Question

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

• A. 1/e
• B. 0
• C. 1/e2
• D. e2

Answer 1

The Correct Option is B

We are given the differential equation: \[ y(x + 1) \, dx - x^2 \, dy = 0, \quad y(1) = e. \] We need to solve this equation and find \( \lim_{x \to 0} f(x) \).
 Step 1: Rearranging the differential equation.
Rearrange the equation to separate the variables: \[ \frac{dy}{dx} = \frac{y(x + 1)}{x^2}. \] Thus, we have: \[ \frac{dy}{y} = \frac{x + 1}{x^2} \, dx. \] Step 2: Integrating both sides.
Integrating both sides:
\[ \int \frac{1}{y} \, dy = \int \frac{x + 1}{x^2} \, dx. \] The integral of \( \frac{1}{y} \) is \( \ln |y| \), and the integral of \( \frac{x + 1}{x^2} \) is: \[ \frac{x + 1}{x^2} = \frac{1}{x} + \frac{1}{x^2}. \] Thus, we integrate: \[ \ln |y| = \int \left( \frac{1}{x} + \frac{1}{x^2} \right) \, dx = \ln |x| - \frac{1}{x} + C. \] 

 Step 3: Solving for \( y \).
Now, exponentiate both sides to solve for \( y \):
\[ |y| = e^{\ln |x| - \frac{1}{x} + C} = e^{\ln |x|} \cdot e^{-\frac{1}{x}} \cdot e^C = C_1 x e^{-\frac{1}{x}}, \] where \( C_1 = e^C \) is a constant. Thus, the solution is: \[ y = C_1 x e^{-\frac{1}{x}}. \] 

Step 4: Applying the initial condition.
We are given that \( y(1) = e \), so substitute \( x = 1 \) into the equation:
\[ e = C_1 \cdot 1 \cdot e^{-\frac{1}{1}} = C_1 e^{-1}. \] Thus: \[ C_1 = e^2. \] 

Step 5: Final solution for \( y \).
The solution for \( y \) is: \[ y = e^2 x e^{-\frac{1}{x}}. \]

Step 6: Finding \( \lim_{x \to 0} f(x) \).
Now, we compute the limit as \( x \to 0 \): \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} e^2 x e^{-\frac{1}{x}}. \] Since \( e^{-\frac{1}{x}} \) tends to 0 very rapidly as \( x \to 0 \), the product \( e^2 x e^{-\frac{1}{x}} \) tends to 0. Thus, \[ \lim_{x \to 0} f(x) = 0. \] Hence, the correct answer is option (2).