Question

Evaluate the following equation:
${{\sec }^{2}}x\tan y\text{ }dy+{{\sec }^{2}}y\tan x\text{ }dx=0$

Answer 1

Hint:First of all, transpose one of the terms to RHS. Then divide the whole equation by tan x tan y to get the term in one variable on each side. Now use the substitution $\tan \theta =t$ and ${{\sec }^{2}}\theta \text{ }d\theta =dt$ to solve the equation.

Complete step by step answer:
Here, we have to solve the differential equation
${{\sec }^{2}}x\tan y\text{ }dy+{{\sec }^{2}}y\tan x\text{ }dx=0$
Let us consider the differential equation given in the question.
${{\sec }^{2}}x\tan y\text{ }dy+{{\sec }^{2}}y\tan x\text{ }dx=0$
By transposing one of the terms to RHS of the above equation, we get,
${{\sec }^{2}}x\tan y\text{ }dy=-{{\sec }^{2}}y\tan x\text{ }dx$
By dividing both the sides of the above equation by ${{\sec }^{2}}x{{\sec }^{2}}y$, we get,
$\dfrac{{{\sec }^{2}}x\tan y\text{ }dy}{{{\sec }^{2}}x{{\sec }^{2}}y}=-\dfrac{{{\sec }^{2}}y\tan x\text{ }dx}{{{\sec }^{2}}x{{\sec }^{2}}y}$
By canceling the like terms in the above equation, we get’
$\dfrac{\tan y\text{ }dy}{{{\sec }^{2}}y}=\dfrac{-\tan x\text{ }dx}{{{\sec }^{2}}x}$
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\sec \theta =\dfrac{1}{\cos \theta }$. By using this in the above equation, we get,
$\dfrac{\dfrac{\sin y}{\cos y}}{\dfrac{1}{\cos y}}dy=\dfrac{\dfrac{-\sin x}{\cos x}dx}{\dfrac{1}{{{\cos }^{2}}x}}dx$
By canceling the like terms in the above equation, we get,
$\sin y\cos ydy=-\sin x\cos xdx$
By multiplying by 2 on both the sides of the above equation, we get,
$2\sin y\cos ydy=-2\sin x\cos xdx$
We know that $2\sin \theta \cos \theta =\sin 2\theta$. By using this in the above equation, we get,
$\sin 2ydy=-\sin 2xdx$
By integrating both the sides of the above equation, we get,
$\int{\left( \sin 2y \right)dy}=-\int{\left( \sin 2x \right)dx}$
We know that $\int{\sin 2\theta d\theta =\dfrac{-\cos 2\theta }{2}+k}$. By using this in the above equation, we get,
$\dfrac{-\cos 2y}{2}+{{k}_{1}}=\dfrac{+\cos 2x}{2}+{{k}_{2}}$
where ${{k}_{1}}$ and ${{k}_{2}}$ are constant
$\Rightarrow \dfrac{\cos 2x+\cos 2y}{2}={{k}_{1}}-{{k}_{2}}$
$\cos 2x+\cos 2y=2\left( {{k}_{1}}-{{k}_{2}} \right)$
By taking $2\left( {{k}_{1}}-{{k}_{2}} \right)=k$, we get,
$\cos 2x+\cos 2y=k$

Note: In these types of questions, always first separate the terms x and y and then only solve the differential equation. Also in this question, some students make this mistake of substituting tan x = t by looking at the terms of tan x and ${{\sec }^{2}}x$ which is not needed here because we generally substitute tan x = t when we have another term like ${{\sec }^{2}}xdx$ which is absent here. So, this must be there in the mind of the students.