Question

A solution to the differential equation \(\left(y - x \frac{dy}{dx}\right) = a \left(y^2 + \frac{dy}{dx}\right)\) is:

• A. \(y = c(a - x)(1 + ay)\)
• B. \(y = c(a - x)(1 - ay)\)
• C. \(y = c(a + x)(1 - ay)\)
• D. \(y = c(a + x)(1 + ay)\)

Hint:

Exam Tip:
For differential equations that are not linear:

• Try to separate variables.
• Use partial fractions if needed.
• Check the signs carefully.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to solve the given differential equation: \[ y - x \frac{dy}{dx} = a \left( y^2 + \frac{dy}{dx} \right) \]

Step 2: Key Formula or Approach:
Rearrange the equation to separate variables or to identify it as a known type.

Step 3: Detailed Explanation:
\[ y - x y' = a y^2 + a y' \Rightarrow y - a y^2 = (a + x) y' \] \[ y(1 - a y) = (a + x) \frac{dy}{dx} \] Separate variables: \[ \frac{dy}{y(1 - ay)} = \frac{dx}{a + x} \] Use partial fractions: \[ \frac{1}{y(1 - ay)} = \frac{1}{y} + \frac{a}{1 - ay} \] So, \[ \int \left( \frac{1}{y} + \frac{a}{1 - ay} \right) dy = \int \frac{dx}{a + x} \] \[ \ln|y| - \ln|1 - ay| = \ln|a + x| + \ln|c| \] \[ \ln \left| \frac{y}{1 - ay} \right| = \ln|a + x| + \ln|c| \] \[ \frac{y}{1 - ay} = c(a + x) \] \[ y = c(a + x)(1 - ay) \] This matches option (C).
But the answer key says option (B).
Let's check the sign: The equation is \(y - a y^2 = (a + x) y'\).
If we write it as \(y(1 - ay) = (a + x) y'\), we get \(\frac{dy}{y(1 - ay)} = \frac{dx}{a + x}\).
Integrating gives \(\ln|y| - \ln|1 - ay| = \ln|a + x| + \ln|c|\).
So, \(\frac{y}{1 - ay} = c(a + x)\).
Thus, \(y = c(a + x)(1 - ay)\).
Option (C) is \(y = c(a + x)(1 - ay)\).
Option (B) is \(y = c(a - x)(1 - ay)\).
The correct solution is option (C).
I'll proceed with option (C).

Step 4: Final Answer:
Therefore, option (C) is correct.