Question

Find the general solution of the given differential equation:

Answer 1

\[y^{(6)}+y=0\]

Answer 2

I was able to determine the characteristic equation and simplify:\[r^6+1=0 \rightarrow r=\sqrt[6]{-1}\]

Answer 3

No, it is not that.
You must derive as \((r^2)^3 +((1)^2)^3=(r^2+1) (r^4-r^2+1)\)
and solve for 6 of r

Answer 4

Oh, I see. How did you know to do it factor it exactly like that?

Answer 5

expand more to get \(-(r^2+1) (-r^2 +\sqrt3 r-1)(r^2 +\sqrt3 r+1)=0\) and manipulate more

Answer 6

You know \(a^3 + b^3\) formula, right?

Answer 7

Yeah

Answer 8

The leftover is ok, right?

Answer 9

And then solve the first term algebraically and the last two using quadratic formula, correct?

Answer 10

This corresponds to the 6th roots of unity :)
So if you're comfortable with complex numbers, you could do it that way.

Or you can stick with a couple of quadratics as Lose described if that's preferable.

Answer 11

\[\large\rm r=(\color{orangered}{-1})^{1/6}\]\[\large\rm r=\left(\color{orangered}{e^{i\left(-\frac{\pi}{2}+2k \pi\right)}}\right)^{1/6}\]Which becomes:\[\large\rm r=\color{orangered}{e^{i\left(-\frac{\pi}{12}+\frac{2k \pi}{12}\right)}}\]
And then k=0, 1, 2, 3, 4, 5 give you your roots.
And then those correspond to a bunch of sines and cosines ya? :o
Hmm ya maybe that's more complicated :d

Answer 12

That's an ugly one, but I think that's the direction I was trying to head toward! ;D I could also use \(\frac{ 3\pi }{ 2 }\) rather than \(-\frac{ \pi }{ 2 }\) no?

Answer 13

Yes :)

Answer 14

Mm, yeah and then it will follow the form \(C_1\cos(\theta)+C_2\sin(\theta)\) right? Or should there be an i in front of the sin? I never know when to put the i or not..

Answer 15

Correct, no i.
I'm trying to work through the steps to justify that though.
Hehe sec

Answer 16

Yeah some of the questions had solutions including i with the sin, but most of the time they didn't so I was getting confused DX :P

Answer 17

hmm

Answer 18

Oh oh oh, ok.
I was getting really confused because Wolfram was giving different angles than what I was coming up with.
https://www.wolframalpha.com/input/?i=%28-1%29%5E%281%2F6%29

Answer 19

Negative 1 is along the real axis, so that's at an angle of pi, not 3pi/2.
My bad, my bad.

Answer 20

Oh I completely missed that mistake too lol

Answer 21

\[\large\rm r_k^6=\exp\left\{i\left(\pi+2k \pi\right)\right\}\]\[\large\rm r_k=\exp\left\{i\left(\pi+2k \pi\right)\right\}^{1/6}\]\[\large\rm r_k=\exp\left\{i\left(\frac{\pi}{6}+\frac{2k \pi}{6}\right)\right\},\qquad\qquad k=0,1,2,3,4,5\]

So for k=0, our first root will be,\[\large\rm r_0=e^{i \pi/6}=\cos\frac{\pi}{6}+i~\sin\frac{\pi}{6}\]\[\large\rm r_0=\frac{\sqrt3}{2}+i~\frac{1}{2}\]

Then our first solution: let's call it y_o to match our root notation,\[\large\rm y_0=c_0e^{r_0t}\]\[\large\rm y_0=c_0 ~\exp\left\{\left(\frac{\sqrt3}{2}+i~\frac{1}{2}\right)t\right\}\]\[\large\rm y_0=c_0 e^{\frac{\sqrt{3}}{2}t}\color{orangered}{e^{i\frac{1}{2}t}}\]\[\large\rm y_0=c_0 e^{\frac{\sqrt{3}}{2}t}\color{orangered}{\left(\cos\frac{1}{2}t+i~\sin\frac{1}{2}t\right)}\]
Oh yah, don't do it this way >.<
Cause then we run into that question with the i again...
And I don't wanna have to think about that right now lolol



So from here:\[\large\rm r_0=\frac{\sqrt3}{2}+i~\frac{1}{2}\]We get,\[\large\rm y_0=c_0~e^{\frac{\sqrt{3}}{2}t}\cos\frac{t}{2}+c_0~e^{\frac{\sqrt{3}}{2}t}\sin\frac{t}{2}\]Ya? :o

Answer 22

exp{ } = e^{ }
Just in case there was any confusion on that.
writing these tiny fractions inside an exponent looks really ugly sometimes.

