ODE: y' + y tan(x) = cos(x) another devious ploy im sure ;)

Hello!

howdy

Im gona answer.

ill brace myself

no giving hints james lol

unless they are cryptic of course

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

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

first order linear - use integrating factor?

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

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

IF = e^INT(tanx dx)

good, what can we do with that?

well, simplify

...by actually evaluating the expression.

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

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

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

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

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

forgot a +C as always tho

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

Yep, that's it.

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

taking diffy qs next term

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

you can, but wheres the excitement in that

Elegance is exciting.

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

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