The solution of the differential equation
`2x^(2)y(dy)/(dx) = tan(x^(2)y^(2))-2xy^(2)`, given `y(1) = sqrt(pi/2)`, is

Options

A. `sin ( x ^(2) y^(2)) -1=0`
B. `cos ((pi)/(2)+x ^(2)y^(2)) +x=0`
C. `sin (x ^(2) y^(2)) =e ^(x-1)`
D. `sin (x ^(2) y ^(2)) =e ^((x-1))`

Answer

To solve the differential equation

\[ 2x^2 y \frac{dy}{dx} = \tan(x^2 y^2) - 2xy^2 \]

with the initial condition \( y(1) = \sqrt{\frac{\pi}{2}} \), we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the given differential equation:

\[ 2x^2 y \frac{dy}{dx} + 2xy^2 = \tan(x^2 y^2) \]

### Step 2: Recognizing the Derivative
Notice that the left-hand side can be recognized as the derivative of \( x^2 y^2 \):

\[ \frac{d}{dx}(x^2 y^2) = 2xy^2 + 2x^2 y \frac{dy}{dx} \]

Thus, we can rewrite the equation as:

\[ \frac{d}{dx}(x^2 y^2) = \tan(x^2 y^2) \]

### Step 3: Integrating Both Sides
Now we integrate both sides:

\[ \int \frac{d}{dx}(x^2 y^2) \, dx = \int \tan(x^2 y^2) \, dx \]

The left-hand side simplifies to:

\[ x^2 y^2 = \int \tan(x^2 y^2) \, dx + C \]

### Step 4: Finding the Integral of \( \tan \)
The integral of \( \tan(u) \) is \( -\log(\cos(u)) \). Thus, we have:

\[ x^2 y^2 = -\log(\cos(x^2 y^2)) + C \]

### Step 5: Applying the Initial Condition
Now we apply the initial condition \( y(1) = \sqrt{\frac{\pi}{2}} \):

Substituting \( x = 1 \) and \( y = \sqrt{\frac{\pi}{2}} \):

\[ 1^2 \left(\sqrt{\frac{\pi}{2}}\right)^2 = -\log(\cos(1^2 \cdot \frac{\pi}{2})) + C \]

This simplifies to:

\[ \frac{\pi}{2} = -\log(0) + C \]

Since \( \cos\left(\frac{\pi}{2}\right) = 0 \), we find that \( C \) must be adjusted accordingly.

### Step 6: Final Form of the Solution
Thus, we can express the solution as:

\[ x^2 y^2 = -\log(\cos(x^2 y^2)) + C \]

After determining \( C \) from the initial condition, we can express the final solution.