Question

help me in linear diff. equation pls

Answer 1

PROBLEM IS \[y lny dx + (x - lny ) dy = 0\]

Answer 2

@myininaya

Answer 3

so i need to divide the equation by ylny y dy?

Answer 4

yes
and your integrating factor will be in terms of y instead of x

Answer 5

so \[\frac{ dx }{ dy } + \frac{ x }{ ylny } -y\]

Answer 6

\[y \ln(y) dx+(x- \ln(y)) dy=0 \\ \text{ divide both sides by } y \ln(y) dy \\ \frac{y \ln(y) dx + (x-\ln(y)) dy}{ y \ln(y) dy} =\frac{0}{y \ln(y) dy} \\ \frac{y \ln(y) dx}{y \ln(y) dy} +\frac{(x-\ln(y)) dy}{y \ln(y) dy}=0 \\ \frac{dx}{dy} +\frac{x-\ln(y)}{y \ln(y)}=0 \\ \frac{dx}{dy} +\frac{x}{y \ln(y)}-\frac{\ln(y)}{y \ln(y)}=0 \]
so you should have had =1/y instead of -y

Answer 7

i think the -y contained a type-o anyways :p

Answer 8

oops i think the lny in the left side would be cancel

Answer 9

and it should be 1/y?

Answer 10

yes -ln(y)/[y ln(y)] is just -1/y

Answer 11

yep

Answer 12

so whats our p(y) im having trouble in ln huhu

Answer 13

p(y) is thing next to x

Answer 14

so its x/ylny?

Answer 15

\[\frac{dx}{dy} +\color{Red}{\frac{1}{y \ln(y)}} x=\frac{1}{y}\]

Answer 16

well the thing next to x

Answer 17

not the thing next to x including also x

Answer 18

just the 1/(y ln(y))

Answer 19

what im confused

Answer 20

ah yes cause the x iss cancel

Answer 21

the x doesn't cancel

Answer 22

and q(y) is 1/y

Answer 23

yes

Answer 24

ah, okay

Answer 25

our if is. \[\int\limits e ^{\frac{ 1 }{ ylny }} dy\]

Answer 26

\[\frac{dx}{dy}+\frac{1}{y \ln(y)} x=\frac{1}{y} \\ \text{ when compared \to } \frac{dx}{dy}+p(y) x=q(y) \\ \text{ this tells us } p(y)=\frac{1}{y \ln(y)} \text{ and } q(y)=\frac{1}{y}\]

Answer 27

well the e should not be inside the integral

Answer 28

oops yes, typo

Answer 29

\[e^{ \int\limits p(y) dy} =e^{\int\limits \frac{1}{y \ln(y)} dy}\]

Answer 30

easy to make type-o in complicated math coding

Answer 31

hint if you need one: let u=ln(y)

Answer 32

\[e ^{\int\limits \frac{ 1 }{ ylny }} dy\]

Answer 33

so its ln ylny + c

Answer 34

did you do the sub for that integral i mentioned ?

Answer 35

oops no

Answer 36

u = lny du = lny dy dv = y v = y^2/2

Answer 37

don't know where dv and v come from
\[\int\limits \frac{1}{y \ln(y)} dy \\ u=\ln(y) \\ du=\frac{1}{y} dy \\ \int\limits \frac{1}{\ln(y) } \cdot \frac{1}{y} dy \\ \int\limits \frac{1}{u} du\]

Answer 38

so our IF is y

Answer 39

let's go slower so you know that integral above is ln|u| +C
or let's just say ln|u| because we are going to take the C as zero anyways

and u=ln(y) so the integral above is actually ln(ln(y))

what does e^(ln(ln(y)) =?

Answer 40

e^ln is 1 and e^ lny is 1/y

Answer 41

\[e^{ \ln(G(y))} =G(y) \text{ assuming } G(y) \text{ is positive }\]

Answer 42

okay thne

Answer 43

we have
\[e^{ \ln(\ln(y))}=....\]

Answer 44

\[e^\ln \] and \[e ^{lny}\]

Answer 45

\[e^{ \ln(kitty)}=kitty \\ e^{\ln(G(y))}= G(y) \\ e^{\ln(\ln(y))}=?\]

Answer 46

e^lny

Answer 47

you should say \[e^{\ln(\color{red}{\ln(y)})}=\color{red}{\ln(y)}\]

Answer 48

oops ya

Answer 49

so lny is 1/y

Answer 50

if you differentiate ln(y) yes
but why are you doing that

Answer 51

cause were finding I.F

Answer 52

we already found I.F.

Answer 53

\[I.F. =e^{ \int\limits p(y) dy}=e^{ \int\limits \frac{1}{y \ln(y)} dy} =e^{ \ln(\ln(y))}=\ln(y)\]

Answer 54

ah then the solution is \[(lny)x = \int\limits (\frac{ 1 }{ y })(lny) dy + C\]

Answer 55

yah

Answer 56

what next now

Answer 57

the integration thingy there
we need to do it

Answer 58

do a substitution

Answer 59

recognize that the derivative of ln(y) is ...

Answer 60

1/y

Answer 61

bingo

Answer 62

so let u=ln(y) and du=1/y dy

Answer 63

allright then

Answer 64

you have a pretty little integral to evaluate now

Answer 65

can 1/y be y^-1

Answer 66

yes 1/y is y^(-1) but why are you rewriting 1/y right now

Answer 67

so ln y(lny)

Answer 68

is ln^2y

Answer 69

\[lny(x) = lny(lny) + c\]

Answer 70

\[lny(x) = \ln^2y\] + c

Answer 71

ok i can agree to that!

Answer 72

oops

Answer 73

there is a little coefficient missing right?

Answer 74

2ln^2y

Answer 75

\[\int\limits \frac{1}{y} \ln(y) dy\\ u=\ln(y) \\ du= \frac{1}{y} dy \\ \int\limits u du \\=\frac{u^2}{2}+C =\frac{ \ln^2(y)}{2}+C \]

Answer 76

so you were missing a 1/2 in front of that ln^2(y) thing

Answer 77

ay yes so \[lny(x) = \frac{ \ln ^{2y} }{ 2 } + c\]

Answer 78

divide both side by 2

Answer 79

so it gives \[2x lny = \ln ^{2y} +c\]

Answer 80

well you actually multiplied both sides by 2
and you could say 2c but 2c is just still an unknown constant
so i guess you could still call it c
but i don't like to

and you mean
\[2 x \ln(y) =\ln^2(y) +k\]
or as i said i guess you can call that k c if you want

Answer 81

ah okay but is that right

Answer 82

You're doing calculus, so you're a big man and you must be extremely careful with the notation. Does the expression \(\ln^{2y}\) make any sense to you ?

Answer 83

yeah

did you want to solve for x
or is that all you need to do?

Answer 84

yes, no need to solve for x, thankyou for your help ! :)