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: Verify that $f(x,y)$ is an integrating factor

Multiply the equation by $f(x,y)$:

$$\frac{1}{x^2y^2}[(y - y^2)dx + xdy] = 0$$

$$\left(\frac{1}{x^2y} - \frac{1}{x^2}\right)dx + \frac{1}{xy^2}dy = 0$$

Step 2: Check exactness

For the equation $Mdx + Ndy = 0$ where:

• $M = \frac{1}{x^2y} - \frac{1}{x^2}$
• $N = \frac{1}{xy^2}$

Compute partial derivatives:

$$\frac{\partial M}{\partial y} = -\frac{1}{x^2y^2}$$

$$\frac{\partial N}{\partial x} = -\frac{1}{x^2y^2}$$

Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is exact, confirming that $f(x,y) = \frac{1}{x^2y^2}$ is an integrating factor.

Step 3: Find the general solution

For an exact equation, the solution $\phi(x,y) = C$ satisfies:

$$\phi = \int M dx = \int \left(\frac{1}{x^2y} - \frac{1}{x^2}\right)dx = -\frac{1}{xy} + \frac{1}{x} + g(y)$$

Using $\frac{\partial \phi}{\partial y} = N$:

$$\frac{1}{xy^2} + g'(y) = \frac{1}{xy^2}$$

Therefore $g'(y) = 0$, giving $\phi = -\frac{1}{xy} + \frac{1}{x} = C$

Step 4: Simplify the solution

$$\frac{1}{x}\left(1 - \frac{1}{y}\right) = C$$

$$1 - \frac{1}{y} = Cx$$

$$\frac{1}{y} = 1 - Cx$$

$$y = \frac{1}{1 - Cx}$$

Setting $k = -C$:

$$y = \frac{1}{1 + kx}$$

This can be written in implicit form as $y(1 + kx) = 1$, which gives:

$$y + kxy = 1$$

$$y = 1 - kxy$$

Or equivalently: $y = 1 + kxy$ (with appropriate sign of $k$)

Conclusion:

The general solution is $y = 1 + kxy$ where $k \in \mathbb{R}$.

Answer: (C) $f$ is an integrating factor and $y = -1 + kxy, k \in \mathbb{R}$ is NOT its general solution