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.