Question

The differential equation representing the family of curves $y^2=2c\left(x+\sqrt{c}\right)$, where $c$ is a positive number, is of:

• A. order $1$, degree $3$
• B. order $2$, degree $2$
• C. degree $3$, order $3$
• D. degree $4$, order $4$

Answer 1

The correct option is A

order $1$, degree $3$

Explanation for correct option

Calculating the degree and order of a differential equation

Given, the equation is:

$$\begin{gathered} y^2=2c\left(x+\sqrt{c}\right) \\ \Rightarrow y^2=2cx+2c^{\frac{3}{2}}....\left(1\right) \end{gathered}$$

On differentiating we get,

$$\begin{gathered} 2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{c} \\ \Rightarrow \mathrm{c}=\mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)....\left(2\right) \end{gathered}$$

Substituting equation $\left(2\right)$ in equation $\left(1\right)$:

$$\begin{gathered} {\mathrm{\gamma }}^2=2\mathrm{xy}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)+2{\left(\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{\frac{3}{2}} \\ \Rightarrow {\left[{\mathrm{\gamma }}^2-2\mathrm{xy}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right]}^2=4{\mathrm{y}}^3{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^3 \end{gathered}$$

Now, we know that order is the highest differential equation and degree is the highest degree of the order.

Therefore, Order $=1$ and Degree $=3$

Hence, the correct answer is Option (A).