`(dy)/(dx)+xsin2y=x^(3)cos^(2)y`

Explanation

To solve the differential equation \(\frac{dy}{dx} + x \sin(2y) = x^3 \cos^2(y)\), we will follow these steps:

### Step 1: Divide the entire equation by \(\cos^2(y)\)
\[ \frac{1}{\cos^2(y)} \frac{dy}{dx} + x \frac{\sin(2y)}{\cos^2(y)} = x^3 \]
Using the identity \(\sin(2y) = 2 \sin(y) \cos(y)\), we can rewrite the equation as:
\[ \sec^2(y) \frac{dy}{dx} + x \cdot 2 \tan(y) = x^3 \]

### Step 2: Substitute \(t = \tan(y)\)
Let \(t = \tan(y)\). Then, we know that:
\[ \frac{dy}{dx} = \frac{1}{\sin^2(y)} \frac{dt}{dx} \]
Thus, we can rewrite \(\sec^2(y) = 1 + \tan^2(y) = 1 + t^2\). Therefore, the equation becomes:
\[ (1 + t^2) \frac{dt}{dx} + 2x t = x^3 \]

### Step 3: Rearranging the equation
Rearranging gives:
\[ \frac{dt}{dx} + \frac{2x}{1 + t^2} t = \frac{x^3}{1 + t^2} \]

### Step 4: Identify \(p\) and \(q\)
Here, \(p = \frac{2x}{1 + t^2}\) and \(q = \frac{x^3}{1 + t^2}\).

### Step 5: Find the integrating factor
The integrating factor \(IF\) is given by:
\[ IF = e^{\int p \, dx} = e^{\int \frac{2x}{1 + t^2} \, dx} \]
Calculating the integral:
\[ IF = e^{x^2} \]

### Step 6: Multiply through by the integrating factor
Multiply the entire differential equation by \(e^{x^2}\):
\[ e^{x^2} \frac{dt}{dx} + 2x t e^{x^2} = x^3 e^{x^2} \]

### Step 7: Solve the left-hand side
The left-hand side can be expressed as:
\[ \frac{d}{dx}(t e^{x^2}) = x^3 e^{x^2} \]

### Step 8: Integrate both sides
Integrating both sides gives:
\[ t e^{x^2} = \int x^3 e^{x^2} \, dx + C \]

### Step 9: Solve the integral
Using integration by parts, let \(u = x^2\) and \(dv = x e^{x^2}dx\):
\[ \int x^3 e^{x^2} \, dx = \frac{1}{2} x^2 e^{x^2} - \frac{1}{4} e^{x^2} + C \]

### Step 10: Substitute back for \(t\)
Substituting back for \(t = \tan(y)\):
\[ \tan(y) e^{x^2} = \frac{1}{2} x^2 e^{x^2} - \frac{1}{4} e^{x^2} + C \]

### Step 11: Solve for \(y\)
Finally, we can express \(y\) in terms of \(x\):
\[ \tan(y) = \frac{1}{2} x^2 - \frac{1}{4} + C e^{-x^2} \]
Thus, the solution to the differential equation is:
\[ y = \tan^{-1}\left(\frac{1}{2} x^2 - \frac{1}{4} + C e^{-x^2}\right) \]