Question

The solution of the differential equation ${{\sec }^{2}}x.\tan ydx+{{\sec }^{2}}y.\tan xdy=0$ is
A)$\tan x.\cot y=C$
B)$\cot x.\tan y=C$
C)$\tan x.\tan y=C$
D)$\sin x.\cos y=C$

Answer 1

Hint: The differential equation in this question can be represented in variable separable form that is,
$f(x)dx=g(y)dy$. After rewriting the given differential equation in variable separable form, the solution can be found out by integrating both sides of the equation.

Complete step-by-step answer:
In this question, the differential equation is given as,
${{\sec }^{2}}x.\tan ydx+{{\sec }^{2}}y.\tan xdy=0$
The above equation is not in variable separable form.
So, we have to change the above differential equation into variable separable form.
This is done by taking all $x$ terms together and $y$ terms together in the LHS and RHS respectively.
This can be done as given below:
$\Rightarrow {{\sec }^{2}}x.\tan ydx=-{{\sec }^{2}}y.\tan xdx......(i)$
Again, rearranging we have,
$\Rightarrow \dfrac{{{\sec }^{2}}x}{\tan x}dx=-\dfrac{{{\sec }^{2}}y}{\tan y}dy......(ii)$
Equation (ii) is the variable separable form.
To solve equation (ii) we have to integrate both sides of the equation (ii),
$\Rightarrow \int{\dfrac{{{\sec }^{2}}x}{\tan x}dx}=-\int{\dfrac{{{\sec }^{2}}y}{\tan y}dy}......(iii)$
We also know that integration of $\tan x$ is ${{\sec }^{2}}xdx$ and integration of $\tan y$ is ${{\sec }^{2}}ydy$.
Therefore, in equation (iii) we can substitute,
$\Rightarrow \tan x=u.....(iv)$
$\Rightarrow \tan y=v.....(v)$
From equation (iv) and equation (v) we know that,
$\begin{aligned} & \Rightarrow {{\sec }^{2}}xdx=du.......(vi) \\\ & \Rightarrow {{\sec }^{2}}ydy=dv......(vii) \\\ \end{aligned}$
Substituting equations (iv), (v), (vi) and (vii) in equation (iii) we get,
$\Rightarrow \int{\dfrac{du}{u}}=-\int{\dfrac{dv}{v}}......(viii)$
Now, we have to integrate equation (viii).
We know that integration of $\dfrac{1}{x}$ is $\log x$.
Thus, applying this concept in equation (viii) for integrating we get,
$\Rightarrow \log u=-\log v+\log C.......(ix)$
Here, $\log C$ is the arbitrary constant of integration.
Now, substitute values for $u$ and $v$ from equations (iv) and (v) in equation (ix).
$\Rightarrow \log |\tan x|=-\log |\tan y|+\log C........(x)$
Now, rearranging equation (x) we have,
$\Rightarrow \log |\tan x|+\log |\tan y|=\log C......(xi)$
We also know,$\log a+\log b=\log ab$. Applying this in equation (xi) we can rewrite the above equation as,
$\Rightarrow \log |\tan x.\tan y|=\log C.......(xii)$
We know that, ${{e}^{\log x}}=x$. Applying this in equation (xii) on both LHS and RHS where $x=\tan x.\tan y$ and $x=C$ respectively we get,
$\begin{aligned} & \Rightarrow {{e}^{\log |\tan x.\tan y|}}={{e}^{\log C}} \\\ & \Rightarrow \tan x.\tan y=C......(xiii) \\\ \end{aligned}$
$\therefore$ the correct answer is $\tan x.\tan y=C$.
Hence, the correct option is (C).

Note: The main idea of this problem is to write the given differential equation in the variable separable form. If this step goes wrong, the solution can’t be evaluated.
Also, there is no need to introduce arbitrary constants of integration on both sides as they can be combined together as $\log C$.