0
$\begingroup$

Hi I've solved the Differential equation:

$$ x ln(x) dy + \sqrt{ 1+y^{2}}dx =0 $$

Domain of the equations in my opinion should be $ x >0 $(becuase of logarithm right?)

I'm dividing equation by $\sqrt{1+y^{2}} \cdot x ln(x)$ and I'm obtaining equation:

$$ \frac{dy}{\sqrt{1+y^{2}}} + \frac{dx}{x ln(x)} $$

Next, because: $$ \int \frac{dy}{\sqrt{1+y^{2}}} = ln\left\lvert \sqrt{1+y^{2}} + y \right\rvert + C $$ And: $$ \int \frac{dx}{x ln(x)} = ln \left\lvert ln(x) \right\rvert + C $$

So the answer in my opinion should be:

$$\left\lvert (\sqrt{1+y^{2}} +y) + ln(x) \right\rvert = C $$

where of course $ x > 0 $

I Don't know why in book from equations come from the answer is:

$$ ln \left\lvert x \right\rvert (y+ \sqrt{1+y^{2}}) = C $$, $ x \neq 0, x \neq 1$ Is that correct answer really?

$\endgroup$
1
$\begingroup$

Error FOUND: $\int \frac{dy}{\sqrt{1+y^2}}=\ln |\sqrt{1+y^2} +y|$ as you said.. but in the next step, you your self wrote it wrong.
It comes out to be $$\\ln |\sqrt{1+y^2} +y|+\ln|\ln(x)|=ln(C)$$Please note that we can write $C$ as $\ln C$ as both are constants.
Thus this now simply gives you:
$$\ln|{x}|(\sqrt{1+y^2}+y)=C$$

Cheers.

$\endgroup$
  • $\begingroup$ Ok thank you for the answer. But I still don't know from here is : $ ln | x | $ ? Instead should not be : $| ln(x) |$ ? $\endgroup$ – Krzysztof Michalski Apr 12 '17 at 23:49
  • $\begingroup$ To maintain logarithmic domain, it is important that $x>0$ thus i think $\ln |x|$ is justified. $\endgroup$ – The Dead Legend Apr 12 '17 at 23:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.