A differential equation

**Tau** · December 9th 2008, 01:23 AM

Hello

$xy' + y^2 = 1$ , $y(1) = \frac{1}{3}$

My thinking:

Rearrange the equation and you will get this:

$\frac{1}{1 - y^{2}}\delta y = \frac{1}{x} \delta x$

Integrate it...

$\frac{1}{2}(\ln|1 - y| + \ln|1 + y|) = \ln|x| + C$

I think that the absolutes can be ignored, in that case I get:

$y = \pm \sqrt{C - x^2}$

But the correct f(x) is something very different.

Thanks for your help.

**Krizalid** · December 9th 2008, 02:01 AM

Originally Posted by Tau

But the correct f(x) is something very different.

Probably 'cause it was set $C=\ln|C|$ for being constant.

**Tau** · December 9th 2008, 02:15 AM

The "answer" to this problem is:

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

According to the book.

Don't ask me what it means or in what context it is a "answer".

**mr fantastic** · December 9th 2008, 02:42 AM

Originally Posted by Tau

Hello

$xy' + y^2 = 1$ , $y(1) = \frac{1}{3}$

My thinking:

Rearrange the equation and you will get this:

$\frac{1}{1 - y^{2}}\delta y = \frac{1}{x} \delta x$

Integrate it...

$\frac{1}{2}({\color{red} - } \ln|1 - y| + \ln|1 + y|) = \ln|x| + C$ Mr F says: The mistake is in this line. I've added the correction (in red).

I think that the absolutes can be ignored, in that case I get:

$y = \pm \sqrt{C - x^2}$

But the correct f(x) is something very different.

Thanks for your help.

After the correction it follows that

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

$\Rightarrow \frac{1 + y}{1 - y} = B x^2$

where $B = e^{2C}$ is also completely arbitrary.

At this stage I'd substitute $y(1) = \frac{1}{3}$ to get the value of B. Then I'd substitute the value for B and make y the subject.

The book's answer is correct.

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