Question

Show that the family of curves for which $\frac{dy}{dx}=\frac{x^2+y^2}{2xy}$, is given by $x^2-y^2=CX$.differential equation is :

Answer 1

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

Put $y=mx$

$\Rightarrow \frac{dy}{dx}=m+x\frac{dm}{dx}$

$\Rightarrow m+x\frac{dm}{dx}=\frac{x^2+m^2{x}^2}{2x.mx}$

$\Rightarrow m+x\frac{dm}{dx}=\frac{m^2+1}{2m}$

$\Rightarrow x\frac{dm}{dx}=\frac{1-m^2}{2m}$

$\Rightarrow \frac{2mdm}{1-m^2}=\frac{dx}{x}$

$\Rightarrow \int \frac{-d(1-m^2)}{1-m^2}=\int \frac{dx}{x}$

$\Rightarrow -ln(1-m^2)=lnx+c_1$

$\Rightarrow x(1-m^2)=c$

$\Rightarrow x(1-\frac{y^2}{x^2})=c$

$\Rightarrow x^2-y^2=cx$.