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I'll state the question from my textbook below:

If $x (1+y)^{1/2} + y(1+x)^{1/2} = 0$, for, $-1<x<1$, prove that

$\frac{dy}{dx} = - \frac{1}{(1+x)^2}$

Here's how I tried proving it:

$x (1+y)^{1/2} + y(1+x)^{1/2} = 0$

Rearranging and squaring we get:

$x^2(1+y) = y^2(1+x)$

Simplifying and factorizing we get:

$(x+y+xy)(x-y) = 0$

Case I:

$x+y+xy = 0$

From here we get the desired equation. Since the proof is irrelevant to my question, I'll skip it. But I note that if I replace the value of $y$ I get from this equation is in harmony with the original equation given in the question.

Case II:

$x-y = 0$

Clearly in this case,

$\frac{dy}{dx} = 1$

This is where the problem arises. Isn't this a valid answer too? Probably not, because the value of $y$ I get from this equation satisfies the original equation only for $x = -1,0$. Why is this? Here are the questions I'm looking answer for:

1) Is something wrong with squaring the equation?

2) Is there a mathematical reason to reject $x-y = 0$ as a possibility?

I am also interested in alternative methods to prove the required equation. So if you have any, feel free to post them as an answer. Any help would be appreciated.

Shinsekai no Kami
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    Your equation has the form $A+B=0$. To get to the form $A^2-B^2=0$ you multiplied it by $A-B=0$ (which you disguised as squaring $A=-B$), adding with it solutions that were not solutions of the original one. –  Apr 15 '18 at 14:20
  • See https://math.stackexchange.com/questions/2727358/if-x-sqrt1yy-sqrt1x-0-find-y – lab bhattacharjee Apr 15 '18 at 14:21
  • @XanderHenderson Finding the derivative is not the point of my question. Is it still considered duplicate? – Shinsekai no Kami Apr 15 '18 at 14:37
  • @SamInuyashaANMF Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Apr 19 '18 at 19:28

3 Answers3

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On putting $y=x$, left side of equation is always positive or negative. Its satisfied only for $(0,0)$. While squaring, there may be some extra conditions introduced, which may not be satisfied by original equation.

As an example, consider the simple straight line $y=x$, squaring will give $y^2 = x^2$ or $y=\pm x$ which is a pair of straight lines.

On the other hand, if you take all things to one side so that there is $0$ on the other side, squaring is justified. That will not introduce any new roots. $y-x = 0 \iff (y-x)^2 = 0$

jeea
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We reject $x-y=0$ because then $y = x$. Now we plug this back into the original equation and find it is not a solution because, as you've stated, we don't get a true statement for all x in (-1, 1). That is the part that is being overlooked by you. The potential solution arising from $x -y =0$ does not satisfy the hypotheses of the original problem and therefore is not a solution.

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Let try to plug in $x=y$ in the original equation to verify that that solution doesn't work.

This means that by squaring we have obtained an extra solution to discard.

user
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