Question

The form of the differential equation of the central conics ${\text{ax}}^2+b{y}^2=1$ is

• A. $\text{x = y (}\frac{\mathrm{dy}}{\mathrm{dx}})$
• B. $\mathrm{x}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2+\mathrm{xy}\left(\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right)-\mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=0$
• C. $\mathrm{x}+\mathrm{y}\left(\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right)=0$
• D. None of these

Answer 1

The correct option is B

$\mathrm{x}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2+\mathrm{xy}\left(\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right)-\mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=0$

Form a differential equation of the given central conic

Given, ${\text{ax}}^2+b{y}^2=1$

Now, differentiate with respect to $x$

$$\begin{gathered} \Rightarrow 2\mathrm{ax}+2\mathrm{by}\frac{\mathrm{dy}}{\mathrm{dx}}=0 \\ \Rightarrow \mathrm{ax}+\mathrm{by}\frac{\mathrm{dy}}{\mathrm{dx}}=0 \\ \Rightarrow \frac{-\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{y}}{\mathrm{x}}\frac{\mathrm{dy}}{\mathrm{dx}}...\left(\mathrm{I}\right) \end{gathered}$$

Again differentiating with respect to $x$ ,

$$\begin{gathered} \Rightarrow \mathrm{a}+\mathrm{b}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2+\mathrm{by}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=0 \\ \Rightarrow \frac{-\mathrm{a}}{\mathrm{b}}={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2+\mathrm{y}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2} \\ \Rightarrow \mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\mathrm{x}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2+\mathrm{xy}\left(\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right) \\ \Rightarrow \mathrm{x}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2+\mathrm{xy}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}-\mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=0 \end{gathered}$$

Hence, the form of the differential equation of the central conics ${\text{ax}}^2+b{y}^2=1$ is $\mathrm{x}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2+\mathrm{xy}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}-\mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=0$.