Interval contained in the domain of definition of non-zero solution of the differential equation `(x-3)^(2)y'+y=0` is:

Options

A. `R-{3}`
B. `((pi)/(2), (3pi)/(2))`
C. `((pi)/(8),(5pi)/(4))`
D. `(-pi, pi)`

Answer

To solve the differential equation \((x-3)^{2}y' + y = 0\) and find the interval contained in the domain of definition of its non-zero solution, we can follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the given equation:
\[ (x-3)^{2}y' + y = 0 \]
This can be rewritten as:
\[ (x-3)^{2} \frac{dy}{dx} = -y \]

### Step 2: Separate Variables
We can separate the variables \(y\) and \(x\):
\[ \frac{dy}{y} = -\frac{dx}{(x-3)^{2}} \]

### Step 3: Integrate Both Sides
Now we integrate both sides:
\[ \int \frac{dy}{y} = \int -\frac{dx}{(x-3)^{2}} \]
The left side integrates to:
\[ \log |y| = -\frac{1}{x-3} + C \]
where \(C\) is the constant of integration.

### Step 4: Solve for \(y\)
Exponentiating both sides to solve for \(y\):
\[ |y| = e^{-\frac{1}{x-3} + C} = e^{C} e^{-\frac{1}{x-3}} \]
Let \(k = e^{C}\), then:
\[ y = k e^{-\frac{1}{x-3}} \]
Since we are looking for non-zero solutions, \(k\) must be non-zero.

### Step 5: Determine the Domain of the Solution
The term \(e^{-\frac{1}{x-3}}\) is defined for all \(x\) except where the denominator is zero. Thus, \(x - 3 = 0\) or \(x = 3\) is a point of discontinuity. Therefore, the solution is defined for all real numbers except \(x = 3\).

### Final Answer
The interval contained in the domain of definition of the non-zero solution of the differential equation is:
\[ (-\infty, 3) \cup (3, \infty) \]