Find the solution of a differential equation \[\left( {1 + y} \right)\tan^{2}x + \tan x\left( {\dfrac{{dy}}{{dx}}} \right) + y = 0\].

Hint: First, simplify the given differential equation in the form \[\dfrac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right)\]. Then calculate the integrating factor of the equation. Solve the differential equation using the integrating factor method for first order to get the solution of the given differential equation.

Formula used:
Integrating factor of a first order differential equation \[\dfrac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right)\] is: \[I.F. = {e^{\int {P\left( x \right)dx} }}\]
The solution of a differential equation is: \[\left( {I.F.} \right)y = \int {\left( {I.F.} \right)Q\left( x \right)} dx + C\]
\[\int {\sec^{2}x} dx = \tan x\]

Complete step by step solution:
The given differential equation is \[\left( {1 + y} \right)\tan^{2}x + \tan x\left( {\dfrac{{dy}}{{dx}}} \right) + y = 0\].
Let’s simplify the above equation in the form \[\dfrac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right)\].
\[\left( {1 + y} \right)\tan^{2}x + \tan x\left( {\dfrac{{dy}}{{dx}}} \right) + y = 0\]
\[ \Rightarrow \tan x\left( {\dfrac{{dy}}{{dx}}} \right) + \tan^{2}x + y\tan^{2}x + y = 0\]
\[ \Rightarrow \tan x\left( {\dfrac{{dy}}{{dx}}} \right) + \left( {1 + \tan^{2}x} \right)y = - \tan^{2}x\]
Divide both sides by \[\tan x\].
\[\left( {\dfrac{{dy}}{{dx}}} \right) + \left( {\dfrac{{1 + \tan^{2}x}}{{\tan x}}} \right)y = \dfrac{{ - \tan^{2}x}}{{\tan x}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} + \left( {\dfrac{1}{{\tan x}} + \tan x} \right)y = - \tan x\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} + \left( {\cot x + \tan x} \right)y = - \tan x\]
Compare the above equation with the general first order differential equation.
We get,
\[P\left( x \right) = \left( {\cot x + \tan x} \right)\], and \[Q\left( x \right) = - \tan x\]

Now calculate the integrating factor of the above first order differential equation.
\[I.F. = {e^{\int {\left( {\cot x + \tan x} \right)dx} }}\]
\[ \Rightarrow I.F. = {e^{ln \sin x + ln \sec x}}\]
Apply the logarithmic identity \[ln\left( a \right) + ln\left( b \right) = ln\left( {ab} \right)\].
 \[I.F. = {e^{ln \left( {\sin x \sec x} \right)}}\]
Use the trigonometric ratio \[\sec x = \dfrac{1}{{\cos x}}\] .
\[I.F. = {e^{ln \left( {sin x \dfrac{1}{{\cos x}}} \right)}}\]
\[ \Rightarrow I.F. = {e^{ln \left( {\tan x} \right)}}\]
Now apply the logarithmic property \[{e^{ln \left( x \right)}} = x\].
\[I.F. = \tan x\]

Multiply both sides of the differential equation by the integrating factor.
\[\tan x\dfrac{{dy}}{{dx}} + \tan x\left( {\cot x + \tan x} \right)y = - \tan^{2} x\]
\[ \Rightarrow \tan x\dfrac{{dy}}{{dx}} + \left( {1 + \tan^{2}x} \right)y = - \tan^{2}x\]
Use the trigonometric identity \[\tan^{2} x = \sec^{2} x - 1\].
\[\tan x\dfrac{{dy}}{{dx}} + \left( {\sec^{2}x} \right)y = - \tan^{2}x\]
\[\dfrac{{d}}{{dx}}\left( {y\tan x} \right) = - \tan^{2}x\]
Now calculate the solution of the given differential equation by integrating both sides.
We get,
\[\int {\dfrac{{d}}{{dx}}\left( {y\tan x} \right)} = - \int {\tan^{2}x} dx\]
\[ \Rightarrow y\tan x = - \int {\tan^{2}x} dx\]
Use the trigonometric identity \[\tan^{2} x = \sec^{2} x - 1\].
\[y\tan x = - \int {\left( {\sec^{2}x - 1} \right)} dx\]
\[ \Rightarrow y\tan x = - \tan x + x + C\] , where \[C\] is the integration constant.

Hence the solution of the differential equation \[\left( {1 + y} \right)\tan^{2}x + \tan x\left( {\dfrac{{dy}}{{dx}}} \right) + y = 0\] is \[y\tan x = - \tan x + x + C\].

Note: Students often get confused about the integrating factor method for the first-order differential equation.
Steps to solve the first order differential equation:
Step 1: Compare differential equation with the general form \[\dfrac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right)\] and find the values of \[P\left( x \right)\] and \[Q\left( x \right)\].
Step 2: Calculate the integrating factor.
Step 3: Multiply both sides of the differential equation by the integrating factor.
Step 4: In the end, integrate the equation and get the solution in the form \[\left( {I.F.} \right)y = \int {\left( {I.F.} \right)Q\left( x \right)} dx + C\].