Step 1: Understanding the Concept:
The given differential equation is of a first-order non-linear type. However, the presence of the term \( x^2 y^2 \) inside the tangent function suggests a substitution technique. If we can find a variable \( v = x^2 y^2 \), we might be able to simplify the equation into a separable form.
Step 2: Key Formula or Approach:
1. Let \( v = x^2 y^2 \).
2. Differentiate \( v \) with respect to \( x \): \( \frac{dv}{dx} = 2xy^2 + x^2 \cdot 2y \frac{dy}{dx} \).
3. Substitute \( x^2 \cdot 2y \frac{dy}{dx} \) from the original equation into the expression for \( \frac{dv}{dx} \).
Step 3: Detailed Explanation:
From \( v = x^2 y^2 \), we have \( \frac{dv}{dx} = 2xy^2 + 2x^2 y \frac{dy}{dx} \).
Rearranging, \( 2x^2 y \frac{dy}{dx} = \frac{dv}{dx} - 2xy^2 \).
Substitute this into the original differential equation:
\[ \frac{dv}{dx} - 2xy^2 = \tan(v) - 2xy^2 \]
The term \( -2xy^2 \) cancels out from both sides, leaving:
\[ \frac{dv}{dx} = \tan(v) \]
This is now a simple variable-separable differential equation. Rearranging the terms:
\[ \frac{dv}{\tan v} = dx \implies \cot v \, dv = dx \]
Integrating both sides:
\[ \int \cot v \, dv = \int dx \]
\[ \ln|\sin v| = x + C \]
Substituting \( v = x^2 y^2 \) back:
\[ \ln|\sin(x^2 y^2)| = x + C \implies \sin(x^2 y^2) = e^{x+C} = A e^x \]
Now, apply the initial condition \( y(1) = \sqrt{\frac{\pi}{2}} \):
At \( x = 1 \), \( x^2 y^2 = (1)^2 (\frac{\pi}{2}) = \frac{\pi}{2} \).
\[ \ln|\sin(\frac{\pi}{2})| = 1 + C \implies \ln(1) = 1 + C \implies 0 = 1 + C \implies C = -1 \]
The final solution is:
\[ \ln(\sin(x^2 y^2)) = x - 1 \implies \sin(x^2 y^2) = e^{x-1} \]
Step 4: Final Answer:
By substitution and integration, the solution is \( \sin(x^2 y^2) = e^{x-1} \).