Question

I'm trying to get the general solution for a higher order differential equation. I think I have the method down correctly, but sadly, I think the factorization is tripping me up. Would appreciate a nudge in the right direction. (Work attached in image).

Answer 1

It's probably simple enough, but I'm supposed to end up with this:

Answer 2

yes! I think you are right!

Answer 3

Ok, so what's confusing me is how to get from the first screenshot to the second.

Answer 4

we have to solve this equation:
\[{\lambda ^2} + \lambda + 1 = 0\]
furthermore each of the roots of that above equation has to be counted 2 times

Answer 5

Which I can get from the calculator easily enough, but doing by hand...I tried this:

Answer 6

Maybe I just stayed up too late last night. :)

Answer 7

I think that we have the subsequent steps:
\[\begin{gathered}
\exp \left( {rx} \right)\left( {{r^4} + 2{r^3} + 3{r^2} + 2r + 1} \right) = 0 \hfill \\
{r^4} + 2{r^3} + 3{r^2} + 2r + 1 = 0 \hfill \\
{r^4} + 2{r^3} + 3{r^2} + 2r + 1 = {\left( {{r^2} + r + 1} \right)^2} \hfill \\
So: \hfill \\
{\left( {{r^2} + r + 1} \right)^2} = 0 \hfill \\
Or: \hfill \\
{r^2} + r + 1 = 0 \hfill \\
\end{gathered} \]

Answer 8

Oh wait, I think I see..

Answer 9

Wolfram's saying to complete the square. Tricky.

Answer 10

It is the same, we will get the same answers

Answer 11

Computing.......... :)

Answer 12

I would use the quadratic formula

Answer 13

Ah, guh! Of course! Sometimes I get so deep in to the hard stuff that I forget the fundamentals.

Answer 14

though you are left with taking the square root of a complex number, which looks painful

Answer 15

I don't think it'll be as bad as the way I was trying to do it.

Answer 16

You can't imagine how much I was looking forward to feeling sheepish over finally seeing the one little thing that was holding me up. :D

Answer 17

Thank you both!

Answer 18

So the vector space of your solution is generate by the set of the subsequent functions:
\[\left\{ {\exp \left( {\frac{{ - 1 + i\sqrt 3 }}{2}} \right),\;\exp \left( {\frac{{ - 1 - i\sqrt 3 }}{2}} \right),\;x\exp \left( {\frac{{ - 1 + i\sqrt 3 }}{2}} \right),\;x\exp \left( {\frac{{ - 1 - i\sqrt 3 }}{2}} \right)} \right\}\]

Answer 19

it looks like after finding the roots of r^2 + r +1=0
we switch to polar coords, we can use
\[ \sqrt{ a \exp(i \theta)} = \sqrt{a} \exp\left(i \frac{\theta}{2} \right) \]

Answer 20

And now Euler's Formula kicks in.

Answer 21

yes, then back to rectangular to get the posted answers

Answer 22

Wow, tricky. I guess that's what I get for doing it ahead of the lecture.

Answer 23

Ok, it's going to take me a bit to make sure I understand what's going on in Euler's in polar coords. I'm reading it in the text now.