Question

Consider the differential equation \( L[y] = (y - y^2)dx + xdy = 0. \) The function \( f(x, y) \) is said to be an integrating factor of the equation if \( f(x, y)L[y] = 0 \) becomes exact. If \( f(x, y) = \dfrac{1}{x^2 y^2}, \) then
 

• A. \( f \) is an integrating factor and \( y = 1 - kxy, \, k \in \mathbb{R} \) is NOT its general solution
• B. \( f \) is an integrating factor and \( y = -1 + kxy, \, k \in \mathbb{R} \) is its general solution
• C. \( f \) is an integrating factor and \( y = -1 + kxy, \, k \in \mathbb{R} \) is NOT its general solution
• D. \( f \) is NOT an integrating factor and \( y = 1 + kxy, \, k \in \mathbb{R} \) is its general solution

Hint:

For an integrating factor, always test exactness using \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). Then integrate step by step to find the potential function.

Answer 1

The Correct Option is C

Step 1: Write the given equation. 
\[ L[y] = (y - y^2)dx + xdy = 0. \] Let \( M = y - y^2 \) and \( N = x. \)

Step 2: Multiply by the integrating factor \( f(x, y) = \dfrac{1}{x^2 y^2}. \)
Then the equation becomes: \[ \dfrac{y - y^2}{x^2 y^2}dx + \dfrac{x}{x^2 y^2}dy = 0. \] \[ \Rightarrow \left( \dfrac{1}{x^2 y} - \dfrac{1}{x^2} \right) dx + \dfrac{1}{x y^2} dy = 0. \]

Step 3: Check for exactness. 
Compute partial derivatives: \[ \frac{\partial M}{\partial y} = -\frac{1}{x^2 y^2}, \frac{\partial N}{\partial x} = -\frac{1}{x^2 y^2}. \] Hence, \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}, \) so the equation is exact.

Step 4: Integrate to find the solution. 
Integrate \( M \) w.r.t \( x \): \[ \psi(x, y) = \int \left( \frac{1}{x^2 y} - \frac{1}{x^2} \right) dx = -\frac{1}{xy} + \frac{1}{x} + h(y). \] Differentiate w.r.t \( y \) and compare with \( N \): \[ \frac{\partial \psi}{\partial y} = \frac{1}{x y^2} + h'(y) = \frac{1}{x y^2} \Rightarrow h'(y) = 0. \] Thus, \( h(y) = \text{constant}. \)

Step 5: General solution. 
\[ -\frac{1}{xy} + \frac{1}{x} = c \Rightarrow y = -1 + kxy, \, k \in \mathbb{R}. \]

Final Answer: \( f \) is an integrating factor and \( y = -1 + kxy \) is its general solution.