Question

Really Tricky Separable Equation
dy/dx=e^(x+y)/(y-1)

Answer 1

\[\frac{ dy }{ dx }=\frac{ e ^{x+y} }{ y-1 }\]

Answer 2

Recall your rule for exponents:\[\Large\bf\sf e^{x+y}\quad=\quad e^x\cdot e^y\]

Answer 3

Do you think you can separate it from there? :o

Answer 4

ln'ing both sides?

Answer 5

what i did was i divided both sides by \[\frac{ e ^{y} }{ y-1 }\] but it gets messy

Answer 6

Yah that seems to be a good approach.\[\Large\bf\sf \frac{dy}{dx}\quad=\quad \frac{e^x\cdot e^y}{y+1}\qquad\to\qquad (y+1)e^{-y}\;dy\quad=\quad e^x\;dx\]

Answer 7

from there integration by parts?

Answer 8

Hmmm ya looks like Parts on the left side.

Answer 9

give me 10000 secs please

Answer 10

lol :3

Answer 11

Oh it was (y-1) :P sorry bout the typo before.

Answer 12

(-y-1)e^-y + e^-y

Answer 13

Hmm I think you should be getting:\[\Large\bf\sf -(y-1)e^{-y}-e^{-y}\]

Answer 14

is that with the y+1 error?

Answer 15

sorry

Answer 16

No I used (y-1) for the factor :o

Answer 17

u=(y-1)
dv=e^{-y}

Answer 18

wait also shouldn't it be respect to x? not y??

Answer 19

im lost im sorry

Answer 20

We're integrating with respect to y, that's what the differential `dy` is telling us.

Answer 21

\[\Large\bf\sf \int\limits(y+1)e^{-y}\;dy\quad=\quad \int\limits e^x\;dx\]So this is where we're at?

Answer 22

So the right side gives us:\[\Large\bf\sf \int\limits\limits(y+1)e^{-y}\;dy\quad=\quad e^x+c\]

Answer 23

yes

Answer 24

should it be y-1 again?

Answer 25

gomenasai

Answer 26

Ohh I copy pasted from the start, my bad lolol

Answer 27

thats cool i still get it

Answer 28

\[\Large\bf\sf \int\limits\limits\limits(y-1)e^{-y}\;dy\quad=\quad e^x+c\]
\[\Large\bf\sf u=(y-1), \qquad\qquad\qquad dv=e^{-y}\;dy\]\[\Large\bf\sf du=\;dy,\qquad\qquad\qquad\qquad v=-e^{-y}\]

Answer 29

\[\Large\bf\sf uv-\int\limits v\;du \qquad\to\qquad (y-1)\left(-e^{-y}\right)-\int\limits \left(-e^{-y}\right)dy\]

Answer 30

1sec im doing it on paper

Answer 31

oaky you're right

Answer 32

Simplify things down a tad:
\[\Large\bf\sf -(y-1)e^{-y}-e^{-y}\quad=\quad -ye^{-y}\]So we have:\[\Large\bf\sf -ye^{-y}\quad=\quad e^{x}+c\]
\[\Large\bf\sf y e^{-y}\quad=\quad C-e^x\]

Answer 33

Which ... unfortunately doesn't have an explicit form that we can get it into :(

Answer 34

:)))))))))))))))))))) so that must be the end of the problem then

Answer 35

as a differential equation professor would you be acceptable of that answer?

Answer 36

Yes unless you were given initial data that you can plug in to solve for c :)

Answer 37

For Diff Eq? Yah that's a fine answer.
There are some weird things that would allow you to write it terms of x, but not anything that you'll learn in DE.

Answer 38

this is just homework, but i worry, that my prof will be dissatisfied and give me an F

Answer 39

bless, thank you very much for you're help!

Answer 40

Np c: hopefully professor is a nice guy lol