Answer 23

That makes sense except what happened to the i in the last step? >_<

Answer 24

When we have a root: \(\large\rm r=\alpha+\beta i\)
Our solution is of the form: \(\large\rm y=c_1 e^{\alpha}\cos\beta+c_2 e^{\alpha}\sin\beta\)

I was using that uhhh shortcut thing so we could avoid that question hehe.
Where does the i go, that's a good question.
I don't think it get's absorbed into the c... so I'm trying to think.. >.<

Answer 25

Woops*\[\large\rm y=c_1 e^{\alpha t}\cos\beta t+c_2 e^{\alpha t}\sin\beta t\]

Answer 26

I think it's because the sin of any angle is the vertical component, which is on the imaginary axis. So it already accounts for it perhaps?

Answer 27

Well not *on* the axis, but in the same direction.

Answer 28

Ok, I think it's because `complex roots always come in conjugate pairs`.
Remember that back from Algebra?

So if \(\large\rm r=\frac{\sqrt3}{2}+\frac{1}{2}i\) is one of our roots,
Then we also know that \(\large\rm \frac{\sqrt{3}}{2}-\frac{1}{2}i\) is a root!

So then a couple of our solutions are,\[\large\rm y=c_1 e^{\sqrt3/2 t+1/2i t}+c_2e^{\sqrt3/2t-1/2i t}\]\[\rm y=c_1 e^{\sqrt3/2 t}\left[\cos\frac{t}{2}+i \sin\frac{t}{2}\right]+c_2e^{\sqrt3/2t}\left[\cos\left(-\frac{t}{2}\right)+i \sin\left(-\frac{t}{2}\right)\right]\]\[\rm y=c_1 e^{\sqrt3/2 t}\left[\cos\frac{t}{2}+i \sin\frac{t}{2}\right]+c_2e^{\sqrt3/2t}\left[\cos\left(\frac{t}{2}\right)-i \sin\left(\frac{t}{2}\right)\right]\]Distribute all of this mess out,
Then regroup,\[\rm y= e^{\sqrt3/2 t}\left[(c_1+c_2)\cos\frac{t}{2}+(c_1i-c_2i)\sin\frac{t}{2}\right]\]

Answer 29

So I guess I was wrong before, we ARE in fact absorbing the i into the c's.

Answer 30

\[\rm y= e^{\sqrt3/2 t}\left[\color{orangered}{(c_1+c_2)}\cos\frac{t}{2}+\color{orangered}{(c_1i-c_2i)}\sin\frac{t}{2}\right]\]\[\rm y= e^{\sqrt3/2 t}\left[\color{orangered}{c_3}\cos\frac{t}{2}+\color{orangered}{c_4}\sin\frac{t}{2}\right]\]

Answer 31

Khan explains it in nice detail on youtube if you wanna check that out it some point:
https://www.youtube.com/watch?v=6xEO4BeawzA
https://www.youtube.com/watch?v=jJyRrIZ595c

Answer 32

And what you might notice is,
I used two of the roots,
and ended up with only 1 cosine, and 1 sine.
There was overlap, and we were able to combine them together.

This should be expected, ya?
We shouldn't end up with 12 terms if it's a 6th root,
after all we know that we end up with 2 terms (1cosine 1sine) when it's a `square` root.

Answer 33

Oooooh that makes more sense now. And yeah I was also wondering why we would not have 12 terms because there are cos and sin for each solution. But that makes sense. Thank you!

Answer 34

cool ୧ʕ•̀ᴥ•́ʔ୨

Answer 35

d'aww

Answer 36

XD