Solution of differential equation `(dy ) /( dx) +(x ) /( 1 - x^2) y= x sqrt(y) ` is

Options

A. `3sqrt(y) =( 1- x^2) + C (1- x^2 ) ^(1//4)`
B. `3 sqrt(y) =- (1-x^2) +c (1-x ^2)`
C. ` 3 sqrt( y) =(1-x^2) + (1- x^(2))`
D. none of these

Answer

To solve the differential equation

\[ \frac{dy}{dx} + \frac{x}{1 - x^2} y = x \sqrt{y}, \]

we will follow these steps:

### Step 1: Rewrite the equation
We can rewrite the equation in the standard form:

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

where \( P(x) = \frac{x}{1 - x^2} \) and \( Q(x) = x \sqrt{y} \).

### Step 2: Identify the integrating factor
The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{x}{1 - x^2} \, dx}. \]

### Step 3: Compute the integral
To compute the integral \( \int \frac{x}{1 - x^2} \, dx \), we can use the substitution \( t = 1 - x^2 \), which gives us \( dt = -2x \, dx \) or \( dx = -\frac{dt}{2x} \). Therefore, we have:

\[ \int \frac{x}{1 - x^2} \, dx = -\frac{1}{2} \int \frac{1}{t} \, dt = -\frac{1}{2} \ln |t| + C = -\frac{1}{2} \ln |1 - x^2| + C. \]

Thus, the integrating factor becomes:

\[ \mu(x) = e^{-\frac{1}{2} \ln |1 - x^2|} = |1 - x^2|^{-\frac{1}{2}}. \]

### Step 4: Multiply through by the integrating factor
Now we multiply the entire differential equation by the integrating factor:

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

### Step 5: Solve the equation
This can be rewritten as:

\[ \frac{d}{dx}\left(y \cdot \frac{1}{\sqrt{1 - x^2}}\right) = \frac{x \sqrt{y}}{\sqrt{1 - x^2}}. \]

Integrating both sides with respect to \( x \):

\[ y \cdot \frac{1}{\sqrt{1 - x^2}} = \int \frac{x \sqrt{y}}{\sqrt{1 - x^2}} \, dx + C. \]

### Step 6: Solve for \( y \)
Now, we can express \( y \) in terms of \( x \):

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

### Final Solution
This gives us the implicit solution of the differential equation. To find an explicit solution, we would need to solve the integral and isolate \( y \).