Solve the following differential equations
`3e^(x) tan y dx +(1-e^(x)) sec^2y dy=0" given that "y= (pi)/(4), " when "x=1`.

Answer

To solve the differential equation \(3e^{x} \tan y \, dx + (1 - e^{x}) \sec^2 y \, dy = 0\) with the initial condition \(y = \frac{\pi}{4}\) when \(x = 1\), we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the given equation to isolate \(\frac{dy}{dx}\):

\[ 3e^{x} \tan y \, dx + (1 - e^{x}) \sec^2 y \, dy = 0 \]

Rearranging gives:

\[ (1 - e^{x}) \sec^2 y \, dy = -3e^{x} \tan y \, dx \]

Dividing both sides by \(dx\) and \((1 - e^{x}) \sec^2 y\):

\[ \frac{dy}{dx} = -\frac{3e^{x} \tan y}{1 - e^{x}} \sec^2 y \]

### Step 2: Separating Variables
Next, we separate the variables \(y\) and \(x\):

\[ \frac{\sec^2 y}{\tan y} \, dy = -\frac{3e^{x}}{1 - e^{x}} \, dx \]

### Step 3: Integrating Both Sides
Now we integrate both sides. The left side can be simplified:

\[ \int \frac{\sec^2 y}{\tan y} \, dy = \int \frac{1}{\sin y \cos y} \, dy = \int \csc y \, dy \]

The integral of \(\csc y\) is:

\[ \int \csc y \, dy = -\ln |\csc y + \cot y| + C_1 \]

For the right side, we have:

\[ \int -\frac{3e^{x}}{1 - e^{x}} \, dx \]

To integrate this, we can use substitution. Let \(u = 1 - e^{x}\), then \(du = -e^{x} \, dx\), or \(dx = -\frac{du}{e^{x}} = -\frac{du}{1 - u}\).

Thus, the integral becomes:

\[ \int \frac{3}{u} \, du = 3 \ln |u| + C_2 = 3 \ln |1 - e^{x}| + C_2 \]

### Step 4: Combining the Results
Combining the results from both integrals, we have:

\[ -\ln |\csc y + \cot y| = 3 \ln |1 - e^{x}| + C \]

### Step 5: Applying Initial Conditions
Now we apply the initial condition \(y = \frac{\pi}{4}\) when \(x = 1\):

First, calculate \(\csc\left(\frac{\pi}{4}\right) + \cot\left(\frac{\pi}{4}\right)\):

\[ \csc\left(\frac{\pi}{4}\right) = \sqrt{2}, \quad \cot\left(\frac{\pi}{4}\right) = 1 \implies \csc\left(\frac{\pi}{4}\right) + \cot\left(\frac{\pi}{4}\right) = \sqrt{2} + 1 \]

Now, substituting \(x = 1\):

\[ 1 - e^{1} = 1 - e \]

Thus, we have:

\[ -\ln |\sqrt{2} + 1| = 3 \ln |1 - e| + C \]

### Step 6: Solving for \(C\)
From the equation above, we can solve for \(C\):

\[ C = -\ln |\sqrt{2} + 1| - 3 \ln |1 - e| \]

### Final Solution
The final implicit solution of the differential equation is:

\[ -\ln |\csc y + \cot y| = 3 \ln |1 - e^{x}| - \ln |\sqrt{2} + 1| - 3 \ln |1 - e| \]