Hi

$\displaystyle \displaystyle y(1+x^2)y'-x(1=y^2)=0\\$

$\displaystyle \displaystyle $$\frac{y}{1+y^2}$$dy=$$\frac{x}{1+x^2}$$dx $

$\displaystyle \displaystyle \int\frac{y}{1+y^2}dy =\int\frac{x}{1+x^2}dx $

$\displaystyle \displaystyle \ln|1+y^2|^\frac{1}{2}=\ln|1+x^2|^\frac{1}{2}+C$

Stuck. I could use a hint as to how the constant C ends up with a natural log.

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

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?