Question

Euler Cauchy equation problem.

Answer 1

Not sure what it means to complex solutions. I assumed it meant the general solution, however that is asked in the next part of the question...

Answer 2

However, I went on try to solve it using the quadratic formula, the characteristic question is \[\lambda ^{2}-4 \lambda+7=0\]

Answer 3

so the \(\lambda\)'s are complex numbers

Answer 4

You mean like 1 +/- 12i?

Answer 5

Assuming if that is what you meant, that's what I worked it out to be.

Answer 6

\[\lambda=\frac{-(-4)\pm \sqrt{(-4)^2-4\times(1)\times(7)}}{2(1)}\]\[\lambda=\frac{4\pm \sqrt{16-28}}{2}\]\[\lambda=\frac{4\pm \sqrt{-12}}{2}\]\[\lambda=2\pm3i\]

Answer 7

Would the answer not be 2 +/- 6i

Answer 8

i.e -12/2?

Answer 9

did you get -12 in the square root?

Answer 10

Yes, I made stupid error by letting b = -2 than -4.


I"m just taking about at the end of your final answer.

Answer 11

\[\lambda=\frac{4\pm \sqrt{-12}}{2}\]\[=\frac 42\pm\frac{\sqrt{-1}\times\sqrt4\times\sqrt 3}{2}\]\[=2\pm\frac{i\times2\times\sqrt 3}{2}\]\[=2\pm \sqrt 3i\]

Answer 12

Ahhh, I see. It's been a while since I worked with complex numbers

Answer 13

so what have you got as solutions?

Answer 14

Let me type this out

Answer 15

It's kinda awkward to type out and I fail at latex!

Answer 16

I am not sure about the whole in terms of \[t^ {\lambda}\] thing

Answer 17

\[\lambda=\mu \pm i\nu\]
\[y(x)=Ae^{\mu x}(A \cos \nu x+B \sin \nu x )\]

Answer 18

Is that not the general solution?

Answer 19

hmm yeah that would be the general solution

Answer 20

i shouldn't have put that first \(A\) before the \(e\)

Answer 21

i guess the general solution is the sum of two solution sets

Answer 22

one where B=0, the other where A=0

Answer 23

ok lets do this as roots are already found by Unklerhaukus

Answer 24

That's the thing, the next part of this question which I haven't uploaded asks for that. Hence it why I confused.

Looking at some text book I found, complex solutions are written in a particular form. like this

\[\lambda _{1,2} =2 \pm 3i = e ^{(2 \pm 3i)lnx}= e ^{2lnx}e ^{\pm 3i lnx}=x ^{2}[Cos(3lnx) \pm iSin(3lnx)]\]

Answer 25

ah i see i found y(t) and called it y(x)

Answer 26

I am so confused hahaha.

Answer 27

Here's the question in it's entire form

Answer 28

\[y(t)=e^{2 t}(A \cos \sqrt 3 t+B \sin \sqrt 3 t )\]
\[t=\ln x\qquad e^t=x\]
\[y(x)=x^2\left[A\cos \left(\sqrt 3 \ln x\right)+B\sin\left(\sqrt 3 \ln x \right)\right]\]

Answer 29

im getting all my letters mixed up

Answer 30

@UnkleRhaukus it should be x(t) not y(t) bcz dependent variable is x also the substitution e^t=x would not work because x is now dependent variable u can use some other variable so that at the end you can use ln(t) thanks!

Answer 31

yeah is it x(t)

Answer 32

@ironictoaster in my response it is solved completely except the roots which already found above

Answer 33

Yeah I was looking through it there, makes sense. I guess I need more practice on these type of question, thanks guys!