt6x+Bsinsqrt6x`

Now in order to calculate `A` and `B` we use initial values. First we find derivation `y'=-Asqrt6sinsqrt6x+Bsqrt6cossqrt6x`

`y(0)=2=>Acos0+Bsin0=2`

`y'(0)=-2=>-Asqrt6sin0+Bsqrt6cos0=-2`

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)`