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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