Question

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

Answer 1

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

$$\begin{gathered} v+x\frac{dv}{dx}=\frac{x^2+v^2{x}^2}{2v{x}^2} \\ \Rightarrow v+x\frac{dv}{dx}=\frac{1+v^2}{2v} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1+v^2}{2v}-v=x\frac{dv}{dx} \\ \Rightarrow \frac{1-v^2}{2v}=x\frac{dv}{dx} \\ \Rightarrow \frac{2v}{1-v^2}dv=\frac{dx}{x} \end{gathered}$$
Integrating on both sides, we get

$$\begin{gathered} \int \frac{2v}{1-v^2}dv=\int \frac{dx}{x} \\ \Rightarrow \int \frac{{\displaystyle -2v}}{{\displaystyle 1-v^2}}dv=-\int \frac{{\displaystyle dx}}{{\displaystyle x}} \\ \Rightarrow \log \left(1-v^2\right)=-\log x+\log C \\ \Rightarrow \log \left(1-v^2\right)+\log x=\log C \end{gathered}$$
$$\begin{gathered} \Rightarrow \log \left(1-v^2\right)x=\log C \\ \Rightarrow \left(1-v^2\right)x=C \\ \Rightarrow \left(1-\frac{y^2}{x^2}\right)x=C \\ \Rightarrow x^2-y^2=Cx \end{gathered}$$
Thus, the family of curves for which $\frac{dy}{dx}$ = $\frac{x^2+y^2}{2xy}$ is given by $x^2-y^2=Cx$.