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$\begingroup$

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

$$\frac{\partial M}{\partial y}=\sqrt{1-y^2}-\frac{-2y^2}{2\sqrt{1-y^2}}=\frac{1-2y^2}{\sqrt{1-y^2}}$$

$$\frac{\partial N}{\partial x}= \sqrt{1-y^2}$$

$$\frac{M_{y}-N_{x}}{M}=\frac{\frac{1-2y^2}{\sqrt{1-y^2}}-\sqrt{1-y^2}}{y\sqrt{1-y^2}}=\frac{-y}{1-y^2}=h(y)$$

$$I=\exp{\left(-\int h(y) dy\right)}= \exp{\left(-\int\frac{-y}{1-y^2}dy\right)}=\exp{\left(\frac{-\ln|1-y^2|}{2}\right)}$$

Is there a way to simplify $$\exp{\left(\frac{-\ln|1-y^2|}{2}\right)}\;?$$

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    according to wolfram alpha it simplifies to sqrt |1-y^2| but i dont know why – 2017-01-31
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    yes, by combining $e^{ln|x|} = ln|x|$ and $a ln(x) = ln(x^a)$. – 2017-01-31
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    We have that $\text{ln}(a^x) = x\text{ln}(a)$ so you can bring the denominator in de logarithm to obtain a square root and use that the logarithm is the inverse of the exponential function. – 2017-01-31
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    didn't you forgive a minus in power $I$. – 2017-01-31
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    @gbox I hope you don't take offense from my edit of your question. It was just very difficult to read the tiny rendering. Of course, you are free to roll back to your original post. – 2017-01-31
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    @MyGlasses Sorry, Edited – 2017-01-31
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    @amWhy offended? not at all, I am learning so much from this site! – 2017-01-31
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    Now $$I=\frac{1}{\sqrt{1-y^2}}$$ – 2017-01-31

2 Answers 2

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$a\ln b=\ln b^a\implies -\frac{1}{2}\ln \left |1-y^2\right |=$ $\ln \frac{1}{\sqrt{1-y^2}}$ and $e^{\ln u}=u$ integrating factor is going to be $\frac{1}{\sqrt{1-y^2}}$

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    So we can say that $-\frac{1}{2}ln(x)=ln(\frac{1}{\sqrt{x}})$? – 2017-01-31
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    @gbox exactly, it's just how fractional/negative exponents are – 2017-01-31
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Apart from your question you can solve this equation by this way as well

$$(y\sqrt { 1-y^{ 2 } } )dx+(x\sqrt { 1-y^{ 2 } } +y)dy=0\\ ydx+xdy+\frac { ydy }{ \sqrt { 1-{ y }^{ 2 } } } =0\\ d\left( xy \right) -\frac { 1 }{ 2 } d\left( \sqrt { 1-{ y }^{ 2 } } \right) =0\\ d\left( xy-\frac { \sqrt { 1-{ y }^{ 2 } } }{ 2 } \right) =0\\ xy-\frac { \sqrt { 1-{ y }^{ 2 } } }{ 2 } =C\\ \\ $$