Question

Integrate:

(dy)/(dx)= (xy)/(x^2-1)

Answer 1

here is the question as written from Diff EQ,
verify that x^2+cy^2=1 where c is an arbitrary nonzero constant, is one parameter family of implicit solutions to
\[\frac{ dy}{ dx }=\frac{ xy }{ x^2-1 }\]
So,
I found that the original equation is
\[y=\sqrt{\frac{ 1-x^2 }{ c }}\]

Answer 2

So how do I begin to integrate from here, It has been a while.

Answer 3

Would y be a arbitrary number and be able to be brought out of the integral?

Answer 4

When you differentiate \(y\) did you get \(dy/dx\)?

Answer 5

The original problem you do not have to differentiate.
Its basically proving it.

Answer 6

It has already been done.

Answer 7

So what exactly is the question asking for? \( \int y dx\)?

Answer 8

Here is where I am at with the integral
\[\int\limits_{}^{}\frac{ xy }{ x^2-1 } \frac{ dy }{ dx}\]
\[Let u=x^2-1 \]
\[du=2x dx\]

Answer 9

Thats why I don't know if I can take the y out of the integral and solve. I have not used this calculus in quite some time.

Answer 10

If you do a \(u\) sub like that, then you'd have to find \(y\) in terms of \(u\).

Answer 11

If I wanted to solve for \(y\), I would separate the function.

Answer 12

the equation I mean.

It's a separable equation.

Answer 13

I need to show the original problem
\[x^2+cy^2=1\]

Answer 14

Okay, you can either use implicit differentiation, or you can just separate the equation and integrate.

Answer 15

With algebra
\[y=\sqrt{\frac{ 1-x^2 }{ c }}\]

Answer 16

Okay sorry, but you have totally confused me now.
You mention integration, you have a derivative, but you only want to use algebra...?

1) What is given exactly?
2) What do we need to prove/show exactly?

Answer 17

problem written as above. Section is Solutions and Initial Values Problems.

Answer 18

How about integration by parts, where \(u = y\)?

Answer 19

and \(dv = x/(x^2-1)\)?

Answer 20

@nestor18 did you try it?

Answer 21

working it had to refresh on ibp

Answer 22

Why are we trying to integrate in one stroke a separable differential equation?

\(\int \dfrac{1}{y}\;dy = \int\dfrac{x}{x^{2}-1}\;dx\)

Answer 23

I am not sure, but when I mentioned separation, @nestor18 didn't want to do it.

Answer 24

@wio Fair enough. I see that on re-reading.
@nestor18 Simplify your life. Solve as a separable differential equation. Seriously.

Answer 25

I am so rusty with this stuff, I forget all rules for integration. This makes more sense now.

Answer 26

\(\ln|y| = \dfrac{1}{2}\ln\left(|x^{2}-1|\right) + C\)

Show us the rest.