Question

Let $y = y(x)$ be the solution of the differential equation $(1 + y^2)e^{\tan x} dx + \cos^2 x (1 + e^{2\tan x}) dy = 0$, $y(0) = 1$. Then $y\left(\frac{\pi}{4}\right)$ is equal to:

• A. $\frac{2}{e}$
• B. $\frac{1}{e^2}$
• C. $\frac{1}{e}$
• D. $\frac{2}{e^2}$

Answer 1

The Correct Option is C

The given differential equation is:
\[ (1 + y^2)e^{\tan^{-1}x}dx + \cos^2x(1 + e^{2\tan^{-1}x})dy = 0. \]
Separate the variables:
\[ \frac{\sec^2x \cdot e^{\tan^{-1}x}dx}{1 + e^{2\tan^{-1}x}} + \frac{dy}{1 + y^2} = 0. \]
Integrating both sides:
\[ \tan^{-1}(e^{\tan^{-1}x}) + \tan^{-1}(y) = C. \]
Using the initial condition \(y(0) = 1\):
\[ \tan^{-1}(e^{\tan^{-1}(0)}) + \tan^{-1}(1) = C. \]
Simplify:
\[ \tan^{-1}(e^0) + \tan^{-1}(1) = C \implies \tan^{-1}(1) + \tan^{-1}(1) = C \implies C = \frac{\pi}{2}. \]
The solution becomes:
\[ \tan^{-1}(e^{\tan^{-1}x}) + \tan^{-1}(y) = \frac{\pi}{2}. \]
At \(x = \frac{\pi}{4}\), substitute into the solution:
\[ \tan^{-1}(e^{\tan^{-1}(\frac{\pi}{4})}) + \tan^{-1}(y) = \frac{\pi}{2}. \]
Rearrange:
\[ \tan^{-1}(y) = \frac{\pi}{2} - \tan^{-1}(e^{\tan^{-1}(\frac{\pi}{4})}). \]
From the properties of \(\tan^{-1}\), substitute:
\[ \tan^{-1}(y) = \cot^{-1}(e). \]
Simplify:
\[ y = \frac{1}{e}. \]
Final Answer: \(\frac{1}{e}\).