Question

The differential equation of all straight lines passing through origin is

• A. $\mathrm{y}=\sqrt{\mathrm{x}}\left(\raisebox{1ex}{$\mathrm{dy}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{dx}$}\right.\right)$
• B. $\raisebox{1ex}{$\mathrm{dy}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{dx}$}\right.=\mathrm{y}+\mathrm{x}$
• C. $\raisebox{1ex}{$\mathrm{dy}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{dx}$}\right.=\mathrm{y}-\mathrm{x}$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option:

Finding the differential equation:

Given that all straight lines are passing through the origin.

The equation is,

$\mathrm{y}=\mathrm{kx}\left(\mathrm{k}=\mathrm{constant}\right)...\left(i\right)$

On differentiating the above equation we get,

$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{k}$

On substituting ‘$k$’ in equation $\left(i\right)$, we get

$$\begin{gathered} y=\left(\frac{dy}{dx}\right)x \\ \frac{dy}{dx}=\frac{y}{x} \end{gathered}$$

$\frac{dy}{dx}=\frac{y}{x}$is the required differential equation

Hence, option (D) is the correct answer.