Second Order Differential Initial Value Problem

[danielu13]

Homework Statement

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

Homework Equations

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

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.

 

[STEMucator]

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.

 

[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.

 

[STEMucator]

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.

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.

 

[danielu13]

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