Question

The differential equation whose solution is ${\mathrm{Ax}}^2+{\mathrm{By}}^2=1$, where $\mathrm{A}$ and $\mathrm{B}$ are arbitrary constant is of:

• A. first-order and second degree
• B. first order and first degree
• C. second-order and first degree
• D. second-order and second degree

Answer 1

The correct option is C

second-order and first degree

Explanation for correct option

Finding the differential equation when the solution is given

Given, the solution of the differential equation is $A{x}^2+B{y}^2=1$
On differentiating with respect to $\mathrm{x}$ we get,
$$\begin{gathered} 2\mathrm{Ax}+2\mathrm{By}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=0 \\ \Rightarrow \frac{\mathrm{y}}{\mathrm{x}}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{A}}{\mathrm{B}} \end{gathered}$$
Differentiating again we get,
$$\begin{gathered} \frac{\mathrm{x}\left\{\mathrm{y}\left(\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right)+{\left({\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}}\right)}^2\right\}-\mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{{\mathrm{x}}^2}=0 \\ \Rightarrow \mathrm{xy}\left(\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right)+\mathrm{x}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2=\mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \end{gathered}$$

We know that,
Order = Highest derivative of equation
Degree = Highest degree of order

Therefore, order $=2$

Degree = $=1$
Hence, the correct answer is Option (C).