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Question

# Show that the family of curves for which $\frac{dy}{dx}$ = $\frac{{x}^{2}+{y}^{2}}{2xy}$, is given by ${x}^{2}-{y}^{2}=Cx$

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Solution

## The given differential equation is $\frac{dy}{dx}=\frac{{x}^{2}+{y}^{2}}{2xy}$ .....(1) This is a homogeneous differential equation. Putting y = vx and $\frac{dy}{dx}=v+x\frac{dv}{dx}$ in (1), we get $v+x\frac{dv}{dx}=\frac{{x}^{2}+{v}^{2}{x}^{2}}{2v{x}^{2}}\phantom{\rule{0ex}{0ex}}⇒v+x\frac{dv}{dx}=\frac{1+{v}^{2}}{2v}$ $⇒\frac{1+{v}^{2}}{2v}-v=x\frac{dv}{dx}\phantom{\rule{0ex}{0ex}}⇒\frac{1-{v}^{2}}{2v}=x\frac{dv}{dx}\phantom{\rule{0ex}{0ex}}⇒\frac{2v}{1-{v}^{2}}dv=\frac{dx}{x}$ Integrating on both sides, we get $\int \frac{2v}{1-{v}^{2}}dv=\int \frac{dx}{x}\phantom{\rule{0ex}{0ex}}⇒\int \frac{-2v}{1-{v}^{2}}dv=-\int \frac{dx}{x}\phantom{\rule{0ex}{0ex}}⇒\mathrm{log}\left(1-{v}^{2}\right)=-\mathrm{log}x+\mathrm{log}C\phantom{\rule{0ex}{0ex}}⇒\mathrm{log}\left(1-{v}^{2}\right)+\mathrm{log}x=\mathrm{log}C$ $⇒\mathrm{log}\left(1-{v}^{2}\right)x=\mathrm{log}C\phantom{\rule{0ex}{0ex}}⇒\left(1-{v}^{2}\right)x=C\phantom{\rule{0ex}{0ex}}⇒\left(1-\frac{{y}^{2}}{{x}^{2}}\right)x=C\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-{y}^{2}=Cx$ Thus, the family of curves for which $\frac{dy}{dx}$ = $\frac{{x}^{2}+{y}^{2}}{2xy}$ is given by ${x}^{2}-{y}^{2}=Cx$.

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