Question

Solution of the differential equation
$tany\, sec^2x\, dx + tan \,x \,sec^2ydy = 0$ is

• A. $tan\,x + tan\,y = k$
• B. $tan\,x- tan\,y - k$
• C. $\frac{tan\,x}{tan\,y}=k$
• D. $tan\,x\cdot tan\, y = k$

Answer 1

Answer: (D)

$tan\,y\,sec^{2}xdx+tanx\,sec^{2}ydy=0$
$\Rightarrow \frac{sec^{2}\,xdx}{tan\,x}=-\frac{sec^{2}\,ydy}{tan\,y}$
On integration, we get $log\, tanx = - log \,tany + log\, k$
$\Rightarrow log \left(tan\,x\cdot tan\,y\right) = log\, k$
$\Rightarrow tanx\cdot tany=k$