Question

In need of calculus help please!!! If dy/dx = (1+x)/(xy), x > 0, and y = –4 when x =1, when x =3, y =
(attachment)

Answer 1

hmm i'm thinking of integral to recover y then evaluate

Answer 2

was this supposed to be after a section of integration? or you haven't done that yet

Answer 3

just want to see the point of those other givens

Answer 4

we have done integration, is that what you're asking? I tried to do the anti-derivative, find C, plug it all in, etc and I utterly failed

Answer 5

I think the y=-4 and x=1 is given in order to find C, I could be totally wrong though

Answer 6

hmm let me try, always try and see heheh

Answer 7

thank you!

Answer 8

have you separated the variables yet?

Answer 9

\[xydy=(1+x)dx \longleftarrow separation\]

Answer 10

\[ydy=(1+x)/x dx \]
\[y^2/2=\ln x +x+c\]

Answer 11

skipped one step of applying integration hopefully that make sense to you

Answer 12

yes - this is perfect!!!

Answer 13

solve for y:
\[y=\sqrt{2(\ln x+x+c)}\]
hmm now i see where you stuck

Answer 14

let plugging first x=1, y=-4 to find that c

Answer 15

\[16=2(1+c) \longrightarrow c=5\]

Answer 16

was the math correct check it lol :)

Answer 17

then back to our equation:
plug in c=5 and x=3

Answer 18

you should get the y

Answer 19

there is one part I would say something about

Answer 20

+ or- yes

Answer 21

y=+ or - 4.2658 ?

Answer 22

hmm did i miss something ?

Answer 23

the final solution you gave those not consider the initial condition given
(yes)
you choose the function f(x)=y where y<0 because of the intial condition that is you should have
\[y=-\sqrt{2(\ln(x)+x+c)}\]

Answer 24

oh yes i did forget that x>0 initial settings

Answer 25

when I plug that in though, the value I get isn't an answer option - am I doing something wrong?

Answer 26

it's closest to -4.711 (I think), but doesn't match...

Answer 27

hmm should have 4.711
@freckles y>0 no?

Answer 28

y is less than 0
the initial given has y is negative

anyways why is C ,5 ?

Answer 29

\[16=2(1+c) \longrightarrow c=7\]

Answer 30

oh yes rechecked it y<0
math error, i read 16 as 12 lol

Answer 31

anyways you get the idea :)

Answer 32

So I should be doing -\[\sqrt{2(\ln(3)+3+7}\]

Answer 33

I always think I see the numbers changing in front of my eyes. Like a 5 turns into a 6 or a 8 turns into a 2 or something else :p

Answer 34

to get -4.711 ?

Answer 35

heheh so that happens to you as well

Answer 36

yea

Answer 37

lol :)

Answer 38

thank you so much!!!!

Answer 39

no problem!

Answer 40

oh shoot... checking the answer options and -4.11 isn't one - can it be 4.711, −4.711?

Answer 41

4.711
or
4.711, −4.711
are my options

Answer 42

did you get 4.711 after plugging c=7 and x=3?

Answer 43

you got the same when you plugged c=5?

Answer 44

shouldn't it be different

Answer 45

no, I didn't get an answer option at all when c=5, when c=7 I get the 4.711
my answer comes out positive if I don't have the negative in front of the radical
and it comes out negative if I do have the negative in front of the radical

Answer 46

so -4.711 isn't an option alone?

Answer 47

if not maybe I was wrong and it is both solutions

Answer 48

unfortunately no :(

Answer 49

Well I think I was wrong then
it is both \[y^2=2(\ln(x)+x+c) \\ y = \pm \sqrt{2(\ln(x)+x+c)}\]

Answer 50

has to be both
if x is positive we still have no control over y
according to the equation

Answer 51

yes 4.711 , -4.711 are the correct

Answer 52

that makes sense!!

Answer 53

okay, now I'm really done hahaha- thank you both again! I have one last question on this assignment - should I make a new post to ask it?

Answer 54

http://www.wolframalpha.com/input/?i=dy%2Fdx%3D%281%2Bx%29%2F%28xy%29%2C+y%281%29%3D-4
though wolfram agrees with me

Answer 55

yessssssss

Answer 56

@SithsAndGiggles Can I have your opinion about the differential equation?

Answer 57

like
we have
\[y'=\frac{1+x}{xy} \text{ given } y(1)=-4 \\ \text{ we get } y^2=2(\ln(x)+x+C) \\ y=- \sqrt{2(\ln(x)+x+7)}\]
shouldn't we just have the negative solution and not both positive and negative solution since our initial condition says to only consider negative y values?

Answer 58

i did agree with you because of the same reason
although i had doubt is y<0 for any x>0?

Answer 59

http://tutorial.math.lamar.edu/Classes/DE/IoV.aspx
like here in example 3
they can either say y=(2/3t)^(3/2) or y=-(2/3t)^(3/2) because either of these include the initial condition
paul gives a little more behind the theorem 2 below the blue box saying we need the solution in order to determine the interval of validity
and it says y_0 cann affect the interval of validity

Answer 60

can*

Answer 61

by y_0 is the number in y(x_0)=y_0

Answer 62

the initial condition point that is