Question

solve DE dy/dx= [(y-x)e^(y/x)]/[xe^(y/x)+x] where y=vx

Answer 1

v = (e^v (v x-x))/(e^v x+x)

Answer 2

\[\frac{ dy }{ dx}=\frac{ (y-x)e ^{\frac{ y }{ x }} }{ xe ^{\frac{ y }{ x }}+x}\]

Answer 3

i am getting A=X(-e^v-v)

Answer 4

Can we please work through it together

Answer 5

Ok start working and then tell me what step u stuck on broham

Answer 6

ok we are given y=vx therefore dy/dx=x(dv/dx)+v

Answer 7

we sub dy/dx and y in the given equation

Answer 8

\[x \frac{ dv }{ dx}+v=\frac{ (v-1)e ^{v} }{ e ^{v}+1 }\]

Answer 9

is that right so far?

Answer 10

That's what I've got too..

Answer 11

i further simplified it \[x \frac{ dv }{ dx }=\frac{ -e ^{v}-v }{ e ^{v}+1}\]

Answer 12

by taking v across to the other side

Answer 13

Yes..

Answer 14

Hmmm, might wanna get the expressions on the other sides through division.

Answer 15

now separate the variables

Answer 16

collect like terms\[\frac{ 1 }{ x} dx=\frac{ e ^{v}+1 }{ -e ^{v}-v }dv\]

Answer 17

I hate when people do it like that, but yeah, that's right.

Answer 18

\[x \frac{ dv }{ dx }=\frac{ -e ^{v}-v }{ e ^{v}+1}\]\[x \frac{ dv }{ dx }=-\frac{ e ^{v}+v }{ e ^{v}+1}\]\[\frac{ e ^{v}+1}{ e ^{v}+v }dv =-\frac{dx}{x}\]\[\frac{1}{ e ^{v}+v }d( e ^{v}+v ) =-\frac{dx}{x}\]

Answer 19

i don"t understand the last step on the LHS what did you do?

Answer 20

He did a u substitution, \(u = e^v+v\)

Answer 21

but instead of writing the result as \(du/u \) he kept the expression as \(e^v+v\)

Answer 22

\(du = (e^v+1) dv\) if you're wondering what happened to it.

Answer 23

yep got it

Answer 24

so intergrating it u get . .

Answer 25

logs

Answer 26

\[In (e ^{v}+v)=-Inx+c\]

Answer 27

And sub. x=y/x into that.

Answer 28

*v=y/x

Answer 29

do i nedd to futher take the exp of both sides and express it in terms of A the arbitary constant or leave it as logs?

Answer 30

You're trying to find \(y\)

Answer 31

No log :\

Answer 32

@wio it would be difficult to factor o

Answer 33

i mean impossible to factor out y

Answer 34

I think you can remove the log first, then sub. v=y/x into what you get. Since we need y, not v.

Answer 35

I mean we need the equation in terms of x and y, not x and v.

Answer 36

i got to this form
\[\ln\left(\frac{e^{x/y}}x+\frac1y\right)=c\]

Answer 37

from In(e^ v +v)=−Inx+c
take the Exp both sides get
e^v+v=A/x

Answer 38

then sub v for

Answer 39

y/x

Answer 40

Yup, continue~~

Answer 41

I dont't understand where @UnkleRhaukus is getting those logs

Answer 42

in the end i am getting xe^(y/x)+y=A

Answer 43

A or c depending on what u choose as the arbitary constant

Answer 44

\[ln (e ^{v}+v)=-lnx+c\]\[ln (e ^{v}+v)+lnx=c\]\[ln [(e ^{v}+v)x]=c\]\[ln [(e ^{\frac{y}{x}}+\frac{y}{x})x]=c\]\[ln (xe ^{\frac{y}{x}}+y)=c\]\[(xe ^{\frac{y}{x}}+y)=c\] \[ln (e ^{v}+v)=-lnx+c\]\[(e ^{v}+v)=cx^{-1}\]\[(e ^{\frac{y}{x}}+\frac{y}{x})=cx^{-1}\]\[xe ^{\frac{y}{x}}+y=c\]

Answer 45

It's not possible to simplify further, is it?

Answer 46

We have a recursive equation.

Answer 47

i donno how u would factor out the y

Answer 48

(i see i made a mistake)

Answer 49

I think just leaving it here is okay ._.!

Answer 50

Thank you very much

Answer 51

Thanks too!!