Question

nonlinear diff eq: dy/dx = y(y-1)/cos(x)

Answer 1

can you figure out how to use separation of variables?

Answer 2

1/cos(x) dx = 1/y(y-1) dy

Answer 3

then integrate and solve for y

Answer 4

yea i got it down to y^2/6(2y-3)=ln(sec(x)+tan(x) +c but i'm not sure how to solve for y from here

Answer 5

oh, you don't solve for y anymore

Answer 6

sorry: ln(y) + ln(y-1) = ln(sec(x)+tan(x) + c*

Answer 7

solve for c

Answer 8

but, your integral, are you sure it's correct?

Answer 9

I'm getting very very close but not exactly

Answer 10

@housey

Answer 11

sorry just got it down to: ln(y-1) - ln(y) = ln|sec(x)+tan(x)| +c, Initial Condition: y(0) = 1/2

Answer 12

uhm well I'm still not getting that exactly, let's check partial fractions on those ys

Answer 13

so Integral(1/y^2-y) dy is the same as integral (-1/y) + integral (1/y-1) using partial fractions...?

Answer 14

I got the opposite

Answer 15

i got -1/y and 1/y-1

Answer 16

sorry nvm dyslexic

Answer 17

haha all good

Answer 18

ok so now, we don't solve for y anymore, we just solve for C

Answer 19

when i used the IC it came out as c = 0, see what u get

Answer 20

or whatever you like to call the constant

Answer 21

y(0)=1/2 right?

Answer 22

yeah

Answer 23

and cause it's 5 am y(0) is the same as y=0 x=0 right?

Answer 24

so ln|-1/2| - ln|1/2| = ln|sec(0)+tan(0)|. meaning 0=ln(1) + c so 0= 0 + 0

Answer 25

nah as in, when x=0, y= 1/2 :)

Answer 26

love a bit of calculus at 5am

Answer 27

ok sorry sped moment

Answer 28

oh goodness

Answer 29

no stress

Answer 30

man I need some sleep haha

Answer 31

watch your subs on the y

Answer 32

should be ln|-2|+ln|1/(1/2-1)|

Answer 33

ends up being ln|-2|+ln|2| right?

Answer 34

not from where i'm sitting haha..: ln|0.5-1|-ln(05) ?
so ln|-0.5| which equals ln(0.5) ?

Answer 35

well ln|-2| but doesn't matter

Answer 36

no it's not a minus

Answer 37

where did you get the minus from

Answer 38

we have ln|-1/y|+ln|1/(y-1)| no?

Answer 39

from partial fractions? ln(y-1) - ln(y)

Answer 40

we just agreed on the opposite I thought

Answer 41

oh ignore that but still it's twoes

Answer 42

not .5

Answer 43

and oh god I need sleep

Answer 44

yes you are correct

Answer 45

100% correct

Answer 46

haha sweet as, just gotta express it in terms of linear y now i guess. cheers, i reckon youve earnt some z's :)

Answer 47

yes I get 0 then provided you did the calculation of ln|tanx+secx| right

Answer 48

no, you just set it = to C

Answer 49

for real, in differential eq. you don't solve for y anymore

Answer 50

so the answer is 0= blah blah blah

Answer 51

real talk, thanks for the help. espec considering the time of day haha

Answer 52

but if you want, combine the logs nd then exp both sides

Answer 53

yea s'what i did, just for aesthetics haha

Answer 54

lnx-lny=ln(x/y) alrighty, night then! well morning I guess

Answer 55

man I'm scatterbrained. Have fun with your dif!

Answer 56

yea thanks enjoy your hard earned rest!