Let `y=f(x)` be the solution of the differential equation `y(x+1)dx-x^(2)dy=0,y(1)=e`.Then `lim_(x rarr0^(-))f(x)` is equal to

Options

A. `(1)/(e^(2))`
B. `(1)/(e)`
C. `0`
D. `e^(2)`

Answer

To solve the differential equation \( y(x+1)dx - x^2 dy = 0 \) with the initial condition \( y(1) = e \), we will follow these steps:

### Step 1: Rearranging the Differential Equation
We start with the given equation:
\[ y(x+1)dx - x^2 dy = 0 \]
Rearranging gives us:
\[ y(x+1)dx = x^2 dy \]
Dividing both sides by \( y(x+1) \) and \( x^2 \):
\[ \frac{dy}{y} = \frac{x+1}{x^2} dx \]

### Step 2: Integrating Both Sides
Now we integrate both sides:
\[ \int \frac{dy}{y} = \int \frac{x+1}{x^2} dx \]
The left side integrates to:
\[ \ln |y| + C_1 \]
For the right side, we can split the fraction:
\[ \int \left( \frac{1}{x} + \frac{1}{x^2} \right) dx = \ln |x| - \frac{1}{x} + C_2 \]
Thus, we have:
\[ \ln |y| = \ln |x| - \frac{1}{x} + C \]
where \( C = C_2 - C_1 \).

### Step 3: Exponentiating to Solve for \( y \)
Exponentiating both sides gives:
\[ y = e^{\ln |x| - \frac{1}{x} + C} = |x| e^{-\frac{1}{x}} e^C \]
Let \( k = e^C \), so:
\[ y = k |x| e^{-\frac{1}{x}} \]

### Step 4: Applying the Initial Condition
Using the initial condition \( y(1) = e \):
\[ e = k \cdot 1 \cdot e^{-1} \]
Thus, we find:
\[ k = e^2 \]
So the function becomes:
\[ y = e^2 x e^{-\frac{1}{x}} \]

### Step 5: Finding the Limit as \( x \to 0^- \)
Now we need to find:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} e^2 x e^{-\frac{1}{x}} \]
As \( x \to 0^- \), \( e^{-\frac{1}{x}} \) approaches \( 0 \) (since \( -\frac{1}{x} \to -\infty \)). Therefore:
\[ \lim_{x \to 0^-} e^2 x e^{-\frac{1}{x}} = e^2 \cdot 0 = 0 \]

### Final Answer
Thus, we conclude:
\[ \lim_{x \to 0^-} f(x) = 0 \]