Differentiate This

**ugkwan** · November 6th 2010, 06:41 PM

Hi

$\displaystyle y(1+x^2)y'-x(1=y^2)=0\\$
$\displaystyle $$\frac{y}{1+y^2}$$dy=$$\frac{x}{1+x^2}$$dx$
$\displaystyle \int\frac{y}{1+y^2}dy =\int\frac{x}{1+x^2}dx$
$\displaystyle \ln|1+y^2|^\frac{1}{2}=\ln|1+x^2|^\frac{1}{2}+C$
Stuck. I could use a hint as to how the constant C ends up with a natural log.

$\displaystyle Ans: 1+y^2=C(1+x^2)$

BTW: How does one indent to the next line without closing the line with [/tex] and repeating a new line with [tex]. Is there a shortcut to save time?

**pickslides** · November 6th 2010, 06:57 PM

Originally Posted by ugkwan

$\displaystyle \ln|1+y^2|^\frac{1}{2}=\ln|1+x^2|^\frac{1}{2}+C$
Stuck. I could use a hint as to how the constant C ends up with a natural log.

You can say $C = \ln{c}$ then add logs on the RHS.

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