# differential equation with separable with Separable Variables

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?

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.
$$\ln|{x}|(\sqrt{1+y^2}+y)=C$$
• Ok thank you for the answer. But I still don't know from here is : $ln | x |$ ? Instead should not be : $| ln(x) |$ ? – Krzysztof Michalski Apr 12 '17 at 23:49
• To maintain logarithmic domain, it is important that $x>0$ thus i think $\ln |x|$ is justified. – The Dead Legend Apr 12 '17 at 23:52