Solve the differential equation
\[x\dfrac{dy}{dx}-y=2{{x}^{2}}\operatorname{cosec}2x\]
(a) y = cx + x ln tan x
(b) y = cx + x ln cot x
(c) y = cx – y ln tan x
(d) y = cx – y ln cot x

Hint: First of all, divide the whole equation by x. Now, it is in the form \[\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)\]. Solve it by solving \[y\left( I.F \right)=\int{Q\left( I.F \right)dx}\] where \[I.F={{e}^{\int{Pdx}}}\]. So, find the I.F and substitute in the above equation and solve it.

Complete step-by-step answer:
We have to solve the differential equation.
\[x\dfrac{dy}{dx}-y=2{{x}^{2}}\operatorname{cosec}2x\]
Let us consider the differential equation given in the question.
\[x\dfrac{dy}{dx}-y=2{{x}^{2}}\operatorname{cosec}2x\]
By dividing x on both the sides of the above equation, we get,
\[\dfrac{dy}{dx}-\dfrac{y}{x}=2x\operatorname{cosec}2x\]
Now, the above equation is of the form \[\dfrac{dy}{dx}+Py=Q\] where P and Q are functions of x. By comparison, we get, \[P=\dfrac{-1}{x}\text{ and }Q=2x\operatorname{cosec}2x\].
To solve this equation, we will first find its integrating factor which is:
\[I.F={{e}^{\int{Pdx}}}\]
By substituting \[P=\dfrac{-1}{x}\], we get,
\[I.F={{e}^{\int{\dfrac{-1}{x}}dx}}\]
We know that
\[\int{\dfrac{1}{x}dx=\ln x}\]
So, we get,
\[I.F={{e}^{-\ln x}}\]
We know that,
\[{{a}^{\log b}}={{b}^{\log a}}\]
So, we get,
\[I.F={{x}^{-\ln e}}\]
We know that ln e = 1, so we get,
\[I.F={{x}^{-1}}=\dfrac{1}{x}\]
Now, the equation is solved by using
\[y\left( I.F \right)=\int{Q\left( I.F \right)dx}\]
By substituting the value of \[I.F=\dfrac{1}{x}\text{ and }Q=2x\operatorname{cosec}2x\], we get,
\[\dfrac{y}{x}=\int{\dfrac{2x\operatorname{cosec}2x}{x}dx}\]
\[\dfrac{y}{x}=\int{2\operatorname{cosec}2x\text{ }dx}\]
\[\dfrac{y}{x}=2\int{\operatorname{cosec}2x\text{ }dx}.....\left( i \right)\]
Now, let us find the value of \[\int{\operatorname{cosec}2x\text{ }dx}\].
We know that \[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }\]. By using this, we get,
\[I=\int{\operatorname{cosec}2x\text{ }dx}=\int{\dfrac{1}{\sin 2x}}dx\]
We know that \[\sin 2\theta =2\sin \theta \cos \theta \]. By using this, we get,
\[I=\int{\dfrac{1}{2\sin x\cos x}}dx\]
By multiplying cos x on both numerator and denominator of the above equation on the RHS, we get,
\[I=\int{\dfrac{1}{2\sin x\cos x}}\times \dfrac{\cos x}{\cos x}dx\]
We know that, \[\dfrac{\sin \theta }{\cos \theta }=\tan \theta \text{ and }\dfrac{1}{\cos \theta }=\sec \theta \]. By using this, we get,
\[I=\int{\dfrac{{{\sec }^{2}}xdx}{2\tan x}....\left( ii \right)}\]
Let us take tan x = t. By differentiating both the sides, we get,
\[{{\sec }^{2}}xdx=dt\]
So, by substituting x in terms of t in equation (ii), we get,
\[I=\int{\dfrac{dt}{2t}}\]
We know that \[\int{\dfrac{1}{x}dx=\ln x+c}\]. By using this, we get,
\[I=\dfrac{\ln t}{2}+c\]
By substituting t = tan x, we get,
\[I=\dfrac{\ln \left( \tan x \right)}{2}+c\]
So, we get,
\[\int{\operatorname{cosec}2xdx=\dfrac{\ln \left( \tan x \right)}{2}}+c\]
Now, by substituting the value of \[\int{\operatorname{cosec}2x}\text{ }dx\] in equation (i), we get,
\[\dfrac{y}{x}=2\int{\operatorname{cosec}2x}\text{ }dx\]
\[\dfrac{y}{x}=2\left[ \dfrac{\ln \left( \tan x \right)}{2}+c \right]\]
\[\dfrac{y}{x}=\ln \left( \tan x \right)+c\]
By multiplying x on both the sides, we get,
\[y=cx+x\ln \left( \tan x \right)\]
Therefore, option (a) is the right answer.

Note: In these types of questions, first of all, it is very important to identify the form and convert it into the standard linear differential equation of the first order that is in \[\dfrac{dy}{dx}+Py=Q\] like we have first divided the whole equation by x to convert it into this form. Also, remember that I.F or integrating factor is \[{{e}^{\int{Pdx}}}\] and not \[{{e}^{\int{Qdx}}}\]. So, always compare the values properly and then only write them.