Question

(y+1)dx+(y-1)(1+x^2)dy = 0

Answer 1

the most frustrating separate variable one in my honest opinion. I did this like 10 times and got stuck

Answer 2

Is this the original equation?

Answer 3

yeah. I'm gonna get an attachment to this

Answer 4

I'd think this is the original equation:\[
(y-1)(1+x^2)y'=-(y+1)
\]

Answer 5

Separating variables\[
-\frac{y-1}{y+1}y'=\frac{1}{1+x^2}
\]

Answer 6

Integrating both sides with respect to \(x\): \[
\int -\frac{y-1}{y+1}y'dx=\int \frac{1}{1+x^2}dx
\]Remember that \(dy=y'dx\)

Answer 7

it's the y that's driving me nutssss...isn't an integral from the tables of integrals or something?

Answer 8

\[
u=y+1,\quad y-1=u-2, \quad du=dy
\]

Answer 9

where did that come from?

Answer 10

simple subsitution?

Answer 11

You want to get rid of an ugly denominator.

Answer 12

My thought process: any linear substitution is cheap and easy.

Answer 13

then I get a 1/u if u = y+1

Answer 14

\[
\int\frac{u-2}{u}\;du
\]