The integrating factor of the differential equation `(dy)/(dx)+(y)/((1-x)sqrt(x))=1-sqrt(x)`, is

Options

A. `(1-sqrt(x))/(1+sqrt(x))`
B. `(1+sqrt(x))/(1-sqrt(x))`
C. `(1-x)/(1+x)`
D. `(sqrt(x))/(1-sqrt(x))`

Explanation

To find the integrating factor of the given differential equation

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

we will follow these steps:

### Step 1: Identify \( p(x) \)
The standard form of a first-order linear differential equation is

\[ \frac{dy}{dx} + p(x)y = q(x). \]

From the given equation, we can identify:

\[ p(x) = \frac{1}{(1-x)\sqrt{x}}. \]

### Step 2: Calculate the Integrating Factor
The integrating factor \( \mu(x) \) is given by the formula:

\[ \mu(x) = e^{\int p(x) \, dx}. \]

Substituting \( p(x) \):

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

### Step 3: Substitute \( \sqrt{x} = t \)
To simplify the integral, we can substitute \( \sqrt{x} = t \). Then, \( x = t^2 \) and \( dx = 2t \, dt \). The expression for \( p(x) \) becomes:

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

### Step 4: Integrate
Now we need to integrate:

\[ \int \frac{2}{1-t^2} \, dt. \]

This can be solved using the formula for the integral of the form \( \frac{1}{1-u^2} \):

\[ \int \frac{1}{1-u^2} \, du = \frac{1}{2} \ln \left| \frac{1+u}{1-u} \right| + C. \]

Thus,

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

### Step 5: Find the Integrating Factor
Now substituting back into the expression for the integrating factor:

\[ \mu(x) = e^{\ln \left| \frac{1+\sqrt{x}}{1-\sqrt{x}} \right|} = \frac{1+\sqrt{x}}{1-\sqrt{x}}. \]

### Final Result
Thus, the integrating factor of the differential equation is:

\[ \mu(x) = \frac{1+\sqrt{x}}{1-\sqrt{x}}. \]

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