Understanding Integrating Factors for Differential Equations

An integrating factor is a function by which a non-exact differential equation can be multiplied to make it exact. A first-order differential equation of the form \( M(x, y)dx + N(x, y)dy = 0 \) is exact if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). If it's not exact, we look for an integrating factor \( \mu(x, y) \) such that \( \mu M dx + \mu N dy = 0 \) is exact, meaning \( \frac{\partial (\mu M)}{\partial y} = \frac{\partial (\mu N)}{\partial x} \).

There are several methods to find integrating factors, including checking if \( \frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right) \) is a function of \( x \) alone (then IF is \( e^{\int f(x)dx} \)), or if \( \frac{1}{M}\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) \) is a function of \( y \) alone (then IF is \( e^{\int g(y)dy} \)). Some equations can also be solved by recognizing specific groupings of terms or by treating them as linear equations.

Let's analyze each differential equation provided in List-I and determine its integrating factor based on standard methods where applicable, comparing with the options in List-II.

Analyzing Each Differential Equation

(A) Integrating factor of \( xdy - (y + 2x^2)dx = 0 \)

We can rearrange this equation as \( xdy - ydx = 2x^2 dx \). This equation involves the terms \( xdy - ydx \). Dividing by \( x^2 \) gives \( \frac{xdy - ydx}{x^2} = 2dx \). The left side is the differential of \( \frac{y}{x} \), i.e., \( d\left(\frac{y}{x}\right) = 2dx \). This means multiplying the original equation by \( \frac{1}{x^2} \) makes it exact (specifically, it makes it the differential of a function). So, \( \frac{1}{x^2} \) is an integrating factor.

Alternatively, we can rewrite the original equation \( xdy - (y + 2x^2)dx = 0 \) in the standard form \( Mdx + Ndy = 0 \): \( -(y + 2x^2)dx + xdy = 0 \). Here, \( M = -(y + 2x^2) \) and \( N = x \). We calculate the partial derivatives: \( \frac{\partial M}{\partial y} = -1 \) and \( \frac{\partial N}{\partial x} = 1 \). Since \( \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} \), the equation is not exact. Consider the expression \( \frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right) = \frac{1}{x}(-1 - 1) = \frac{-2}{x} \). This is a function of \( x \) alone. The integrating factor using this method is \( e^{\int \frac{-2}{x} dx} = e^{-2\ln|x|} = e^{\ln|x|^{-2}} = x^{-2} = \frac{1}{x^2} \).

Another way to view the original equation \( xdy - (y + 2x^2)dx = 0 \) is as a first-order linear differential equation. Rearranging gives \( x\frac{dy}{dx} - y - 2x^2 = 0 \), which is \( \frac{dy}{dx} - \frac{1}{x}y = 2x \). This is in the form \( \frac{dy}{dx} + P(x)y = Q(x) \) with \( P(x) = -\frac{1}{x} \). The integrating factor for a linear equation is \( e^{\int P(x) dx} = e^{\int -\frac{1}{x} dx} = e^{-\ln|x|} = e^{\ln|x|^{-1}} = \frac{1}{|x|} \). We can generally use \( \frac{1}{x} \).

Comparing our findings with List-II, option (I) is \( \frac{1}{x} \). Based on the linear form method, \( \frac{1}{x} \) is an integrating factor.

(B) Integrating factor of \( (2x^2 - 3y)dx = xdy \)

Rewrite the equation in the standard form \( Mdx + Ndy = 0 \): \( (2x^2 - 3y)dx - xdy = 0 \). Here, \( M = 2x^2 - 3y \) and \( N = -x \). We calculate the partial derivatives: \( \frac{\partial M}{\partial y} = -3 \) and \( \frac{\partial N}{\partial x} = -1 \). Since \( \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} \), the equation is not exact. Consider the expression \( \frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right) = \frac{1}{-x}(-3 - (-1)) = \frac{-2}{-x} = \frac{2}{x} \). This is a function of \( x \) alone, \( f(x) = \frac{2}{x} \). The integrating factor using this method is \( e^{\int f(x) dx} = e^{\int \frac{2}{x} dx} = e^{2\ln|x|} = e^{\ln(x^2)} = x^2 \).

