Question

Hello, can someone check variation of parameters (2nd order, DE) problem?

Answer 1

there.

Answer 2

I entered \(x\) there. It is an accidental mistake, it was supposed to be \(\theta\). Just ignore this, and consider ever \(x\) I used in this document to be \(\theta\).

Answer 3

https://www.youtube.com/watch?v=fKP-qFY-L9g

Answer 4

http://prntscr.com/a948bo

Answer 5

now, I am trying to figure out what exactly I did incorrectly
to get \(-\cos\ln|\cos\theta|\) instead of \(+\cos\ln|\cos\theta|\).

Answer 6

just do it on wolfram

Answer 7

redo ur wronskins

Answer 8

I haven't heard this term really. Well, heard of it, but the way I seen this worked out is using Cramer's rule.

Answer 9

Basically, that restriction for \(v_1'y_1+v_2'y_2=0\), and
the \(v_1'y_1'+v_2'y_2'=f/a\) which is just a direct
consequence of plugging \(y_p\) into the initial equation.

I have shown my work in the document, how I solved for
\(\displaystyle v_1'\) and \(\displaystyle v_2'\), and integrated, after which I just substituted
them into \(y_p\). and then the general solution was easy.

I am curious to know what was wrong about my application of Cramer's Rule.

Answer 10

do u know the proof to cramers rule

Answer 11

No

Answer 12

I only know (well, I think I know,) how to apply this rule.

Answer 13

and the only ones that I tried is a 2×2 matrix. This is my first time/

Answer 14

\(~\)

Answer 15

https://www.desmos.com/calculator/setu61b8xc
it seems as tho, that they are the same, or just about the same thing.

Answer 16

the only explanation i can come up with, is that really for any choice of \(c_1\) and \(c_2\) in my version of solution, you can find an alternative choice of \(c_1\) and \(c_2\) such that the graphs are exactly the same, since the constants may vary because they are arbitrary.

Well, this is probably incorrect, but this is my first impression. Otherwise, I don't know what my error is,.

Answer 17

I mean an alternative choice of \(c_1\) and \(c_2\) in the "correct" solution, such that ...