Solution of the diff. eqn. `(1-x^(2)) (dy)/(dx) +xy = xy^(2)` is ………..

Answer

To solve the differential equation \((1 - x^2) \frac{dy}{dx} + xy = xy^2\), we will follow these steps:

### Step 1: Rearrange the Equation
First, we rearrange the equation to isolate \(\frac{dy}{dx}\):

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

### Step 2: Divide by \((1 - x^2)\)
Next, we divide the entire equation by \((1 - x^2)\):

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

### Step 3: Factor the Right-Hand Side
Now, we can factor out \(xy\) from the right-hand side:

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

### Step 4: Separate Variables
To separate the variables, we rewrite the equation:

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

### Step 5: Integrate Both Sides
Now, we will integrate both sides. The left side requires partial fraction decomposition:

\[ \frac{1}{y(y - 1)} = \frac{A}{y} + \frac{B}{y - 1} \]

Solving for \(A\) and \(B\):

\[ 1 = A(y - 1) + By \implies A + B = 0, -A = 1 \implies A = -1, B = 1 \]

Thus,

\[ \frac{1}{y(y - 1)} = -\frac{1}{y} + \frac{1}{y - 1} \]

Now we integrate:

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

The left side integrates to:

\[ -\ln|y| + \ln|y - 1| = \ln\left|\frac{y - 1}{y}\right| \]

For the right side, we use the substitution \(u = 1 - x^2\), \(du = -2x dx\):

\[ \int \frac{x}{1 - x^2} dx = -\frac{1}{2} \ln|1 - x^2| \]

### Step 6: Combine Results
Combining both sides, we have:

\[ \ln\left|\frac{y - 1}{y}\right| = -\frac{1}{2} \ln|1 - x^2| + C \]

### Step 7: Exponentiate to Solve for \(y\)
Exponentiating both sides gives:

\[ \frac{y - 1}{y} = K(1 - x^2)^{-1/2} \]

Where \(K = e^C\). Rearranging yields:

\[ y - 1 = Ky(1 - x^2)^{-1/2} \]

### Step 8: Solve for \(y\)
Rearranging gives:

\[ y(1 + K(1 - x^2)^{-1/2}) = 1 \]

Thus,

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

### Final Solution
The final solution of the differential equation is:

\[ y = \frac{1}{1 + K\sqrt{\frac{1}{1 - x^2}}} \]