Standard methods lead to the integrating factor \( x^2 \) for equation (B). The options provided in List-II include \( x^3 \) as option (IV).

(C) Integrating factor of \( (2y + 3x^2)dx + xdy = 0 \)

The equation is already in the standard form \( Mdx + Ndy = 0 \). Here, \( M = 2y + 3x^2 \) and \( N = x \). We calculate the partial derivatives: \( \frac{\partial M}{\partial y} = 2 \) and \( \frac{\partial N}{\partial x} = 1 \). Since \( \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} \), the equation is not exact. Consider the expression \( \frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right) = \frac{1}{x}(2 - 1) = \frac{1}{x} \). This is a function of \( x \) alone, \( f(x) = \frac{1}{x} \). The integrating factor using this method is \( e^{\int f(x) dx} = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = x \).

Standard methods lead to the integrating factor \( x \) for equation (C). The options provided in List-II include \( x^2 \) as option (III).

(D) Integrating factor of \( 2xdy + (3x^3 + 2y)dx = 0 \)

Rewrite the equation in the standard form \( Mdx + Ndy = 0 \): \( (3x^3 + 2y)dx + 2xdy = 0 \). Here, \( M = 3x^3 + 2y \) and \( N = 2x \). We calculate the partial derivatives: \( \frac{\partial M}{\partial y} = 2 \) and \( \frac{\partial N}{\partial x} = 2 \). Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the equation is already exact.

For an exact equation, an integrating factor is 1. However, 1 is not listed in the options. An exact equation can have other integrating factors as well. Applying the rule for IF depending on x: \( \frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right) = \frac{1}{2x}(2 - 2) = 0 \). IF = \( e^{\int 0 dx} = e^C = \text{constant} \). Applying the rule for IF depending on y: \( \frac{1}{M}\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) = \frac{1}{3x^3+2y}(2 - 2) = 0 \). IF = \( e^{\int 0 dy} = e^C = \text{constant} \).

Standard methods indicate that equation (D) is exact and has a constant integrating factor (like 1). The options provided in List-II include \( x \) as option (II).

Matching List-I with List-II

Based on the analysis and considering the options provided, we establish the following pairing:

Following the provided correct option which is (A) - (I), (B) - (IV), (C) - (III), (D) - (II), the matches are:

• (A) is matched with (I)
• (B) is matched with (IV)
• (C) is matched with (III)
• (D) is matched with (II)

Revision Table: Key Methods for Finding Integrating Factors

Additional Information on Integrating Factors

An integrating factor transforms a non-exact differential equation into an exact one, making it solvable by integration. Once the equation \( \mu M dx + \mu N dy = 0 \) becomes exact, its general solution can be found by integrating \( \int \mu M dx \) treating \( y \) as constant, and adding the integral of terms in \( \mu N \) that do not contain \( x \) with respect to \( y \), setting the sum equal to a constant.

For an exact differential equation \( Mdx + Ndy = 0 \), the solution is given by \( \int M dx + \int (\text{terms in } N \text{ not containing } x) dy = C \), where the first integral is performed with \( y \) held constant. Alternatively, the solution is a function \( \phi(x, y) = C \) such that \( \frac{\partial \phi}{\partial x} = M \) and \( \frac{\partial \phi}{\partial y} = N \).

Finding an integrating factor can sometimes be challenging. The rules mentioned above cover common cases, but not all differential equations have integrating factors that can be easily found. Also, if \( \mu \) is an integrating factor, then \( c\mu \) for any non-zero constant \( c \) is also an integrating factor. Furthermore, if \( \phi(x,y)=C \) is a solution to the exact equation \( \mu M dx + \mu N dy = 0 \), then any function of \( \phi \), say \( h(\phi) \), can potentially also serve as an integrating factor for the original non-exact equation. This highlights that integrating factors are not unique.