The order and degree of the differential equation \(\sqrt{\sin x}\,(dx+dy)=\sqrt{\cos x}\,(dx-dy)\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1, 1
Opening the brackets and collecting the dx and dy terms:
\(\sqrt{\sin x}\,dx + \sqrt{\sin x}\,dy = \sqrt{\cos x}\,dx - \sqrt{\cos x}\,dy\)
\(\left(\sqrt{\sin x} + \sqrt{\cos x}\right)dy = \left(\sqrt{\cos x} - \sqrt{\sin x}\right)dx\)
\( \displaystyle \frac{dy}{dx} = \frac{\sqrt{\cos x} - \sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \)
The highest derivative present is the first derivative dy/dx, so the order is 1.
The equation is a polynomial in dy/dx, and dy/dx appears only to the power 1, so the degree is 1.
The square roots involve only x, not the derivative, so they do not affect the degree.