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

1. Give the general equation of the curves that satisfy this equation.

2. Show that the straight lines y=1 and y=-1 are also solutions.

3. Do any of the curves you found in 1) intersect y=1?

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

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- Nov 4th 2006, 08:12 AM413Separable equations
dy/dx= (y^2 -1)/x

1. Give the general equation of the curves that satisfy this equation.

2. Show that the straight lines y=1 and y=-1 are also solutions.

3. Do any of the curves you found in 1) intersect y=1?

i started by

dy/(y^2 -1)= dx/x - Nov 4th 2006, 12:20 PMtopsquark
1. $\displaystyle \frac{dy}{dx} = \frac{y^2 - 1}{x}$

Thus:

$\displaystyle \frac{dy}{y^2-1} = \frac{dx}{x}$

We aren't given points on the solution curve, so do indefinite integrals:

$\displaystyle \int \frac{dy}{y^2-1} = \int \frac{dx}{x}$

The RHS is easy: $\displaystyle \int \frac{dx}{x} = ln|x| + C'$.

For the LHS use the method of partial fractions: $\displaystyle y^2 - 1 = (y+1)(y-1)$, so:

$\displaystyle \frac{1}{y^2 - 1} = \frac{A}{y+1} + \frac{B}{y-1}$ for some constant A, B.

$\displaystyle \frac{1}{y^2 - 1} = \frac{A(y-1) + B(y+1)}{y^2 - 1} = \frac{(A+B)y + (-A + B)}{y^2 - 1}$

So

$\displaystyle A + B = 0$

$\displaystyle -A + B = 1$

Any way you solve it, I get A = -1/2 and B = 1/2.

Thus the LHS from above is:

$\displaystyle \int \frac{dy}{y^2-1} = \int \left ( \frac{-\frac{1}{2}}{y+1} + \frac{\frac{1}{2}}{y-1} \right ) dy$ = $\displaystyle -\frac{1}{2} ln|y+1| + \frac{1}{2} ln|y-1| + C''$

Equating the LHS and the RHS:

$\displaystyle -\frac{1}{2} ln|y+1| + \frac{1}{2} ln|y-1| + C'' = ln|x| + C'$

$\displaystyle ln \sqrt{ \left | \frac{y - 1}{y + 1} \right | } = ln|x| + C$ where C = C' - C'' is a new integration constant.

Thus

$\displaystyle \sqrt{ \left | \frac{y - 1}{y + 1} \right | }= a|x|$ where $\displaystyle a = e^C$ stands as the new integration constant.

This is kind of ugly as it stands. But if we square both sides we get rid of the ugly absolute values. The trouble is that once we do this we might be adding extra solutions to the equation. We need to check these at the end.

$\displaystyle \left | \frac{y - 1}{y + 1} \right | = a^2|x|^2$

$\displaystyle \left | \frac{y - 1}{y + 1} \right | = a^2x^2$

Square again:

$\displaystyle \frac{(y-1)^2}{(y+1)^2} = a^4x^4$

$\displaystyle (y-1)^2 = a^4x^4(y+1)^2$ Expanding and moving everything to the LHS:

$\displaystyle (1 - a^4x^4)y^2 - (2 + 2a^4x^4)y + (1 - a^4x^4) = 0$

This looks terrible, but is simply a quadratic in y. So use the quadratic formula:

$\displaystyle y = \frac{(2+2a^4x^4) \pm \sqrt{(-2-2a^4x^4)^2 - 4(1-a^4x^4)}}{2(1 - a^4x^4)}$

After some simplification:

$\displaystyle y = \frac{1 + a^4x^4 \pm 2a^2x^2}{(1 - a^2x^2)(1+a^2x^2)} = -\frac{(a^2x^2 \pm 1)^2}{(a^2x^2-1)(a^2x^2 +1)}$

So

$\displaystyle y = -\frac{a^2x^2+1}{a^2x^2-1}$ or $\displaystyle y = -\frac{a^2x^2-1}{a^2x^2+1}$

As it happens, both solutions check in the original differential equation.

So pick one, say $\displaystyle y = -\frac{a^2x^2+1}{a^2x^2-1}$. (The choice between the two is merely a matter of picking an "a" value.)

-Dan

PS I just wanted to be kind of geeky and say that I really enjoyed solving this one. I haven't seen one quite like this before. :) - Nov 4th 2006, 12:29 PMtopsquark
2.$\displaystyle y = -\frac{a^2x^2+1}{a^2x^2-1}$

This one is a little dicier. The y = 1 seems easy enough:

If we set a = 0 then y = 1. The problem is that "a" can't be zero. We had earlier defined $\displaystyle a = e^C$ where C is an arbitrary constant. Now, if we let $\displaystyle C \to -\infty$ then $\displaystyle a \to 0$ and then y = 1 as a limit.

We can "set" $\displaystyle a \to \infty$ and then y = -1 appears as another limiting solution. This one is about as bizzare as the other one since what we are really doing is letting $\displaystyle C \to \infty$.

(Someone check me on this: These are "envelope" solutions? The y = +/- 1 solutions don't seem to be envelopes to me, but the limiting behavior is at least similar.)

