You visited us 1 times! Enjoying our articles? Unlock Full Access!

Orthogonal Trajectories

The different...

Question

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

\frac{dy}{dx} = - \frac{y}{2 x}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{dy}{dx} = - \frac{2 x}{y}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{dy}{dx} = \frac{2 x}{y}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{dy}{dx} = \frac{y}{x}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{dy}{dx} = - \frac{2 x}{y}$
The given differential equation is $\frac{dy}{dx} = \frac{y}{2 x}$ . The slope of the tangent of given family of curves is $\frac{y}{2 x}$ . So, the orthogonal trajectory will have slope $\frac{- 1}{(\frac{y}{2 x})}$ as $m_{1} m_{2} = - 1$ .
Thus the differential equation of the required orthogonal trajectory is $\frac{dy}{dx} = - \frac{2 x}{y}$ .

Suggest Corrections

Similar questions

\frac{dy}{dx} = \frac{y}{2 x} .

\frac{d y}{d x} + \frac{y}{x}

\frac{d y}{d x} + y = \frac{1 + y}{x} is_____________________.

\frac{d^{2}}{{dx}^{2}} (\frac{dy}{dx}) + 2 \frac{d^{2} y}{{dx}^{2}} + \frac{dy}{dx} = sinx

\frac{d y}{d x} + 3 y = e^{m x}

\frac{d y}{d x} - y = \cos 2 x

x \frac{d y}{d x} - y = (x + 1) e^{- x}

x \frac{d y}{d x} + y = x^{4}

(x \log x) \frac{d y}{d x} + y = \log x

\frac{d y}{d x} - \frac{2 x y}{1 + x^{2}} = x^{2} + 2

\frac{d y}{d x} + y \cos x = e^{\sin x} \cos x

(x + y) \frac{d y}{d x} = 1

\frac{d y}{d x} \cos^{2} x = \tan x - y

e^{- y} \sec^{2} y d y = d x + x d y

x \log x \frac{d y}{d x} + y = 2 \log x

x \frac{d y}{d x} + 2 y = x^{2} \log x

Join BYJU'S Learning Program