Differential equation solutions | A complex number became a real number?

Oscillation

Oscillation

Community Member
MBTI
INFJ
Hi!

I figured some of you might have got into the mathematics at some point. Good for you(!), and good for me. I happen to have one simple question on one step they take in my textbook.
The diff.equation y'' - 4y' + 13y = 0 has the solution y = C[SUB]1[/SUB]e[SUP](2+3i)x[/SUP] + C[SUB]2[/SUB]e[SUP](2-3i)x[/SUP], so far so good. Then we proceed to rewrite the solution in a different form, okay.

y = C[SUB]1[/SUB]e[SUP](2+3i)x[/SUP] + C[SUB]2[/SUB]e[SUP](2-3i)x[/SUP] =
= C[SUB]1[/SUB]e[SUP]2x[/SUP](cos 3x + isin 3x) + C[SUB]2[/SUB]e[SUP]2x[/SUP](cos 3x - isin 3x) =
= e[SUP]2x[/SUP]((C[SUB]1[/SUB] + C[SUB]2[/SUB])cos 3x + i(C[SUB]1[/SUB] - C[SUB]2[/SUB])sin 3x) =
=| C[SUB]1[/SUB] + C[SUB]2[/SUB] = A, i(C[SUB]1[/SUB] - C[SUB]2[/SUB]) = B |=
= e[SUP]2x[/SUP](Acos 3x + Bsin 3x)​

My concern is that they handle the constant B as if it was a real number later on, but isn't it a complex number?! The whole reason why they rewrote the solution in the first place was to "write the solutions in real form". What?! So they just said that i(C[SUB]1[/SUB] - C[SUB]2[/SUB]) = B, and then pretended that there was no imaginary unit in it?

I'm confused. Do you have any explanations to give?
Thank you!


P.S. I took the course about 2 years ago, so I'm a little bit rusty. That's why I've picked up the textbook again D.S.
 
I wish I could add something real to this. I am very interested in maths but haven't gotten around to studying it properly yet

Can you suggest any sources?
 
I feel like Euler's formula or some rule in a similar vein is at play here. I'm bad at remembering all the particulars.
 
Well, in differential equations like that, you have to do a proof. y should not be the solution, but the function you differentiate ie. y'' is the second derivative of y.

A and B might be used instead of more complex (pun not intended) expressions, as substitution, so later during the calculations you would have the imaginary part be returned. I have not done diff. equations with complex numbers so I could be wrong.
 
Poetic Justice said:
I wish I could add something real to this. I am very interested in maths but haven't gotten around to studying it properly yet

Can you suggest any sources?
Click to expand...
Oh, sorry. I know only some swedish sites.
Wolfram Alpha though... that's a really handy site to visit.

Wyote said:
I feel like Euler's formula or some rule in a similar vein is at play here. I'm bad at remembering all the particulars.
Click to expand...
Yes! Right on!


So to simplify the problem: How can iD = B, if both B and D are real numbers?
B should be a complex number for the equation to work, but I guess it has something to do with the nature of diff.equations, that you're not that interested in the distinction of imaginary and real numbers - perhaps it's just the value/"length" of it that's interesting?

I hope someone has a clue, but it's not the whole world if there's not :P
 
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Peppermint said:
Well, in differential equations like that, you have to do a proof. y should not be the solution, but the function you differentiate ie. y'' is the second derivative of y.

A and B might be used instead of more complex (pun not intended) expressions, as substitution, so later during the calculations you would have the imaginary part be returned. I have not done diff. equations with complex numbers so I could be wrong.
Click to expand...

It goes the same as with real numbers: you set y = Ce[SUP]rx[/SUP] (or wich letters you'd like), and then solve the equation for r. If it happen's to be a complex or real number doesn't matter. It works the same way.

"Solution" might be the wrong word, yes, but I'll stick with it. To solve a diff.equation is to find the function y. In this case I just skipped the steps and gave the solution right away, since it wasn't the main focus of my concerns.

I could show it if that would be interesting for a larger group?
 
=| C[SUB]1[/SUB] + C[SUB]2[/SUB] = A, i(C[SUB]1[/SUB] - C[SUB]2[/SUB]) = B |=
= e[SUP]2x[/SUP](Acos 3x + Bsin 3x)
Click to expand...

