Question

Initial Value Problem: dy/dx=y/ln y; y(0)=e
I got it all wrong and used integration by parts. Please help...

Answer 1

is it \[y'=\frac{y}{\ln y}\]

Answer 2

y'=y/ln(y)
dy/dx=y/ln(Y)
ln(y)/ydy=dx

Answer 3

NOw you can see the d/dy(ln(y) is sitting in there so its simply 1/2(ln(y))^2=x+C
after integrating both sides
(ln(y))^2=2x+C
ln(y)=SQRT(2x+C)
y=e^SQRT(2x+C)
e=e^(sqrt(0+c)
c=1
y=e^SQRT(2x+1)
do you agree? james wanna help out?

Answer 4

It seems steps were left out. From the beginning, what happened to the dx?

Answer 5

ok want me to go line by line with you?

Answer 6

Please, so frustrated here.

Answer 7

haha alright jus relax these are confusing at 1st.

ok y'=dy/dx do you agree

Answer 8

If there was no dx, yes. But there is, so what happened to it?

Answer 9

an example y=x^2 the y' or the derivative of y with respect to x =2x OR dy/dx=y's

Answer 10

ok

I guess it would make sense if it was d/dy=, rather than dy/dx. I don't see x's role.

Answer 11

when you see y' program it in your brain to see dy/dx

Answer 12

so dy/dx=y/ln(y) do you agree up to there

Answer 13

yes

Answer 14

ok dy/dx=y/ln(y)
dy(ln(y))=ydx yes?
ln(y)/ydy=dx yes?

Answer 15

now thiis should look familiar as a separable equation

Answer 16

The x disappeared, and is back again?

Answer 17

no, i simply multiplied the dx to the other side and divided by ln(y)/y to get all the dy and y's on one side and isolate the x

Answer 18

isolated the DX

Answer 19

ok I did that. Still confused how y' would equal dy/dx. Seems like it would equal just dy

Answer 20

youll see in the end why its dy/dx and not just dy

Answer 21

ok so we agree up to ln(y)/ydy=dx?

Answer 22

So if you mult. bo sides by dx, wouldn't that leave you with just dy on left side?

Answer 23

yes dy on the other side but then i multiplies the right side by the inverse of y/ln(y) so it would so away on the right and appear on the left with the dy as ln(Y)/y

\[\int\limits_{}^{}\ln(y)/ydy=\int\limits_{}^{}dx\]
then we integrate both sides

Answer 24

ok. Did that, not sure how the x is magic, but ok.

Answer 25

just take it as truth right now which it is.

so can you tell me what the integral of both sides is?

Answer 26

[ln y]^2/2=x+c

Answer 27

CORRECT!
now can you isolate the y?

Answer 28

I just plugged 0 in for y for IVP and got 1/2=x+c

Answer 29

:)

Answer 30

no you need to isolate y before you plug in your y(0)

Answer 31

so 1/2(ln(y))^2=x+c
(ln(y))^2=2x+C
can you finish it?

Answer 32

sqrt both sides, now I have sqrt(2x+c). Where does the IVP come into play here?

Answer 33

Should I raise both sides to base e? Or am I ready to plug in 0 for y?

Answer 34

no because you still have ln(y)=y
so you need to e^SQRT(2x+C)

Answer 35

ln(Y)=SQRT(2x+C) i meant

Answer 36

NOW YOU HAVE y=e^SQRT(2x+C)
now in calc one and 2 the teach said take the derivative of this function
so you took the derivative of y w.r.t x
dy/dx?

Answer 37

I have y on left by iteself now. I plug in 0 for y, and e on right for IVP. I get 0=e to some power..impossible.

Answer 38

yes so Y(x)=e^SQRT(2x+C)
y(0)=e=e^sqrt(0+C)
now solve for c

Answer 39

y is on left side, so I put 0 in for it. I have e^something equalling the zero on left?

Answer 40

no y(0)=e
you have to have the e^SQRT(2x+C)=y(0) which is e

Answer 41

:)

Answer 42

e=e^sqrt(0+C)
take ln of both sides
1=SQRT(0+C)
1^2=C
C=1
so now your final answer is y=e^SQRT(2x+1)

Answer 43

How did you get 1 on the left side? It was y(0) before.

Answer 44

I had e^ (ln y), so it turned into just y. Now it is just 1?

Answer 45

your IVP states y(0)=e which means y(0) must =e on the right hand side
so you need to solve for C to satisfy this

Answer 46

so any IVP set you y(0)= to your solved eq and then solve for C

Answer 47

because when you have y(x)=e^SQRT(2x+c)
when x=0 you get e thats the point of it

Answer 48

when c=1

Answer 49

It looks good to me, thank you!!

Answer 50

i am happy i helped you, have a great day

Answer 51

You too. It's nice to see there are patient, friendly people out there