-Dan - Nov 4th 2006, 12:34 PMtopsquark
$\displaystyle y = -\frac{a^2x^2+1}{a^2x^2-1}$

We are trying to find at least one "a" such that:

$\displaystyle 1 = -\frac{a^2x^2+1}{a^2x^2-1}$

$\displaystyle a^2x^2 - 1 = -a^2x^2 -1 $

We MUST choose an "a" such that the x's cancel out here. The only "a" value that will do is a = 0:

$\displaystyle 0^2x^2 - 1 = -0^2x^2 -1 $

$\displaystyle -1 = -1$

Since this is true, y for a = 0 will intersect the line y = 1.

BUT, note that a = 0 gives y(x) = 1. So I would say that the answer is "yes" because y = 1 solves the differential equation, but "no" because this isn't really an intersection as much as it is an identity.

-Dan - Nov 4th 2006, 02:59 PMtopsquark
Oops! Been drinking too much "firewater" again I guess. I screwed this one up.

We are trying to find at least one "x" (not a!) such that:

$\displaystyle 1 = -\frac{a^2x^2+1}{a^2x^2-1}$

$\displaystyle a^2x^2 - 1 = -a^2x^2 -1 $

$\displaystyle a^2x^2= -a^2x^2$

Thus x = 0 when a is not zero. So there is always a point of intersection with the curve y(x) and y = 1: the point (0,1). And, as pointed out before, when a = 0, y(x) = 1 so the curve IS the line y = 1.

-Dan - Nov 4th 2006, 07:14 PMThePerfectHacker
For simplicity sake, $\displaystyle x>0$.

What I cannot understand (similar to my other post) is why can you divide by $\displaystyle y^2-1$.

If you assume that $\displaystyle y^2-1\not = 0$ for any point $\displaystyle x>0$. Then you can solve it as a seperable

However, if you assume that there is a point where it is zero then you have two cases.

1)$\displaystyle y^2-1$ is always zero for any point $\displaystyle x>0$. In that case, $\displaystyle y=\pm 1$ are solutions.

2)$\displaystyle y^2-1$ is zero and non-zero at some point. Then there exists an open interval where the function is non-zero. Thus, a solution as in the first example exists for this interval. In the other interval where it is zero is something else. Thus, the full solution is the piecewise defined function between these two, but that cannot be because that would imply non-differenciability, right?? Thus, that case is impossible.

Last comment I want to make is that, $\displaystyle y=\pm 1$ is not a solution! You should say a function defined as $\displaystyle y=\pm 1$ for $\displaystyle x>0$. - Nov 5th 2006, 04:15 AMtopsquark
I'll agree that I should have been more careful in my derivation of a solution. I completely missed the $\displaystyle y = \pm 1$ problem. However, as the curves $\displaystyle y = \pm 1$ for $\displaystyle x \neq 0$ actually DO satisfy the differential equation I am not concerned. And, as I pointed out, $\displaystyle y = \pm 1$ does not really come out of my derivation since I needed to take limits of my solution to obtain them. (This is why I was wondering if they qualified as "envelopes.")

However, your point (as usual) is quite valid: The answer to the last part should be that there is no intersection of any solution curve with the line y = 1 (unless you take the solution y(x) = 1) because the differential equation stipulates that x can't be 0, and the only possible intersection point is (0, 1).

-Dan - Nov 7th 2006, 11:29 AM413
the question i have is why x can't be zero?

- Nov 7th 2006, 01:06 PMtopsquark
- Nov 7th 2006, 02:46 PM413
and also i don't really get the second part, because you guys talked about "envelope" solutions" and ThePerfectHacker said that y=1,-1 are not solutions, it really confuses me. Is there a more clear or easy way to prove that y=1 and -1 are solutions?

PS. Sorry i have so many questions. - Nov 7th 2006, 04:36 PMtopsquark
Don't worry about asking questions. That's how you learn. :)

The simplest way to show that y = (+/-) 1 are solutions is to directly substitute them into the differential equation. In the end, that is the only criterion for a function to be a solution.

What I meant by "envelopes" is this: When you plot several solutions to a differential equation in many cases you note that the solutions are bounded by one or more curves. (The solutions for different values of the arbitrary constant, that is.) These curves are called "envelopes" and tend to be solutions of the differential equation. (I want to say they always are, but I don't remember for sure.)

In the graph below I have given an example of a family of curves $\displaystyle y(x) = \sqrt{2cx-c^2}$ for several values of c (red curves.) The envelope (dotted blue line) is y = x. As you can see it touches each solution curve as a tangent. Many differential equations have such envelopes.

The solutions y = (+/-) 1 have some similarities to an envelope solution, but have differences as well. I was wondering if one of the more knowledgable mentors here could tell me if they are envelopes or not.

-Dan - Nov 7th 2006, 06:57 PMThePerfectHacker
Do not worry about that so much. My remark was based on the theory of differencial equations not how to solve them. Indeed the function $\displaystyle y=1$ is not a solution. But this implies $\displaystyle x=0$ is in its domain (it is a constant function). What you should have said is the function $\displaystyle y=\pm 1$ on any open interval excluding the origin. Then it is a solution.