Differential equation: solving for the General Solution

[jenniferAOI]

Homework Statement

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

Homework Equations

Separating the variables I arrive with:
dy/sqrt(1-y^2) = dx/sqrt(1-x^2)

By integration on both sides by trigonometric substitution and putting it in a general solution:
Arcsin y - Arcsin x = C

The Attempt at a Solution

If I want to simplify more. Would it be right if I write:
Arcsin (y-x) = C
Multiply sin on both sides:
y-x = C would be my final answer.

 

[Mark44]

Homework Statement

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

Homework Equations

Separating the variables I arrive with:
dy/sqrt(1-y^2) = dx/sqrt(1-x^2)

This isn't quite right. Here's an example that is similar to what you did:
x² = 4
Taking the square root of both sides, we get x = 2.
Do you see what's wrong with that?

By integration on both sides by trigonometric substitution and putting it in a general solution:
Arcsin y - Arcsin x = C

The Attempt at a Solution

If I want to simplify more. Would it be right if I write:
Arcsin (y-x) = C

No, this is wrong. Arcsin(y) - Arcsin(x) ≠ Arcsin(x - y)

You can't "distribute" across arcsin(), any more than $\sqrt{a} + \sqrt{b} = \sqrt{a + b}$. I hope this looks wrong to you.

Multiply sin on both sides:
y-x = C would be my final answer.

 

[Ray Vickson]

Homework Statement

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

Homework Equations

Separating the variables I arrive with:
dy/sqrt(1-y^2) = dx/sqrt(1-x^2)

By integration on both sides by trigonometric substitution and putting it in a general solution:
Arcsin y - Arcsin x = C

The Attempt at a Solution

If I want to simplify more. Would it be right if I write:
Arcsin (y-x) = C
Multiply sin on both sides:
y-x = C would be my final answer.

I see four separate DEs here:
(1) $dy/\sqrt{1-y^2} = dx/\sqrt{1-x^2},$ with $x^2, y^2 < 1$.
(2) $dy/\sqrt{1-y^2} = - dx/\sqrt{1-x^2},$ with $x^2, y^2 < 1$.
(3) $dy/\sqrt{y^2-1} = dx/\sqrt{x^2-1},$ with $x^2, y^2 > 1$.
(4) $dy/\sqrt{y^2-1} = -dx/\sqrt{x^2-1},$ with $x^2, y^2 > 1$.

 

[epenguin]

Anyway if you just look at the question

(dy/dx)² = (1 - y²)/(1 - x²) ...(1)

or in general any equation f(x) dx = f(y) dy (yes I know there is more to it in this case)

then it is I think obvious that x = y is a solution.

Is it obvious that x = -y is also solution to (1)? Can you argue this? Always try to get conclusions by minimal arguments if you can.

The first solution you give reduces to this when C = 0 .

Does your solution really work for any other C? Does it have the right symmetry?

Whether there are other solutions to those as well as those outlined I cannot see in my present state of tiredness. But if there are, they would have to have certain symmetries.

 

[jenniferAOI]

No, this is wrong. Arcsin(y) - Arcsin(x) ≠ Arcsin(x - y)

You can't "distribute" across arcsin(), any more than √a+√b=√a+ba+b=a+b\sqrt{a} + \sqrt{b} = \sqrt{a + b}. I hope this looks wrong to you.

Thank you. I get it.

however,

↑

Homework Statement

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

Homework Equations

Separating the variables I arrive with:
dy/sqrt(1-y^2) = dx/sqrt(1-x^2)
This isn't quite right. Here's an example that is similar to what you did:
x2 = 4
Taking the square root of both sides, we get x = 2.
Do you see what's wrong with that?

I don't see it. Sorry. Can you explain it further? I do not see anything wrong with it, though.

 

[jenniferAOI]

No, this is wrong. Arcsin(y) - Arcsin(x) ≠ Arcsin(x - y)

So,
arcsin (y) - arcsin (x) = C

would now be my final general solution, right?

 

[Mark44]

(dy/dx)^2 = (1-y^2) / (1-x^2)
Separating the variables I arrive with:
dy/sqrt(1-y^2) = dx/sqrt(1-x^2)

Here's an example that is similar to what you did:
x² = 4
Taking the square root of both sides, we get x = 2.
Do you see what's wrong with that?

I don't see it. Sorry. Can you explain it further? I do not see anything wrong with it, though.

The equation x² = 4 has two solutions: x = 2, x = -2. Taking the square root of both sides gives you only one of the solutions.

You did essentially this same thing in going from your first equation above to the second one you show.
Ray Vickson's and epenguin's comments are related to the idea in my example.

 

[epenguin]

I have recovered my senses enough to be clear that y = ± x is the solution only for the particular integration constant 0. Your solution is a general one. Well part of it, I think you haven't yet spelt out the general solution totally, as has been pointed out.

At the same time this seems to me a classical case of "when you have an answer it's not finished" (Polya).

Firstly what do the solutions even look like? I think it is quite hard to tell from your formulae. But you might be able to sketch them from the differential equation itself. And also if you have a curve drawing program you can use that. We already know that everything is going to happen with x and y in the range -1 to 1.

You can also get the solution in a more intelligible form y = ...
Try it.

The practice and habit of seeing a solution through to the end with an active approach will turn out useful for the mathematical parts, and not only those, of your chemical engineering studies.