Question

ODE: y' + y tan(x) = cos(x)

another devious ploy im sure ;)

Answer 1

Hello!

Answer 2

howdy

Answer 3

Im gona answer.

Answer 4

ill brace myself

Answer 5

no giving hints james lol

Answer 6

unless they are cryptic of course

Answer 7

Have you learnt integrating factors yet? If so, find it! That's maybe cryptic.

Answer 8

y'(x)+y(x) tan(x) = cos(x) or y'[x] + y[x] Tan[x] == Cos[x]

Answer 9

first order linear - use integrating factor?

Answer 10

i know the e^int(...) stuff yes

Answer 11

If you do, this ODE is fun and quite easy.

Answer 12

IF = e^INT(tanx dx)

Answer 13

good, what can we do with that?

Answer 14

well, simplify

Answer 15

...by actually evaluating the expression.

Answer 16

essentially i believe we are turning it into a reversal of the product rule

Answer 17

yes, that's the idea of the integrating factor. So stop stalling and just do it. :-)

Answer 18

ok, if we can turn this into a x'y + xy' by utilizing the e^x stuff; multiply this contraption by: \(e^{\int tan(x)dx}\)
\[y'e^{ln|sec(x)|}+tan(x)\ e^{ln|sec(x)|}=cos(x)\ e^{ln|sec(x)|}\]
\[ye^{ln|sec(x)|}=\int cos(x)\ e^{ln|sec(x)|}dx\]

Answer 19

since e^(ln|sec(x)|) = sec(x) ...

Answer 20

\[ye^{ln|sec(x)|}=\int cos(x)\ sec(x)dx\]
\[ye^{ln|sec(x)|}=\int \frac{cos(x)}{cos(x)}dx\]
\[ye^{ln|sec(x)|}=\int dx\]
\[ye^{ln|sec(x)|}=x\]
right? so far

Answer 21

forgot a +C as always tho

Answer 22

\[ye^{ln|sec(x)|}=x+C\]
\[y=\frac{x+C}{e^{ln|sec(x)|}}\]
\[y=\frac{x+C}{sec(x)}\]
\[y=x\cos(x)+C\cos(x)\] maybe

Answer 23

Yep, that's it.

Answer 24

yay!! i knew i could do it ... again lol

Answer 25

taking diffy qs next term

Answer 26

I'd just say that you can simplify the integrating factor to sec x much sooner rather than later.

Answer 27

you can, but wheres the excitement in that

Answer 28

Elegance is exciting.

Answer 29

i dropped a "y" at the start but i picked it up in the end