t^(-1)(dy/dt) = 2cos^2(y) ; y(0) = pi/4

Question

t^(-1)(dy/dt) = 2cos^2(y)   ; y(0) = pi/4

anonymous · Answer

yes, yes.... come my minions

anonymous · Answer

this is LGB's turn now :P

anonymous · Answer

oh, sorry, what?

lgbasallote · Answer

hmm $$\large t^{-1} \frac{dy}{dt} = 2\cos^2 y$$
$$\large \implies  \frac{dy}{tdt} = 2\cos^2y$$
so you have to cross multiply...

anonymous · Answer

sooooo:  1/(2cos^2(y))dy = tdt?

lgbasallote · Answer

...well if i were you i wont move the 2...

lgbasallote · Answer

but if you wish...that works too

anonymous · Answer

i think it is more complicated than it looks like :(((

lgbasallote · Answer

actually...it's not @sami-21 lol

anonymous · Answer

no, no, it looks easier if i leave the 2

lgbasallote · Answer

okay let's leave the 2
$$\large \implies\frac{dy}{\cos^2 y} = 2tdt$$
now...recall your trig identities...what is the value of $$\frac{1}{\cos^2 y}$$

anonymous · Answer

ooooone sec

anonymous · Answer

is this an inverse trig func?

lgbasallote · Answer

nope

lgbasallote · Answer

i think it's called reciprocal trig function or something...

lgbasallote · Answer

http://www.regentsprep.org/Regents/math/algtrig/ATT1/trigsix.htm scroll down to the part it says "reciprocal functions"

anonymous · Answer

sec^2(y)?

lgbasallote · Answer

yup.
$$\large \implies \sec^2 y dy = 2tdt$$
now integrate both sides

anonymous · Answer

you got it :)))

anonymous · Answer

dam, so i still need to integrate...

lgbasallote · Answer

me? oh stop it youuu :""">

lgbasallote · Answer

hah cant escape integration in differential equations @IStutts :p hahaha

lgbasallote · Answer

except for higher order linear de...yeah...that has few integrations

anonymous · Answer

so tan y = t^2 + c?

lgbasallote · Answer

yup

anonymous · Answer

okay, now how would I continue from there?

anonymous · Answer

or at least solve for y

lgbasallote · Answer

...dont you stop there?

lgbasallote · Answer

that's rarely done in these cases...but...if you insist...

lgbasallote · Answer

take the arctan of both sides

anonymous · Answer

so y = arctan(t^2 + c)

lgbasallote · Answer

yup

anonymous · Answer

and yea, my teacher is a jerk

lgbasallote · Answer

they all are =))

anonymous · Answer

so how would i solve for the initial value y(0) = pi/4?

lgbasallote · Answer

that means when t = 0 then y = pi/4

anonymous · Answer

so how did the back of the book get y = arctan ( t^2 + 1 )?

lgbasallote · Answer

because c = 1

anonymous · Answer

yea, that's how picky my teacher is, and we spent one day one separable equations, this section at least

anonymous · Answer

just by chance?  or is there math i a missing?

lgbasallote · Answer

y(0) = pi/4
$$\tan y = t^2 + c$$
$$\tan (\pi/4) = 0^2 + c$$
$$\1 = c$$

anonymous · Answer

....oookay

lgbasallote · Answer

you didnt get it..did you...

anonymous · Answer

i never memorized the unit circle....

lgbasallote · Answer

oh lol

lgbasallote · Answer

that's gonna be your undoing now :p mwahaha

anonymous · Answer

yea, it's sad to blame my teacher, but i can only blame myself.  in high school, my math teachers said it was rare they would use it, then in college....well, here i am...

anonymous · Answer

i see....  ....said the blind man

lgbasallote · Answer

well it's not too late now

anonymous · Answer

eh, i dont think my teacher will be that big of a douche. there will only be 2 of 8 questions from this chapter...

anonymous · Answer

true true

lgbasallote · Answer

maybe this can help you openstudy.com/updates/4f9be55ee4b000ae9ed11844

anonymous · Answer

anyways, onto the next one, i think i got 2.2 now

lgbasallote · Answer

http://openstudy.com/updates/4f9be55ee4b000ae9ed11844

anonymous · Answer

thank you