Solve the following differential equations
`(1-x^2)(dy)/(dx)-xy=x^2" given that "x=0,y=2`.

Answer

To solve the differential equation \((1-x^2)\frac{dy}{dx} - xy = x^2\) given the initial condition \(x=0, y=2\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ (1-x^2)\frac{dy}{dx} - xy = x^2 \]
We can rearrange this to isolate \(\frac{dy}{dx}\):
\[ (1-x^2)\frac{dy}{dx} = xy + x^2 \]
Now, divide both sides by \(1-x^2\):
\[ \frac{dy}{dx} = \frac{xy + x^2}{1-x^2} \]

### Step 2: Identify \(p\) and \(q\)
This is a linear first-order differential equation of the form \(\frac{dy}{dx} + p(x)y = q(x)\). Here, we can identify:
\[ p(x) = -\frac{x}{1-x^2}, \quad q(x) = \frac{x^2}{1-x^2} \]

### Step 3: Find the integrating factor
The integrating factor \(\mu(x)\) is given by:
\[ \mu(x) = e^{\int p(x) \, dx} = e^{\int -\frac{x}{1-x^2} \, dx} \]
To solve the integral, we can use substitution. Let \(t = 1 - x^2\), then \(dt = -2x \, dx\) or \(-\frac{1}{2} dt = x \, dx\):
\[ \int -\frac{x}{1-x^2} \, dx = \int \frac{1}{2t} \, dt = \frac{1}{2} \ln |t| + C = \frac{1}{2} \ln |1-x^2| + C \]
Thus, the integrating factor is:
\[ \mu(x) = e^{\frac{1}{2} \ln |1-x^2|} = \sqrt{1-x^2} \]

### Step 4: Multiply through by the integrating factor
Multiply the entire equation by the integrating factor:
\[ \sqrt{1-x^2} \frac{dy}{dx} - \frac{x}{\sqrt{1-x^2}} y = \frac{x^2}{\sqrt{1-x^2}} \]

### Step 5: Rewrite the left side as a derivative
The left side can be expressed as the derivative of a product:
\[ \frac{d}{dx} \left( y \sqrt{1-x^2} \right) = \frac{x^2}{\sqrt{1-x^2}} \]

### Step 6: Integrate both sides
Integrate both sides:
\[ \int \frac{d}{dx} \left( y \sqrt{1-x^2} \right) \, dx = \int \frac{x^2}{\sqrt{1-x^2}} \, dx \]
The left side simplifies to:
\[ y \sqrt{1-x^2} \]
For the right side, we can use integration by parts or a trigonometric substitution. The integral \(\int \frac{x^2}{\sqrt{1-x^2}} \, dx\) can be solved using the substitution \(x = \sin(\theta)\):
\[ \int \sin^2(\theta) \, d\theta = \frac{1}{2} \theta - \frac{1}{4} \sin(2\theta) + C = \frac{1}{2} \sin^{-1}(x) - \frac{x}{2}\sqrt{1-x^2} + C \]

### Step 7: Combine results
Thus, we have:
\[ y \sqrt{1-x^2} = \frac{1}{2} \sin^{-1}(x) - \frac{x}{2} \sqrt{1-x^2} + C \]

### Step 8: Solve for \(y\)
Rearranging gives:
\[ y = \frac{1}{2} \frac{\sin^{-1}(x)}{\sqrt{1-x^2}} - \frac{x}{2} + \frac{C}{\sqrt{1-x^2}} \]

### Step 9: Apply initial condition
Now we apply the initial condition \(x=0, y=2\):
\[ 2 = \frac{1}{2} \frac{\sin^{-1}(0)}{\sqrt{1-0^2}} - \frac{0}{2} + \frac{C}{\sqrt{1-0^2}} \implies 2 = 0 + 0 + C \implies C = 2 \]

### Final Solution
Substituting \(C\) back into the equation gives:
\[ y = \frac{1}{2} \frac{\sin^{-1}(x)}{\sqrt{1-x^2}} - \frac{x}{2} + \frac{2}{\sqrt{1-x^2}} \]