Question

The differential equation which represents the family of curves given by $\tan y=c\left( 1-{{e}^{x}} \right)$ is
A. ${{e}^{x}}\tan ydx+\left( 1-{{e}^{x}} \right)dy=0$
B. ${{e}^{x}}\tan ydx+\left( 1-{{e}^{x}} \right){{\sec }^{2}}ydy=0$
C. ${{e}^{x}}\left( 1-{{e}^{x}} \right)dx+\tan ydy=0$
D. ${{e}^{x}}\tan ydy+\left( 1-{{e}^{x}} \right)dx=0$

Answer 1

Hint : The one interesting fact about differential equations is that if we differentiate the equation of curve and get rid of the constants, we get the differential equation of the curve. So, for this problem, we need to differentiate the curve equation. We first take differentials on both sides and then substitute the value of c from the equation $\tan y=c\left( 1-{{e}^{x}} \right)$ as $c=\dfrac{\tan y}{\left( 1-{{e}^{x}} \right)}$ , and get ${{\sec }^{2}}ydy=\dfrac{\tan y}{\left( 1-{{e}^{x}} \right)}\left( -{{e}^{x}}dx \right)$ . Rearranging it, we get the correct option.

Complete step-by-step solution:
We can express a family of curves or a condition in two forms. One is the cartesian equation and the other is the differential equation. We deduce the former equation using the locus of the point in Cartesian coordinates and the latter one using the study of an elementary particle. The one interesting fact about differential equations is that if we differentiate the equation of curve and get rid of the constants, we get the differential equation of the curve. So, for this problem, we need to differentiate the curve equation.
The equation is,
$\tan y=c\left( 1-{{e}^{x}} \right)$
Taking differentials on both sides, we get,
$\Rightarrow {{\sec }^{2}}ydy=c\left( -{{e}^{x}}dx \right)$
Substituting the value of c from the equation $\tan y=c\left( 1-{{e}^{x}} \right)$ as $c=\dfrac{\tan y}{\left( 1-{{e}^{x}} \right)}$ , we get,
$\begin{aligned} & \Rightarrow {{\sec }^{2}}ydy=\dfrac{\tan y}{\left( 1-{{e}^{x}} \right)}\left( -{{e}^{x}}dx \right) \\\ & \Rightarrow {{e}^{x}}\tan ydx+\left( 1-{{e}^{x}} \right){{\sec }^{2}}ydy=0 \\\ \end{aligned}$
Thus, we can conclude that the differential equation which represents the family of curves given by $\tan y=c\left( 1-{{e}^{x}} \right)$ is ${{e}^{x}}\tan ydx+\left( 1-{{e}^{x}} \right){{\sec }^{2}}ydy=0$ that is option B.

Note: We can also solve the problem in another way. Instead of differentiating the equation of the curve, we integrate the options. The second differential equation is,
${{e}^{x}}\tan ydx+\left( 1-{{e}^{x}} \right){{\sec }^{2}}ydy=0$
We take ${{e}^{x}}\tan ydx$ to the RHS, and get,
$\Rightarrow \left( 1-{{e}^{x}} \right){{\sec }^{2}}ydy=-{{e}^{x}}\tan ydx$
We take $\tan y$ to the LHS, $\left( 1-{{e}^{x}} \right)$ to the RHS, and get,
$\Rightarrow \dfrac{{{\sec }^{2}}ydy}{\tan y}=\dfrac{-{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx$
Integrating both sides, we get,
$\begin{aligned} & \Rightarrow \int{\dfrac{{{\sec }^{2}}ydy}{\tan y}}=\int{\dfrac{-{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx} \\\ & \Rightarrow \ln \left| \tan y \right|=\ln \left| 1-{{e}^{x}} \right|+\ln c \\\ & \Rightarrow \tan y=c\left( 1-{{e}^{x}} \right) \\\ \end{aligned}$
Which is the original family of curves that we are given.