Second Order Linear Homogeneous DE

[QuarkCharmer]

Homework Statement

[tex]3\frac{d^{2}y}{dx^{2}} + 2\frac{dy}{dx} + y = 0[/tex]

Homework Equations

The Attempt at a Solution

[tex]3y'' + 2y' +y = 0[/tex]

I know the solution is going to be in the form of [tex]y=Ce^{mx}+De^{nx}+...[/tex]
(Unless there is a multiplicity, in which case I understand that too)

So I'll just skip to the Aux Equation:
[tex]3m^{2} + 2m + 1 = 0[/tex]
[tex]m = \frac{-1+\sqrt{2}i}{3} or \frac{-1-\sqrt{2}i}{3}[/tex]

Thus, a general form of the solution is:
[tex]y(x) = C_{1}e^{\frac{-1}{3}}e^{\frac{\sqrt{2}ix}{3}} + C_{2}e^{\frac{-1}{3}}e^{\frac{-\sqrt{2}ix}{3}}[/tex]

So now I just rename the constants..

[tex]Ae^{\frac{\sqrt{2}ix}{3}} + Be^{\frac{-\sqrt{2}ix}{3}}[/tex]

and since...

[tex]e^{ix} = cos(x) + isin(x)[/tex]

[tex]A(cos(\frac{\sqrt{2}x}{3})+isin(\frac{\sqrt{2}x}{3})) + B(cos(\frac{-\sqrt{2}x}{3})+isin(\frac{-\sqrt{2}x}{3}))[/tex]

Since cosine is even and sine is odd...

[tex]Acos(\frac{\sqrt{2}x}{3}) + iAsin(\frac{\sqrt{2}x}{3}) + Bcos(\frac{\sqrt{2}x}{3}) -iBsin(\frac{\sqrt{2}x}{3})[/tex]

Usually here is where the imaginary stuff cancels out, but it's just not happening this time and I think I am making a mistake somewhere but I can't see it. What do I do??

 

[micromass]

Homework Statement

[tex]3\frac{d^{2}y}{dx^{2}} + 2\frac{dy}{dx} + y = 0[/tex]

Homework Equations

The Attempt at a Solution

[tex]3y'' + 2y' +y = 0[/tex]

I know the solution is going to be in the form of [tex]y=Ce^{mx}+De^{nx}+...[/tex]
(Unless there is a multiplicity, in which case I understand that too)

So I'll just skip to the Aux Equation:
[tex]3m^{2} + 2m + 1 = 0[/tex]
[tex]m = \frac{-1+\sqrt{2}i}{3} or \frac{-1-\sqrt{2}i}{3}[/tex]

Thus, a general form of the solution is:
[tex]y(x) = C_{1}e^{\frac{-1}{3}}e^{\frac{\sqrt{2}ix}{3}} + C_{2}e^{\frac{-1}{3}}e^{\frac{-\sqrt{2}ix}{3}}[/tex]

Samll mistake, but you also want [itex]e^{-\frac{1}{3}x}[/itex] instead of [itex]e^{-\frac{1}{3}}[/itex]. Indeed, your general solution requires you to multiply entire m with x. This will not save you from the trouble below however.

So now I just rename the constants..

[tex]Ae^{\frac{\sqrt{2}ix}{3}} + Be^{\frac{-\sqrt{2}ix}{3}}[/tex]

and since...

[tex]e^{ix} = cos(x) + isin(x)[/tex]

[tex]A(cos(\frac{\sqrt{2}x}{3})+isin(\frac{\sqrt{2}x}{3})) + B(cos(\frac{-\sqrt{2}x}{3})+isin(\frac{-\sqrt{2}x}{3}))[/tex]

Since cosine is even and sine is odd...

[tex]Acos(\frac{\sqrt{2}x}{3}) + iAsin(\frac{\sqrt{2}x}{3}) + Bcos(\frac{\sqrt{2}x}{3}) -iBsin(\frac{\sqrt{2}x}{3})[/tex]

Usually here is where the imaginary stuff cancels out, but it's just not happening this time and I think I am making a mistake somewhere but I can't see it. What do I do??

Now you found all complex solutions, which is good. But you want all real solutions. Do you see choices of A and B that will get you real solutions?? (you can choose A and B complex if you want to).

The hardest part is proving that you got all real solutions.

 

[QuarkCharmer]

Thanks Micromass. I see what I did wrong. I'll try to re-derive it. I just don't want to blindly use the formula to solve this type of DE without understanding where it's coming from.

 

[Mark44]

For diff. equations of this type, if the roots of the characteristic equation are a [itex]\pm[/itex] bi, a pair of solutions is y₁ = e^atsin(bt) and y₂ = e^atcos(bt).

 

[QuarkCharmer]

I figured it out.

What I was doing wrong was this:

Once the equation turned into the form with (or without) the exponential and then the sum of sines and cosines, I needed to group the two functions with the complex variable in front and factor them out into the constant. So it's essentially possible that the constant contains a complex variable.

I don't know what happened in the above work, I lost some things in translation to Latex and now looking over it, it doesn't make much sense! I fully understand how an equation of that form with an aux equation containing both complex and real part solutions works now, and simplifies into exp(x)(sin+cos) (for second order at least).

Today we went over how the discriminant determines which parts of the solution you get, the three cases (d>0, d<0, d=0) and what that means exactly. Now it all makes sense.