`dy/dx +tany/(1+x) = (1+x)e^x secy`

Answer

To solve the differential equation

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

we will follow a systematic approach.

### Step 1: Rewrite the equation
We start with the given equation:

\[ \frac{dy}{dx} + \frac{\tan y}{1+x} = (1+x)e^x \sec y. \]

### Step 2: Isolate \(\frac{dy}{dx}\)
We can isolate \(\frac{dy}{dx}\) by rearranging the equation:

\[ \frac{dy}{dx} = (1+x)e^x \sec y - \frac{\tan y}{1+x}. \]

### Step 3: Divide by \(\sec y\)
Next, we divide the entire equation by \(\sec y\):

\[ \frac{dy}{dx} \cdot \frac{1}{\sec y} + \frac{\tan y}{(1+x) \sec y} = (1+x)e^x. \]

Since \(\frac{1}{\sec y} = \cos y\), we can rewrite this as:

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

### Step 4: Substitute \(v = \sin y\)
Let \(v = \sin y\). Then, we have:

\[ \frac{dy}{dx} = \frac{dv}{dx} \cdot \frac{1}{\cos y}. \]

Substituting this into our equation gives:

\[ \frac{dv}{dx} + \frac{v}{1+x} = (1+x)e^x. \]

### Step 5: Identify \(p\) and \(q\)
Now, we can identify \(p\) and \(q\) from the standard form of a linear differential equation:

\[ \frac{dv}{dx} + p v = q. \]

Here, \(p = \frac{1}{1+x}\) and \(q = (1+x)e^x\).

### Step 6: Find the integrating factor
The integrating factor \(I\) is given by:

\[ I = e^{\int p \, dx} = e^{\int \frac{1}{1+x} \, dx} = e^{\ln(1+x)} = 1+x. \]

### Step 7: Multiply through by the integrating factor
Multiply the entire equation by the integrating factor:

\[ (1+x) \frac{dv}{dx} + v = (1+x)^2 e^x. \]

### Step 8: Integrate both sides
Now, we can integrate both sides:

\[ \int \left[(1+x) \frac{dv}{dx} + v\right] dx = \int (1+x)^2 e^x \, dx. \]

Using integration by parts for the right-hand side, we will have:

\[ (1+x)v = \int (1+x)^2 e^x \, dx + C. \]

### Step 9: Solve for \(v\)
Now, we can solve for \(v\):

\[ v = \frac{1}{1+x} \left( \int (1+x)^2 e^x \, dx + C \right). \]

### Step 10: Substitute back for \(y\)
Recall that \(v = \sin y\), so we have:

\[ \sin y = \frac{1}{1+x} \left( \int (1+x)^2 e^x \, dx + C \right). \]

### Final Solution
Thus, the solution to the differential equation is:

\[ y = \arcsin\left( \frac{1}{1+x} \left( \int (1+x)^2 e^x \, dx + C \right) \right). \]