Find the equation of the curve passing throug | vedclass

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(D) The given differential equation is: $\sin x \cos y \, dx + \cos x \sin y \, dy = 0$. Dividing both sides by $\cos x \cos y$,we get: $\frac{\sin x}

Vedclass · Accepted Answer

$\cos y = \frac{\sec x}{\sqrt{2}}$

Vedclass · Answer

(D) The given differential equation is: $\sin x \cos y \, dx + \cos x \sin y \, dy = 0$. Dividing both sides by $\cos x \cos y$,we get: $\frac{\sin x}{\cos x} \, dx + \frac{\sin y}{\cos y} \, dy = 0$. This simplifies to: $	an x \, dx + 	an y \, dy = 0$. Integrating both sides,we get: $\int 	an x \, dx + \int 	an y \, dy = C_1$. $-\ln |\cos x| - \ln |\cos y| = C_1$,which can be written as $\ln |\cos x \cos y| = -C_1 = C$. So,$\cos x \cos y = e^C = K$. The curve passes through the point $\left(0, \frac{\pi}{4}ight)$. Substituting $x = 0$ and $y = \frac{\pi}{4}$: $\cos(0) \cos\left(\frac{\pi}{4}ight) = K \Rightarrow 1 	imes \frac{1}{\sqrt{2}} = K \Rightarrow K = \frac{1}{\sqrt{2}}$. Thus,the equation of the curve is $\cos x \cos y = \frac{1}{\sqrt{2}}$,which implies $\cos y = \frac{1}{\sqrt{2} \cos x} = \frac{\sec x}{\sqrt{2}}$.

Find the equation of the curve passing through the point $\left(0, \frac{\pi}{4}\right)$ whose differential equation is $\sin x \cos y \, dx + \cos x \sin y \, dy = 0$.

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