If the solution of differential equation ` ( dy )/(dx) =1 + x +y^2 + xy ^2 ` where Y (0) =0 is ` Y= tan ( x + ( x^2)/( a))`, then a is _______

Answer

To solve the differential equation given by

\[ \frac{dy}{dx} = 1 + x + y^2 + xy^2 \]

with the initial condition \( Y(0) = 0 \), and to find the value of \( a \) in the solution

\[ Y = \tan\left(x + \frac{x^2}{a}\right), \]

we will follow these steps:

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

\[ \frac{dy}{dx} = 1 + x + y^2 + xy^2. \]

We can factor out \( y^2 \):

\[ \frac{dy}{dx} = 1 + x + y^2(1 + x). \]

### Step 2: Separate variables
We can rearrange the equation to separate variables:

\[ \frac{dy}{1 + y^2} = (1 + x)dx. \]

### Step 3: Integrate both sides
Now we integrate both sides:

\[ \int \frac{dy}{1 + y^2} = \int (1 + x)dx. \]

The left side integrates to:

\[ \tan^{-1}(y) + C_1, \]

and the right side integrates to:

\[ x + \frac{x^2}{2} + C_2. \]

Combining the constants \( C_1 \) and \( C_2 \) into a single constant \( C \), we have:

\[ \tan^{-1}(y) = x + \frac{x^2}{2} + C. \]

### Step 4: Apply the initial condition
Using the initial condition \( Y(0) = 0 \):

\[ \tan^{-1}(0) = 0 + \frac{0^2}{2} + C \implies 0 = C. \]

Thus, we have:

\[ \tan^{-1}(y) = x + \frac{x^2}{2}. \]

### Step 5: Solve for \( y \)
Taking the tangent of both sides gives:

\[ y = \tan\left(x + \frac{x^2}{2}\right). \]

### Step 6: Compare with the given solution
The problem states that the solution is

\[ Y = \tan\left(x + \frac{x^2}{a}\right). \]

From our derived solution, we see that:

\[ \frac{x^2}{2} = \frac{x^2}{a}. \]

### Step 7: Solve for \( a \)
Equating the coefficients gives:

\[ a = 2. \]

### Final Answer
Thus, the value of \( a \) is

\[ \boxed{2}. \]