Let y(x) be the solution of differential equation `(1 + y^2) e^(tan x) dx + cos^2x(1 + e^(2 tan x)) dy = 0` and `y(0) = 1` then `y(pi/4)` is

Options

A. e
B. 2e
C. `1/e`
D. `1/(2e)`

Answer

To solve the differential equation

\[ (1 + y^2) e^{\tan x} \, dx + \cos^2 x (1 + e^{2 \tan x}) \, dy = 0 \]

with the initial condition \( y(0) = 1 \), we can follow these steps:

### Step 1: Rearranging the Equation
We can rearrange the equation to separate the variables \( y \) and \( x \):

\[ (1 + y^2) e^{\tan x} \, dx = -\cos^2 x (1 + e^{2 \tan x}) \, dy \]

Dividing both sides by \( \cos^2 x (1 + e^{2 \tan x}) \):

\[ \frac{dy}{1 + y^2} = -\frac{e^{\tan x}}{\cos^2 x (1 + e^{2 \tan x})} \, dx \]

### Step 2: Integrating Both Sides
Now we integrate both sides. The left side integrates to:

\[ \int \frac{dy}{1 + y^2} = \tan^{-1}(y) + C_1 \]

For the right side, we need to simplify and integrate:

\[ \int -\frac{e^{\tan x}}{\cos^2 x (1 + e^{2 \tan x})} \, dx \]

Let \( t = e^{\tan x} \), then \( dt = e^{\tan x} \sec^2 x \, dx \). Thus, we can rewrite \( dx \):

\[ dx = \frac{dt}{t \sec^2 x} \]

Substituting this back into the integral gives:

\[ \int -\frac{t}{1 + t^2} \cdot \frac{dt}{t \sec^2 x} = -\int \frac{1}{1 + t^2} \, dt = -\tan^{-1}(t) + C_2 \]

### Step 3: Combining Integrals
Combining the results from both integrals gives us:

\[ \tan^{-1}(y) = -\tan^{-1}(e^{\tan x}) + C \]

### Step 4: Applying Initial Condition
We apply the initial condition \( y(0) = 1 \):

\[ \tan^{-1}(1) = -\tan^{-1}(e^{\tan(0)}) + C \]

Since \( \tan^{-1}(1) = \frac{\pi}{4} \) and \( \tan(0) = 0 \) implies \( e^{\tan(0)} = 1 \):

\[ \frac{\pi}{4} = -\tan^{-1}(1) + C \implies \frac{\pi}{4} = -\frac{\pi}{4} + C \implies C = \frac{\pi}{2} \]

### Step 5: Final Solution
Thus, the particular solution is:

\[ \tan^{-1}(y) = -\tan^{-1}(e^{\tan x}) + \frac{\pi}{2} \]

### Step 6: Finding \( y(\frac{\pi}{4}) \)
Now we need to find \( y\left(\frac{\pi}{4}\right) \):

\[ \tan^{-1}(y) = -\tan^{-1}(e^{\tan(\frac{\pi}{4})}) + \frac{\pi}{2} \]

Since \( \tan\left(\frac{\pi}{4}\right) = 1 \):

\[ \tan^{-1}(y) = -\tan^{-1}(e) + \frac{\pi}{2} \]

Using the identity \( \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2} \):

\[ \tan^{-1}(y) = \tan^{-1}\left(\frac{1}{e}\right) \]

Thus, we find:

\[ y = \frac{1}{e} \]

### Final Answer
Therefore, the value of \( y\left(\frac{\pi}{4}\right) \) is:

\[ \boxed{\frac{1}{e}} \]