Question

the differetial equation dy/dx=(x^2-y^2)/2xy is

Answer 1

Answer:

$x^{3}=\frac{c}{1-3t^{2} }$

Step-by-step explanation:

Given ,   $\frac{dy}{dx}=\frac{x^{2}-y^{2} }{2xy}$

              $\frac{dy}{dx}=\frac{1- [\frac{y}{x}]^{2} }{2[\frac{y}{x}] }$

            Let y/x=t , then dy/dx= t +x(dt/dx)

substituting,   $t+x\frac{dt}{dx} =\frac{1-t^{2} }{2t}$

After solving,    $\frac{dt}{dx} =\frac{1-3t^{2} }{2xt}$

                        $\int\frac{2t}{1-3t^{2} }dt =\int\frac{dx}{x}$

                         $\frac{ln(1-3t^{2}) }{-3}=ln(x) +c$   (c is integral constant)

                      $ln(x) +\frac{ln(1-3t^{2}) }{3}+c =0$

                     $ln[(x^{3})*(1-3t^{2})]=c$

                      $x^{3}=\frac{c}{1-3t^{2} }$