Question

The particular solution of the differential equation \( x \, dy + 2y \, dx = 0 \), when \( x = 2 \) and \( y = 1 \) is

• A. \( x y^2 = 4 \)
• B. \( x^2 y = 4 \)
• C. \( x^2 y = -4 \)
• D. \( x y^2 = -4 \)

Hint:

When solving differential equations, always check the initial conditions to find the particular solution.

Answer 1

The Correct Option is B

Step 1: Solve the differential equation.
The given equation is: \[ x \, dy + 2y \, dx = 0 \] We can separate the variables and solve: \[ \frac{dy}{y} = -\frac{2}{x} \, dx \]
Step 2: Integrate both sides.
Integrating both sides gives: \[ \int \frac{dy}{y} = \int -\frac{2}{x} \, dx \] This leads to: \[ \ln |y| = -2 \ln |x| + C \] Simplifying: \[ \ln |y| = \ln |x^{-2}| + C \] \[ y = \frac{C}{x^2} \]
Step 3: Apply the initial conditions.
Given \( x = 2 \) and \( y = 1 \), we substitute these values into the equation: \[ 1 = \frac{C}{2^2} \quad \Rightarrow \quad C = 4 \]
Step 4: Final solution.
Thus, the particular solution is: \[ y = \frac{4}{x^2} \] Multiplying both sides by \( x^2 \), we get: \[ x^2 y = 4 \] Therefore, the correct answer is (B).