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
$\sqrt{\sin x}(dx+dy)=\sqrt{\cos x}(dx-dy)$
so $\sqrt{\sin x}(1+\frac{dy}{dx})=\sqrt{\cos x}(1-\frac{dy}{dx})$
so $\sqrt{\sin x}+\sqrt{\sin x}\frac{dy}{dx}=\sqrt{\cos x}-\sqrt{\cos x}\frac{dy}{dx}$
so $\frac{dy}{dx}=\frac{\sqrt{\cos x}-\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}$
order → 1, degree → 1