Question

Find the general solution of the differential equation
3ln(y)y′−ty=0

Answer 1

last one!

Answer 2

do you happen to know the integrating factor method?

Answer 3

unless it has a different name, no.

Answer 4

this is a separable diffeq...that is easy to integrate

Answer 5

so far i'm at
\[\frac{ln(y)}{y}dy=\frac{t}{3}dt\]

Answer 6

yep

Answer 7

zarkon is right, use u substitution on ln(y)

Answer 8

\[\frac{ln(y)^2}{2}=\frac{t^2}{6}\]

Answer 9

+c

Answer 10

oh ty

Answer 11

awesome :)

Answer 12

\[ln(2y)=\frac{t^2}{3}+C\]

Answer 13

?

Answer 14

edit: \[2ln(y)\]

Answer 15

no

Answer 16

only if it was \[\ln(y^2)=2\ln(y)\]

but it is not

Answer 17

ah

Answer 18

just take the sqrt of both sides

Answer 19

\[ln(y)=(sqrt{\frac{t^2}{3}+C})\]

Answer 20

sqrt syntax?

Answer 21

\sqrt{}

Answer 22

ok

Answer 23

this is getting harder and harder to type lol
\[y = e^{\sqrt{\frac{t^2}{3}+C}}\]

Answer 24

now you see why i like to write it out on paper lolol

Answer 25

lol

Answer 26

lol!

Answer 27

lol!

Answer 28

Hes a catch. The general solution is different if y is initially greater than or less than 1. what does that mean?

Answer 29

when you took the sqrt...you introduced two solution + and -

so it was possible the ln was negative

Answer 30

lol!

Answer 31

is it pos/neg on the sqrt?

Answer 32

yes..depending on the initial condition

Answer 33

lol!

Answer 34

bingo.

Answer 35

lol!

Answer 36

Thanks guys! good night.

Answer 37

lates :P

Answer 38

do you have the answer from the book?