Let `x(1-x)\ dy/dx = x-y`

Options

A. a) General solution of given differential equation is `xy = (1-x) ln|1-x| + 1+c(1-x)`
B. b) General solution of given differential equation is `xy = (1-x) ln|1-x|-1+cx(1-x)`
C. c) If `y = f(x)` is a solution of given differential equation, then `underset(x rarr1)lim f(x)` does not exist
D. d) If `y = f(x)` is solution of given differential equation then `underset(x rarr 1)lim f(x) = 1`

Answer

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

### Step 1: Rewrite the equation
We start with the given equation:
\[ x(1-x) \frac{dy}{dx} = x - y \]
We can rearrange this to isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{x - y}{x(1-x)} \]

### Step 2: Separate variables
Next, we can separate the variables \(y\) and \(x\):
\[ \frac{dy}{dx} + \frac{y}{x(1-x)} = \frac{1}{1-x} \]
This is now in the standard form of a linear first-order differential equation:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
where \(P(x) = \frac{1}{x(1-x)}\) and \(Q(x) = \frac{1}{1-x}\).

### Step 3: Find the integrating factor
The integrating factor \(IF\) is given by:
\[ IF = e^{\int P(x) \, dx} = e^{\int \frac{1}{x(1-x)} \, dx} \]
To compute this integral, we can use partial fraction decomposition:
\[ \frac{1}{x(1-x)} = \frac{A}{x} + \frac{B}{1-x} \]
Solving for \(A\) and \(B\), we find:
\[ 1 = A(1-x) + Bx \]
Setting \(x = 0\) gives \(A = 1\), and setting \(x = 1\) gives \(B = 1\). Thus:
\[ \frac{1}{x(1-x)} = \frac{1}{x} + \frac{1}{1-x} \]
Now we can integrate:
\[ \int \frac{1}{x} \, dx + \int \frac{1}{1-x} \, dx = \ln|x| - \ln|1-x| = \ln\left|\frac{x}{1-x}\right| \]
Thus, the integrating factor is:
\[ IF = e^{\ln\left|\frac{x}{1-x}\right|} = \frac{x}{1-x} \]

### Step 4: Multiply through by the integrating factor
Now we multiply the entire differential equation by the integrating factor:
\[ \frac{x}{1-x} \frac{dy}{dx} + \frac{y}{1-x} = \frac{x}{(1-x)^2} \]

### Step 5: Integrate both sides
The left-hand side can be rewritten as:
\[ \frac{d}{dx}\left(\frac{xy}{1-x}\right) = \frac{x}{(1-x)^2} \]
Integrating both sides gives:
\[ \frac{xy}{1-x} = \int \frac{x}{(1-x)^2} \, dx \]
Using integration by parts or substitution, we find:
\[ \int \frac{x}{(1-x)^2} \, dx = -\frac{1}{1-x} + C \]
Thus we have:
\[ \frac{xy}{1-x} = -\frac{1}{1-x} + C \]

### Step 6: Solve for \(y\)
Multiplying through by \(1-x\) gives:
\[ xy = -1 + C(1-x) \]
Rearranging gives:
\[ xy = C(1-x) - 1 \]

### Final Form
Thus, the general solution of the differential equation is:
\[ xy = (C - 1)(1-x) \]