Question

Solve the differential equation: $\sec^{2}x\cdotp\tan ydx+\sec^{2}y\cdotp\tan xdy=0$

Answer 1

To solve the differential equation:

$\sec^{2}x\cdotp\tan ydx+\sec^{2}y\cdotp\tan xdy=0$,

first separate the variables than integrate both sides,

$\sec^{2}x\cdotp\tan ydx+\sec^{2}y\cdotp\tan xdy=0 \\ \\ \sec^{2}x\cdotp\tan ydx = - \sec^{2}y\cdotp\tan xdy \\ \\ \frac{\sec^{2}x}{tan \: x} dx = \frac{ - \sec^{2}y}{tan \: y} dy$
As we know that
$\frac{d \: tan \: x}{dx} = {sec}^{2} x$

now integrate it with substitute method

$\int\frac{\sec^{2}x}{tan \: x} dx = \int\frac{ - \sec^{2}y}{tan \: y} dy \\ \\ \\ let \: tan \: x = t \\ \\ {\sec^{2}x} \: dx = dt \\ \\ \: substitute \: in \: both \: sides \\ \\ \int\frac{1}{t} dt =- \int \frac{1}{z} dz \\ \\ log \: t = - log z + c \\ \\ log \: tan \: x = - log \: tan \: y + log \: c \\ \\ \\ log \: tan \: x = log \: cot \: y + log \: c \\ \\ log \: tan \: x = log \: c (\: cot \: y) \\ \\ tan \: x = c \: (cot \: y) \\ \\ cot \: y = \frac{tan \: x}{c}$

is the solution.