Question

The differential equation for a family of curves is $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{y}{2x}.$ What is the differential equation for the orthogonal trajectory of the curves?

• A. $\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{y}{2x}$
• B. $\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{2x}{y}$
• C. $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2x}{y}$
• D. $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{y}{x}$

Answer 1

The correct option is B $\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{2x}{y}$

The given differential equation is $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{y}{2x}$. The slope of the tangent of given family of curves is $\frac{y}{2x}$. So, the orthogonal trajectory will have slope $\frac{-1}{\left(}$ as $m_1{m}_2=-1$.
Thus the differential equation of the required orthogonal trajectory is $\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{2x}{y}$.