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.