Let `y = y(x)` be the solution of the differential equation `(1 + y^2)e^(tan x) dx + cos^2x(1 + e^(2tanx)dy = 0, y(0) = 1`. Then `y(pi/4)` is equal to

Options

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

Answer

To solve the differential equation given by

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

with the initial condition \( y(0) = 1 \) and find \( y\left(\frac{\pi}{4}\right) \), we will follow these steps:

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

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

### Step 2: Integrating Both Sides
Now, we will 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 the expression before integrating. We can rewrite it as:

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

Using the substitution \( t = e^{\tan x} \), we have \( dt = e^{\tan x} \sec^2 x \, dx \) or \( dx = \frac{dt}{t \sec^2 x} \). Thus, we can express \( \sec^2 x \) in terms of \( t \):

\[ \sec^2 x = 1 + \tan^2 x = 1 + \left(\ln(t)\right)^2 \]

Now, substituting this into our integral, we can simplify and integrate.

### Step 3: Finding the General Solution
After performing the integration, we will have:

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

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

\[ \tan^{-1}(1) = -\tan^{-1}(e^{\tan(0)}) + C \implies \frac{\pi}{4} = -\tan^{-1}(1) + C \]

Since \( \tan^{-1}(1) = \frac{\pi}{4} \):

\[ \frac{\pi}{4} = -\frac{\pi}{4} + C \implies C = \frac{\pi}{2} \]

Thus, the particular solution is:

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

### Step 5: Finding \( y\left(\frac{\pi}{4}\right) \)
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 have:

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

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

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