Question

For the differential equations , find the general solution: \(sec^2x tan\ y \ dx+sex^2y tan\  x \ dy = 0\)

Answer 1

\(sec^2x. tan\ y \ dx+sex^2y. tan\  x \ dy = 0\)

\(\frac {sec^2x. tan\ y \ dx+sex^2y .tan\  x \ dy}{tan \ x .tan \ y} = 0\)

\(⇒\)\(\frac {sec^2x}{tan\ x }dx\) + \(\frac {sec^2y}{tan\ y }dy\) = 0

\(⇒\)\(\frac {sec^2x}{tan\ x }dx\) = -\(\frac {sec^2y}{tan\ y }dy\)

Integrating both sides of this equation, we get:

\(∫\)\(\frac {sec^2x}{tan\ x }dx\) = -\(∫\)\(\frac {sec^2y}{tan\ y }dy\)     ...(1)

\(Let\  tan\ x=t\)

∴\(\frac {d}{dx}(tanx)=\frac {dt}{dx}\)

\(⇒sec^2x = \frac {dt}{dx}\)

\(⇒sec^2x dx = dt\)

Now, \(∫\)\(\frac {sec^2x}{tan\ x }dx\) = \(∫\frac {1}{t}dt\) =\( log\  t\) = \(log\ (tan\ x)\)

Similarly, \(∫\)\(\frac {sec^2x}{tan\ x }dy\) = \(log\ (tan\ y)\)

Substituting these values in equation (1), we get:

\(log\ (tan\ x)=-log\ (tan\ y)+log\ C\)

\(⇒log\ (tan\ x)=log\ (\frac {C}{tan\ y})\)

\(⇒tan\ x=\frac {C}{tan\ y}\)

\(⇒tan\ x tan\ y=C\)

This is the required general solution of the given differential equation.