Question

Let \( \varphi : \mathbb{R} \to \mathbb{R} \) be the solution of the differential equation \[ x \frac{dy}{dx} = (y - 1)(y - 3), \] satisfying \( \varphi(0) = 2 \). Then, which one of the following is TRUE?

• A. \( \lim_{x \to \infty} \varphi(x) = 0 \)
• B. \( \lim_{x \to \ln \sqrt{2}} \varphi(x) = 1 \)
• C. \( \lim_{x \to -\infty} \varphi(x) = 3 \)
• D. \( \lim_{x \to \ln \frac{1}{\sqrt{2}}} \varphi(x) = 6 \)

Hint:

When solving a separable differential equation, always use partial fractions to simplify the integrals. Use initial conditions to find constants of integration.

Answer 1

The Correct Option is C

Step 1: Rewrite the Differential Equation.
We are given the equation: \[ x \frac{dy}{dx} = (y - 1)(y - 3) \] Rewriting this, we separate the variables: \[ \frac{dy}{(y - 1)(y - 3)} = \frac{dx}{x} \] Step 2: Integrate both sides.
To integrate the left-hand side, use partial fractions: \[ \frac{1}{(y - 1)(y - 3)} = \frac{A}{y - 1} + \frac{B}{y - 3} \] Solving for \( A \) and \( B \), we get: \[ \frac{1}{(y - 1)(y - 3)} = \frac{1}{2(y - 1)} - \frac{1}{2(y - 3)} \] Thus, the integral becomes: \[ \int \left( \frac{1}{2(y - 1)} - \frac{1}{2(y - 3)} \right) dy = \int \frac{dx}{x} \] Integrating both sides: \[ \frac{1}{2} \ln \left| \frac{y - 1}{y - 3} \right| = \ln |x| + C \] Step 3: Solve for \( \varphi(x) \).
Exponentiating both sides gives: \[ \left| \frac{y - 1}{y - 3} \right| = Cx \] This simplifies to: \[ \frac{y - 1}{y - 3} = Cx \] Solving for \( y \), we get: \[ y = \frac{3Cx + 1}{Cx + 1} \] Step 4: Use the initial condition.
Using the initial condition \( \varphi(0) = 2 \), we substitute \( x = 0 \) and \( y = 2 \): \[ 2 = \frac{1}{1} \] Thus, \( C = 1 \). Therefore, the solution is: \[ y = \frac{3x + 1}{x + 1} \] Step 5: Find the limit as \( x \to -\infty \).
As \( x \to -\infty \), the limit of \( y \) is: \[ \lim_{x \to -\infty} \varphi(x) = 3 \] Final Answer: \[ \boxed{\lim_{x \to -\infty} \varphi(x) = 3} \]