The order and degree of the differential equation $\sqrt{sin\, x}(dx+dy)=\sqrt{cos\, x}(dx-dy)$ is :

1, 1

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