Question

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

Show that the family of curves for which the slope of the tangent at any point (x, y) on it is \[\frac{x^2 + y^2}{2xy}\] is given by x² − y² = 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

\[v + x\frac{dv}{dx} = \frac{x^2 + v^2 x^2}{2v x^2}\]

\[ \Rightarrow v + x\frac{dv}{dx} = \frac{1 + v^2}{2v}\]

\[\Rightarrow \frac{1 + v^2}{2v} - v = x\frac{dv}{dx}\]

\[ \Rightarrow \frac{1 - v^2}{2v} = x\frac{dv}{dx}\]

\[ \Rightarrow \frac{2v}{1 - v^2}dv = \frac{dx}{x}\]

Integrating on both sides, we get

\[\int\frac{2v}{1 - v^2}dv = \int\frac{dx}{x}\]

\[ \Rightarrow \int\frac{- 2v}{1 - v^2}dv = - \int\frac{dx}{x}\]

\[ \Rightarrow \log\left( 1 - v^2 \right) = - \log x + \log C\]

\[ \Rightarrow \log\left( 1 - v^2 \right) + \log x = \log C\]

\[\Rightarrow \log\left( 1 - v^2 \right)x = \log C\]

\[ \Rightarrow \left( 1 - v^2 \right)x = C\]

\[ \Rightarrow \left( 1 - \frac{y^2}{x^2} \right)x = C\]

\[ \Rightarrow x^2 - y^2 = Cx\]

Thus, the family of curves for which \[\frac{dy}{dx}\] \[\frac{x^2 + y^2}{2xy}\] is given by \[x^2 - y^2 = Cx\].