What is the integrating factor of the differential equation `(1-y^(2))(dx)/(dy) +yx=ay` `(-1 lt ylt1)`?

Explanation

To find the integrating factor of the differential equation \((1 - y^2) \frac{dx}{dy} + yx = ay\) for \(-1 < y < 1\), we can follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the given differential equation:
\[ (1 - y^2) \frac{dx}{dy} + yx = ay \]
We can rewrite this in a standard form by dividing through by \(1 - y^2\):
\[ \frac{dx}{dy} + \frac{yx}{1 - y^2} = \frac{ay}{1 - y^2} \]
This is now in the form:
\[ \frac{dx}{dy} + P(y)x = Q(y) \]
where \(P(y) = \frac{y}{1 - y^2}\) and \(Q(y) = \frac{ay}{1 - y^2}\).

### Step 2: Identify \(P(y)\)
From the rewritten equation, we have:
\[ P(y) = \frac{y}{1 - y^2} \]

### Step 3: Find the Integrating Factor
The integrating factor \(\mu(y)\) is given by:
\[ \mu(y) = e^{\int P(y) \, dy} \]
Substituting \(P(y)\):
\[ \mu(y) = e^{\int \frac{y}{1 - y^2} \, dy} \]

### Step 4: Evaluate the Integral
To evaluate the integral \(\int \frac{y}{1 - y^2} \, dy\), we can use the substitution:
\[ t = 1 - y^2 \quad \Rightarrow \quad dt = -2y \, dy \quad \Rightarrow \quad dy = -\frac{dt}{2y} \]
Thus, we rewrite \(y\) in terms of \(t\):
\[ y = \sqrt{1 - t} \]
Now substituting:
\[ \int \frac{y}{1 - y^2} \, dy = \int \frac{y}{t} \left(-\frac{dt}{2y}\right) = -\frac{1}{2} \int \frac{dt}{t} = -\frac{1}{2} \ln |t| + C \]
Substituting back for \(t\):
\[ -\frac{1}{2} \ln |1 - y^2| + C \]

### Step 5: Write the Integrating Factor
Thus, we have:
\[ \mu(y) = e^{-\frac{1}{2} \ln |1 - y^2|} = \frac{1}{\sqrt{1 - y^2}} \]

### Final Result
The integrating factor for the given differential equation is:
\[ \mu(y) = \frac{1}{\sqrt{1 - y^2}} \]