Question

Which of the following is the solution of the differential equation \[3{e^x}\tan ydx + \left( {1 - {e^x}} \right){\sec ^2}ydy = 0\].
A \[\tan y = c{\left( {1 - {e^x}} \right)^3}\]
B \[\left( {1 - {e^x}} \right)\tan y = c\]
C \[\tan y = c\left( {1 - {e^x}} \right)\]
D \[\left( {1 - {e^x}} \right)\tan y = c\]

Answer 1

Hint: To find the solution of the differential equation use variable separation method. First separate x, y variables. Then integrate the equation. Use the substitution method to solve the integration. The solution is in the form of a general solution.

Formula used:
\[\int {\dfrac{1}{x}dx} = \log x + c\]
Here, c is a constant.

Complete step by step solution:
The given differential equation is \[3{e^x}\tan ydx + \left( {1 - {e^x}} \right){\sec ^2}ydy = 0\].
Simplify the equation.
\[3{e^x}\tan ydx = - \left( {1 - {e^x}} \right){\sec ^2}ydy\]
Separate the variables as follows,
\[\begin{array}{l}3{e^x}\tan ydx = - \left( {1 - {e^x}} \right){\sec ^2}ydy\\ - 3\dfrac{{{e^x}}}{{1 - {e^x}}}dx = \dfrac{{{{\sec }^2}y}}{{\tan y}}dy\end{array}\]
Further integrate the equation.
\[\int { - 3\dfrac{{{e^x}}}{{1 - {e^x}}}dx} = \int {\dfrac{{{{\sec }^2}y}}{{\tan y}}dy} \]
Substitute \[1 - {e^x} = u \Rightarrow - {e^x}dx = du\]and \[\tan y = v \Rightarrow {\sec ^2}ydy = dv\]
\[\begin{array}{l}\int { - 3\dfrac{{{e^x}}}{{1 - {e^x}}}dx} = \int {\dfrac{{{{\sec }^2}y}}{{\tan y}}dy} \\3\int {\dfrac{1}{u}du} = \int {\dfrac{1}{v}dv} \\3\log \left| u \right|+\log c = \log \left| v \right|\end{array}\]
Substitute back \[u = 1 - {e^x}\]and \[v = \tan y\]. Rewrite the equation.
\[\begin{array}{l}\log \left| v \right| = 3\log \left| u \right|+\log c\\\log \left| {\tan y} \right| = 3\log \left| {1 - {e^x}} \right| + \log c\end{array}\]
Further simplify as follows to find general solution,
\[\tan y = c{\left( {1 - {e^x}} \right)^3}\]
Hence option A is the correct option.

Note:The common mistake happens while solving this kind of question is substitution of \[\sec x = \dfrac{1}{{\cos x}}\]and \[\dfrac{1}{{\tan x}} = \dfrac{{\cos x}}{{\sin x}}\]. Which is the wrong approach to solving the question. This makes the question complicated.