Question

The differential equation is \[ \frac{dy}{dx}+\frac{y}{x}=0 \] and \(y(1)=2\). Then the value of \(y(3)\) is

• A. \(2\)
• B. \(3\)
• C. \(\frac{2}{3}\)
• D. \(1\)

Hint:

For equations of the form \(\frac{dy}{dx}+\frac{y}{x}=0\), the solution is usually of the form \(xy=C\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem involves solving a first-order ordinary differential equation and then applying a given initial condition to find a specific value of the function.
The equation is \(\frac{dy}{dx} + \frac{y}{x} = 0\), which can be solved using the variable separable method.
Step 2: Key Formula or Approach:
Rearrange the differential equation to separate variables \(x\) and \(y\).
Integrate both sides to find the general solution.
Use the initial condition \(y(1) = 2\) to find the constant of integration.
Substitute \(x = 3\) into the particular solution to find \(y(3)\).
Step 3: Detailed Explanation:

Given differential equation: \(\frac{dy}{dx} = -\frac{y}{x}\).

Separate the variables:
\[ \frac{1}{y} dy = -\frac{1}{x} dx \]

Integrate both sides:
\[ \int \frac{1}{y} dy = -\int \frac{1}{x} dx \]
\[ \ln|y| = -\ln|x| + \ln|c| \]

Using logarithmic properties (\(\ln a + \ln b = \ln(ab)\)):
\[ \ln|y| + \ln|x| = \ln|c| \]
\[ \ln|xy| = \ln|c| \]
\[ xy = c \] ... (1)

Apply the initial condition: \(y(1) = 2\), which means when \(x = 1\), \(y = 2\).
\[ (1)(2) = c \implies c = 2 \]

The particular solution is:
\[ xy = 2 \implies y = \frac{2}{x} \]

To find \(y(3)\), substitute \(x = 3\) into the particular solution:
\[ y(3) = \frac{2}{3} \]

Step 4: Final Answer:
The solution to the differential equation with the given initial condition is \(y = 2/x\). Evaluating at \(x=3\) gives \(2/3\).