Question

id i do this correctly
solve the differential equation
dy/dx= (y-x)/(y+x)

Answer 1

my answer was 1/2(y/x)^2+(y/x)=lnx+c

Answer 2

replacing v=y/x

Answer 3

no dice ?

Answer 4

Ok ok i finally got it.
Sorry a bit rusty, had to go through it step by step.

Answer 5

its cool

Answer 6

Umm I came up with something quite a bit different, let's see what's going on here..

Answer 7

ok

Answer 8

\[\Large\rm v=\frac{y}{x}\qquad\to\qquad vx=y\]Differentiating with respect to x gives us,\[\Large\rm y'=v'x+v\]--------------------
Plugging in all of the pieces,\[\Large\rm v'x+v=\frac{v-1}{v+1}\]Look ok so far?

Answer 9

ur right i think i kn where i went wrong

Answer 10

Combining the fractions maybe? :O
Very easy place to mess up.

Answer 11

yea

Answer 12

If you're able to do your separation of variables correctly,
then before integrating, you should get:\[\Large\rm \frac{v+1}{v^2+1}~dv=-\frac{1}{x}~dx\]

Answer 13

how did you get the v^2+1 in the bottom not thinking or over thinking lol

Answer 14

I skipped a bunch of steps, wasn't sure if you wanted to see them all :) lol

Answer 15

\[\Large\rm v'x+v=\frac{v-1}{v+1}\]subtracting v from each side, then looking for a common denominator gives us:\[\Large\rm v'x=\frac{v-1}{v+1}-\frac{v(v+1)}{v+1}\]

Answer 16

Distributing the negative v to each term in the second fraction,\[\Large\rm v'x=\frac{v-1}{v+1}+\frac{-v^2-v}{v+1}\]Combine the fractions,\[\Large\rm v'x=\frac{v-1-v^2-v}{v+1}\]Cancel the v's,\[\Large\rm v'x=\frac{-v^2-1}{v+1}\]Err I guess we don't want the negative distributed, that doesn't really help us,\[\Large\rm \frac{dv}{dx}x=-\frac{v^2+1}{v+1}\]

Answer 17

Then separate variables,

Answer 18

so thats fine if we divide over and leave the negative with the X variable

Answer 19

Yah it ends up working better if we leave the negative with the x side since we'll have to cut the v's into two fractions to integrate them.

Answer 20

for the integral we use the by parts or can i use U subs

Answer 21

\[\Large\rm \frac{v+1}{v^2+1}~dv=-\frac{1}{x}~dx\]Can be written as,\[\Large\rm \left(\frac{v}{v^2+1}+\frac{1}{v^2+1}\right)~dv=-\frac{1}{x}~dx\]

Answer 22

The first fraction is a simple u-substitution.
The second fraction should look familiar!! :)

Answer 23

today is not math day for me

Answer 24

yea got it right now thx for ur help

Answer 25

cool