Discussion Forum

The solution of $\frac{dy}{dx}=\frac{x^{2}+y^{2}+1}{2xy}$ , satisfying y(1)=0 is given by 

Answer:

Explanation:

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

 $\Rightarrow$   $2xy dy =(x^{2}+1) dx+y^{2}dx$

 $\Rightarrow$    $\frac{xd(y^{2})-y^{2}dx}{x^{2}}=\left(\frac{x^{2}+1}{x^{2}}\right)dx$

  $\Rightarrow$   $\int  d\left(\frac{y^{2}}{x}\right)=\int \left(1+\frac{1}{x^{2}}\right)dx$

$\Rightarrow$   $\frac{y^{2}}{x}=x-\frac{1}{x}+C$

$\Rightarrow$   $y^{2}=(x^{2}-1+Cx)$

where x=1;y=0

$\Rightarrow$     0=1-1+C

$\Rightarrow$  C=0

$\therefore$  The solution is $x^{2}-y^{2}=1$ i.e, hyperbola