Question

The solution of the differential equation $x\frac{dy}{dx} + y = 0$ passing through the point (1,1) is $y =$

• A. $x^2$
• B. $x^{-1}$
• C. $x^{-2}$
• D. $x$

Hint:

$x\frac{dy}{dx} + y = 0$ is just the expansion of $\frac{d}{dx}(xy) = 0$. This means $xy = c$ immediately!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To solve this first-order differential equation, we can use the method of separation of variables. This involves rearranging the equation so that all \(y\) terms are on one side and all \(x\) terms are on the other.
Step 2: Key Formula or Approach:
1. Separate variables: \(\frac{dy}{y} = -\frac{dx}{x}\).
2. Integrate both sides: \(\int \frac{1}{y} \, dy = -\int \frac{1}{x} \, dx\).
3. Use the given point (1,1) to find the constant of integration \(C\).
Step 3: Detailed Explanation:
Starting with \( x \frac{dy}{dx} = -y \): \[ \frac{dy}{y} = -\frac{dx}{x} \] Integrating both sides: \[ \ln|y| = -\ln|x| + \ln|C| \] \[ \ln|y| + \ln|x| = \ln|C| \] \[ \ln|xy| = \ln|C| \implies xy = C \] Substitute the point (1,1): \[ (1)(1) = C \implies C = 1 \] Thus, the specific solution is \( xy = 1 \), which can be written as \( y = \frac{1}{x} \) or \( y = x^{-1} \).
Step 4: Final Answer:
The solution is \( y = x^{-1} \).