If integrating factor of `x(1-x^2)dy+(2x^2y-y-a x^3)dx=0` is `e^(intp dx ,)` then `P` is equal to

Options

A. `(2x^(2)-ax^(2))/(x(1-x^(2))`
B. `2x^(3)-1`
C. `(2x^(2)-a)/(ax^(3))`
D. `(2x^(2)-1)/(x(1-x^(2))`

Explanation

To find the value of \( P \) in the given differential equation, we start with the equation:

\[ x(1-x^2)dy + (2x^2y - y - ax^3)dx = 0 \]

### Step 1: Rearranging the Equation
We can rewrite the equation in the form of \( \frac{dy}{dx} \):

\[ x(1-x^2) \frac{dy}{dx} + (2x^2y - y - ax^3) = 0 \]

Dividing through by \( dx \):

\[ \frac{dy}{dx} = -\frac{2x^2y - y - ax^3}{x(1-x^2)} \]

### Step 2: Simplifying the Right Side
Now we simplify the right-hand side:

\[ \frac{dy}{dx} = -\frac{(2x^2 - 1)y + ax^3}{x(1-x^2)} \]

### Step 3: Identifying \( P \) and \( Q \)
This equation can be compared to the standard form of a linear differential equation:

\[ \frac{dy}{dx} + P(x)y = Q(x) \]

From our equation, we can identify:

\[ P(x) = \frac{2x^2 - 1}{x(1-x^2)} \]

### Step 4: Finding the Integrating Factor
The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int P(x) \, dx} \]

### Step 5: Conclusion
Thus, we need to find \( P \):

\[ P = \frac{2x^2 - 1}{x(1-x^2)} \]

### Final Answer
The value of \( P \) is:

\[ P = \frac{2x^2 - 1}{x(1-x^2)} \]

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