Question

The order and degree of the differential equation $\sqrt{\sin x}\left(\mathrm{d}x+\mathrm{d}y\right)=\sqrt{\cos x}\left(\mathrm{d}x-\mathrm{d}y\right)$ are

• A. $\left(1,2\right)$
• B. $\left(2,2\right)$
• C. $\left(1,1\right)$
• D. $\left(2,1\right)$
• E. $\left(0,1\right)$

Answer 1

The correct option is C

$\left(1,1\right)$

Explanation for the correct option:

Given $\sqrt{\sin x}\left(\mathrm{d}x+\mathrm{d}y\right)=\sqrt{\cos x}\left(\mathrm{d}x-\mathrm{d}y\right)$

The order of a differential equation is the highest derivative involved in the equation i.e., the $n$ in $\frac{d^{n}y}{d{x}^n}$
The degree of a differential equation is the power of the highest order of differential i.e., the $p$ in ${\left(\frac{d^{n}y}{d{x}^n}\right)}^p$

Simplifying the given equation,
$\begin{array}{cc}& \sqrt{\sin x}\left(\mathrm{d}x+\mathrm{d}y\right)=\sqrt{\cos x}\left(\mathrm{d}x-\mathrm{d}y\right)\\ \Rightarrow & \sqrt{\sin x}\mathrm{d}x+\sqrt{\sin x}\mathrm{d}y=\sqrt{\cos x}\mathrm{d}x-\sqrt{\cos x}\mathrm{d}y\\ \Rightarrow & \sqrt{\sin x}\mathrm{d}x-\sqrt{\cos x}\mathrm{d}x=-\sqrt{\cos x}\mathrm{d}y-\sqrt{\sin x}\mathrm{d}y\\ \Rightarrow & \left(\sqrt{\sin x}-\sqrt{\cos x}\right)\mathrm{d}x=-\left(\sqrt{\cos x}+\sqrt{\sin x}\right)\mathrm{d}y\\ \Rightarrow & \frac{dx}{dy}=-\frac{\sqrt{\cos x}+\sqrt{\sin x}}{\sqrt{\sin x}-\sqrt{\cos x}}\end{array}$

So we see that the order of the differential is $1$ and its degree is also $1$.

Hence, option(C) is correct.