So, the issue is noting the absolute value of a complex number. i.e. The "i" is not accounted for in the solution.

I researched a few sites (because I'm no expert :D), but the only explanation that I came up with was that "i" can't be simplified so, since it remains a constant, it's a part of B.

Or since B is not determined, the imaginary can't be paired with the constant in the final equation.
 
Gist said:
So, the issue is noting the absolute value of a complex number. i.e. The "i" is not accounted for in the solution.

I researched a few sites (because I'm no expert :D), but the only explanation that I came up with was that "i" can't be simplified so, since it remains a constant, it's a part of B.

Or since B is not determined, the imaginary can't be paired with the constant in the final equation.
Click to expand...

Gaaaah! I know, it's frustrating! The imaginary unit i can't just vanish! If it's part of B, then B is complex. So why are they handling B as if it were real? Sigh.
But, again, I suspect it's because of some basic nature to linear diff.equations, 'cause when you at first set y = Ce[SUP]rx[/SUP] you don't define wheather C is real or complex, and... I guess it works both ways? Humm.
 
maybe you should layout the entire problem from beginning to end. That may help to better understand why i is not accounted for.
 
Gist said:
maybe you should layout the entire problem from beginning to end. That may help to better understand why i is not accounted for.
Click to expand...

They just give you the equation y'' - 4y' + 13 = 0, and goes ahead giving you the solution for it. This passage isn't that interested in the method in itself, but various forms of writing a solution. The real problem lies in the frase "We prefer, however, to write the solutions to the real form" just before they go ahead and simplify it to the final form y = e[SUP]2x[/SUP](Acos 3x + Bsin 3x). So that means there is no complex part of it.

I could just ask some of my old teachers, I guess.
 
So far you haven't determined if B is real or complex, but when you solve for constants A and B you will find that B is always real, which implies C1-C2 is imaginary.
 
BAM! According to a video I just watched (in swedish, sorry) the C[SUB]1[/SUB] and C[SUB]2[/SUB] are COMPLEX (real numbers are just complex without any imaginary part), so that means B would be REAL. Finally!


So, the answer, according to this video and my own deduction, is that B is complex, and just happen to be interpreted as real numbers in this case. I have dwelled upon this problem this late hours, and found out myself that it seems right, because when you set y = Ce[SUP]rx[/SUP] you get a constant C that doesn't intervene with the equation, but rather just vanish... so in other words it doesn't matter whether it's real or complex!
 
There is actually a helpful reddit thread on this very topic!
https://www.reddit.com/r/askscience/comments/2qxe1u/why_can_we_ignore_the_imaginary_part_of_the/
EDIT: I think when you are dealing with a real world problem the initial conditions will force B to be real, which for once the rabbit hole of reddit was pretty helpful in explaining. EDIT2: EX: the typical examples would be like a mass-spring-damper system or an RLC circuit, and you'll find that the coefficients are always real when you're using real physical values. In other words, you don't need to just ignore the imaginary number because it will disappear on its own.
 
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PintoBean said:
There is actually a helpful reddit thread on this very topic!
https://www.reddit.com/r/askscience/comments/2qxe1u/why_can_we_ignore_the_imaginary_part_of_the/
EDIT: I think when you are dealing with a real world problem the initial conditions will force B to be real, which for once the rabbit hole of reddit was pretty helpful in explaining. EDIT2: EX: the typical examples would be like a mass-spring-damper system or an RLC circuit, and you'll find that the coefficients are always real when you're using real physical values. In other words, you don't need to just ignore the imaginary number because it will disappear on its own.
Click to expand...

So I figured. It becomes a whole other issue when it comes to more complex physics though, such as Schrödinger equation, I think, as it is a complex differential equation. But nevertheless, thank you! I will read at reddit tomorrow.
 
Oscillation said:
So I figured. It becomes a whole other issue when it comes to more complex physics though, such as Schrödinger equation, I think, as it is a complex differential equation. But nevertheless, thank you! I will read at reddit tomorrow.
Click to expand...

Yeah, if you know from the physics of the problem and the initial conditions that you're applying that B must be real. In other cases, you can consider a solution in the complex domain. So it all depends on the problem you're looking at.
 
Dang, I was just about to excitedly answer this thread with an explanation, but I think everyone's got it covered.
 
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