Question

Find the equation of the curve passing through the point \((0,\frac \pi4) \)whose differential equation is, \(sin\ x cos \ y\ dx+cos\ xsin\  y\ dy=0\)

Answer 1

The differential equation of the given curve is:

\(sin\ x cos \ y\ dx+cos\ xsin\  y\ dy=0\)

⇒\(\frac {sin\ x cos \ y\ dx+cos\ xsin\  y\ dy}{cos\ x cps\ y}=0\)

⇒\(tan\ x\  dx+tan \ y\ dy=0\)

Integrating both sides, we get:

\(log\ (sec\ x)+log\ (sec\ y)=log\ C\)

\(log\ (sec\ x.sec\ y)=log\ C\)

⇒\(sec\ x.sec\ y=C  \)       ...(1)

The curve passes through point \((0,\frac \pi4)\).

\(∴1×\sqrt2=C\)

⇒\(C=\sqrt 2\)

On substituting \(C=\sqrt 2\) in equation (1), we get:

\(sec\ x.sec \ y=\sqrt 2\)

⇒\(secx.\frac {1}{cos\ y}=\sqrt 2\)

⇒\(cos\ y=\frac {sec\ x}{\sqrt 2}\)

Hence, the required equation of the curve is \(cos\ y=\frac {sec\ x}{\sqrt 2}\).