Solve : `3e^xtany dx+(1-e^x)sec^2y\ dy=0`

Explanation

To solve the differential equation \(3e^x \tan y \, dx + (1 - e^x) \sec^2 y \, dy = 0\), we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation to separate the variables:
\[ 3e^x \tan y \, dx = - (1 - e^x) \sec^2 y \, dy \]
This can be rewritten as:
\[ \frac{dx}{\sec^2 y} = -\frac{(1 - e^x)}{3e^x \tan y} \, dy \]

### Step 2: Separating Variables
We can now separate the variables \(x\) and \(y\):
\[ \frac{dx}{(1 - e^x)} = -\frac{3 \tan y}{\sec^2 y} \, dy \]
Since \(\tan y = \frac{\sin y}{\cos y}\) and \(\sec^2 y = \frac{1}{\cos^2 y}\), we have:
\[ \frac{dx}{(1 - e^x)} = -3 \sin y \cos y \, dy \]

### Step 3: Integrating Both Sides
Now we integrate both sides:
\[ \int \frac{dx}{(1 - e^x)} = -3 \int \sin y \cos y \, dy \]
The left-hand side can be integrated using substitution. Let \(t = 1 - e^x\), then \(dt = -e^x \, dx\) or \(dx = -\frac{dt}{e^x} = -\frac{dt}{1-t}\):
\[ \int \frac{dx}{(1 - e^x)} = -\ln |1 - e^x| + C_1 \]
The right-hand side can be simplified using the identity \(\sin(2y) = 2 \sin y \cos y\):
\[ -3 \int \sin y \cos y \, dy = -\frac{3}{2} \int \sin(2y) \, dy = \frac{3}{4} \cos(2y) + C_2 \]

### Step 4: Combining Results
Combining the results from both integrals, we have:
\[ -\ln |1 - e^x| = \frac{3}{4} \cos(2y) + C \]
where \(C = C_2 - C_1\).

### Step 5: Exponentiating Both Sides
To eliminate the logarithm, we exponentiate both sides:
\[ |1 - e^x| = e^{-\frac{3}{4} \cos(2y) - C} \]
Let \(A = e^{-C}\), then:
\[ 1 - e^x = \frac{A}{e^{\frac{3}{4} \cos(2y)}} \]

### Step 6: Final Rearrangement
Rearranging gives us:
\[ e^x = 1 - \frac{A}{e^{\frac{3}{4} \cos(2y)}} \]
This is the general solution of the given differential equation.