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Mathematics 16 Online
OpenStudy (anonymous):

Euler Cauchy equation problem.

OpenStudy (anonymous):

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...

OpenStudy (anonymous):

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

OpenStudy (unklerhaukus):

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

OpenStudy (anonymous):

You mean like 1 +/- 12i?

OpenStudy (anonymous):

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

OpenStudy (unklerhaukus):

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

OpenStudy (anonymous):

Would the answer not be 2 +/- 6i

OpenStudy (anonymous):

i.e -12/2?

OpenStudy (unklerhaukus):

did you get -12 in the square root?

OpenStudy (anonymous):

Yes, I made stupid error by letting b = -2 than -4. I"m just taking about at the end of your final answer.

OpenStudy (unklerhaukus):

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

OpenStudy (anonymous):

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

OpenStudy (unklerhaukus):

so what have you got as solutions?

OpenStudy (anonymous):

Let me type this out

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (unklerhaukus):

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

OpenStudy (anonymous):

Is that not the general solution?

OpenStudy (unklerhaukus):

hmm yeah that would be the general solution

OpenStudy (unklerhaukus):

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

OpenStudy (unklerhaukus):

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

OpenStudy (unklerhaukus):

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

OpenStudy (anonymous):

ok lets do this as roots are already found by Unklerhaukus

OpenStudy (anonymous):

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

OpenStudy (anonymous):

OpenStudy (unklerhaukus):

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

OpenStudy (anonymous):

I am so confused hahaha.

OpenStudy (anonymous):

Here's the question in it's entire form

OpenStudy (unklerhaukus):

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

OpenStudy (unklerhaukus):

im getting all my letters mixed up

OpenStudy (anonymous):

@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!

OpenStudy (unklerhaukus):

yeah is it x(t)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

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