# Second Order Differential Initial Value Problem

1. Oct 11, 2012

### danielu13

1. The problem statement, all variables and given/known data

y''+4y'+6y
y(0) = 2; y'(0) = 4

2. Relevant equations

$\alpha ± β = e^{x\alpha}(cosβx + sinβx)$

3. The attempt at a solution

Auxilary equation is $r^2+4r+6$, which solves for $-2 ± i$

I get the general solution:

$e^{-2x}(c$1$cosx + c$2$sinx)$

$y' = -2e^{-2x}(c$1$cosx + c$2$sinx) + e^{2x}(c$2$cosx - c$1$sinx)$

$= c$1$(cosx-sinx) + c$2$(cosx+sinx) = 4$

I also have:

c1 + c2 = 2 from the initial value.

I now have a system of equations, but don't really know how to solve it without using a computer.

2. Oct 11, 2012

### Zondrina

No no, the system of equations produces infinitely many solutions, but the initial value given gives a specific solution curve when you solve for the constants.

3. Oct 11, 2012

### danielu13

That's what I was thinking, but I just can't think of how to solve for the constants without using a system of equations at the moment. I know how to solve it when $\alpha = 0$, but not sure how to do this.

4. Oct 11, 2012

### Zondrina

You have two equations. You get one from solving the initial value problem for y and one for solving the initial value problem for y'.

They will give you a system of 2 equations in 2 unknowns c1, c2.

Solving for both unknowns will give you the particular equation for y which will satisfy your initial values.

5. Oct 11, 2012

### danielu13

Oh, I completely overlooked inputting the x values into the equations for some reason. Thanks though!