## Differential Equations and Linear Algebra (4th Edition)

$\frac{dy}{dx}=\frac{2xy}{x^2-y^2}$
given that $x^2+y^2=2cx$ $c=\frac{x^2+y^2}{2x}$ differentiate the above term with respect to x $2x+2y\frac{dy}{dx}=2c$ $\frac{dy}{dx}=\frac{c-x}{y}$ $\frac{dy}{dx}=\frac{\frac{x^2+y^2}{2x}-x}{y}$ $\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$ for orthogonal trajectory $\frac{dy}{dx}=\frac{-1}{f(x,y)}$ $\frac{dy}{dx}=\frac{2xy}{x^2-y^2}$ thus,the final answer $\frac{dy}{dx}=\frac{2xy}{x^2-y^2}$