To solve the given differential equation and find the value of \( (AB)^2 \), we will follow these steps:
### Step 1: Rewrite the Differential Equation
The given differential equation is:
\[
\sec^2 y \frac{dy}{dx} + 2x \tan y = x^3
\]
We can rewrite this as:
\[
\sec^2 y \frac{dy}{dx} = x^3 - 2x \tan y
\]
### Step 2: Substitute \( \tan y = t \)
Let \( t = \tan y \). Then, we differentiate both sides:
\[
\frac{dt}{dx} = \sec^2 y \frac{dy}{dx}
\]
Thus, we can substitute in our equation:
\[
\frac{dt}{dx} = x^3 - 2x t
\]
### Step 3: Rearrange into Standard Form
We can rearrange the equation into the standard form of a linear differential equation:
\[
\frac{dt}{dx} + 2xt = x^3
\]
### Step 4: Identify \( p \) and \( q \)
From the standard form \( \frac{dt}{dx} + pt = q \), we identify:
- \( p = 2x \)
- \( q = x^3 \)
### Step 5: Find the Integrating Factor
The integrating factor \( \mu(x) \) is given by:
\[
\mu(x) = e^{\int p \, dx} = e^{\int 2x \, dx} = e^{x^2}
\]
### Step 6: Multiply the Equation by the Integrating Factor
Multiply the entire differential equation by \( e^{x^2} \):
\[
e^{x^2} \frac{dt}{dx} + 2x e^{x^2} t = x^3 e^{x^2}
\]
### Step 7: Integrate Both Sides
The left-hand side can be expressed as the derivative of a product:
\[
\frac{d}{dx}(e^{x^2} t) = x^3 e^{x^2}
\]
Integrating both sides gives:
\[
e^{x^2} t = \int x^3 e^{x^2} \, dx
\]
### Step 8: Solve the Integral
To solve the integral \( \int x^3 e^{x^2} \, dx \), we use integration by parts. Let:
- \( u = x^2 \) and \( dv = x e^{x^2} dx \)
Then, \( du = 2x dx \) and \( v = \frac{1}{2} e^{x^2} \).
Using integration by parts:
\[
\int x^3 e^{x^2} \, dx = \frac{1}{2} x^2 e^{x^2} - \frac{1}{2} \int e^{x^2} (2x) \, dx
\]
This simplifies to:
\[
\int x^3 e^{x^2} \, dx = \frac{1}{2} x^2 e^{x^2} - \int x e^{x^2} \, dx
\]
### Step 9: Final Expression for \( t \)
After integrating and simplifying, we find:
\[
t e^{x^2} = \frac{1}{2} x^2 e^{x^2} + C
\]
Thus,
\[
t = \frac{1}{2} x^2 + C e^{-x^2}
\]
Substituting back \( t = \tan y \):
\[
\tan y = \frac{1}{2} x^2 + C e^{-x^2}
\]
### Step 10: Compare with the Given Expression
The given expression is:
\[
\tan y = c e^{-x^A} + B(x^2 - 1)
\]
From our derived expression, we can compare:
- \( A = 2 \)
- \( B = \frac{1}{2} \)
### Step 11: Calculate \( (AB)^2 \)
Now we calculate \( (AB)^2 \):
\[
A = 2, \quad B = \frac{1}{2} \implies AB = 2 \cdot \frac{1}{2} = 1
\]
Thus,
\[
(AB)^2 = 1^2 = 1
\]
### Final Answer
\[
(AB)^2 = 1
\]
