Question

Diff EQ: (x+y)y' = x-y

I'm guessing I need to substitute y=vx but I dont see what that gets me

Answer 1

If all is Ok, then you should have something like that,
\[y=vx\Rightarrow y'=v'x+v\Rightarrow y'=\frac{x-vx}{x+vx}\\
v'x+v=\frac{x-vx}{x+vx}\Rightarrow v'x=\frac{(v-1)^2}{1+v}\]Then,
\[\int{\frac{1+v}{(v-1)^2}}dv=\int xdx\] And solve it. At the end you can substitute v=y/x.

Answer 2

thanks I'm working it now

Answer 3

I have a mistake in one step, it should be,
\[v'x=\frac{1-2v-v^2}{1+v}\Rightarrow\int{\frac{1+v}{1-2v-v^2}}dv=\int{\frac{1}{x}dx}\]

Answer 4

oh thank goodness --- that -v^2 was causing me trouble

Answer 5

;)

Answer 6

ok now i see how that got inverted - v' = dv/dx....is that common for these kinds of problems?

Answer 7

Yes, it is usual.

Answer 8

You mean,
\[v'=\frac{dv}{dx}\]

Answer 9

yes and then to get the dv on the side of the equation with the other v's you end up with 1/dv...

Answer 10

sorry so slow that integration should be simpler.....ah factor out the -2 then the v+1 cancels and the u sub can work

Answer 11

Yes, at the end, you need a side with dv and the other with dx, in order to get the integrals.

Answer 12

I don't understand the term "cancels" in the way you use it.

Answer 13

If the u substitution is u=-v^2 - 2v +1 then du=-2v-2 then dv= 1/-2v-2 but the other term in the integral is (v+1)....could not figure out (v+1)/(-2v-2) for a moment or two

Answer 14

Ok, so you obtain the integral as,
\[-\frac{1}{2} \ln \left(1-v^2-2 v\right)\]

Answer 15

thats what I have yes = ln x +c

Answer 16

Perfect ;)

Answer 17

then i looks like my book gets rid of the -1/2 by multiplying by -2 both sides then exp for both sides and solve for c then substitute out the v's...ok cool thanks for getting me started....not sure how this becomes a general process but got the first one

Answer 18

Hope it goes well. ;)