Integrating Factors for Non-Exact Differential Equations
An integrating factor (I.F.) is a non-zero function \( u(x, y) \) such that, when a non-exact differential equation is multiplied by \( u(x, y) \), the equation becomes exact.
Example: Consider \( y dx - x dy = 0 \).
Since \( \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} \), the given equation is not exact.
Multiplying the given equation with \( \frac{1}{x^2} \), we get
\[ \frac{y}{x^2} dx - \frac{1}{x} dy = 0 \]
which is an exact equation.
Thus, \( \frac{1}{x^2} \) is an integrating factor of \( y dx - x dy = 0 \).
It can be easily verified that \( \frac{1}{y^2}, \frac{1}{xy}, \frac{1}{x^2+y^2}, \ldots \) are also integrating factors of \( y dx - x dy = 0 \).
Note: A non-exact differential equation can have more than one integrating factors.
Methods to Find Integrating Factors
Method I: I.F. Found by Inspection
Sometimes, the integrating factor can be found after regrouping the terms of the given equation and recognizing each group as being part of an exact differential equation. In this connection, the following exact differentials are listed below for ready reference:
- \( y dx + x dy = d(xy) \)
- \( \frac{x dy - y dx}{x^2} = d\left(\frac{y}{x}\right) \)
- \( \frac{y dx - x dy}{y^2} = d\left(\frac{x}{y}\right) \)
- \( \frac{x dy - y dx}{xy} = d\left[\log \left(\frac{y}{x}\right)\right] \)
- \( \frac{y dx - x dy}{xy} = d\left[\log \left(\frac{x}{y}\right)\right] \)
- \( \frac{x dy - y dx}{x^2 + y^2} = d\left[\tan^{-1}\left(\frac{y}{x}\right)\right] \)
- \( \frac{y dx - xdy}{x^2 + y^2} = d\left[\tan^{-1}\left(\frac{x}{y}\right)\right] \)
- \( \frac{2xy dx - x^2 dy}{y^2} = d\left(\frac{x^2}{y}\right) \)
- \( \frac{y e^x dx - e^x dy}{y^2} = d\left(\frac{e^x}{y}\right) \)
1. Solve \( ydx - xdy + 3x^2 y^2 e^{x{^3}} dx = 0 \)
Solution: Given equation is \( ydx - xdy + 3x^2 y^2 e^{x{^3}} dx = 0 \)
Multiply the given equation with \( \frac{1}{y^2} \), we get
\[ \frac{ydx - xdy}{y^2} + \frac{3x^2 y^2 e^{x{^3}} dx}{y^2} = 0 \]i.e., \[ d \left( \frac{x}{y} \right) + d \left( e^{x{^3}} \right) = 0 \]
Integrating, we get
\[ \frac{x}{y} + e^{x{^3}} = c \]Method II: If \( Mdx + Ndy = 0 \) be a non-exact homogeneous equation and \( Mx + Ny \neq 0 \), then \( \frac{1}{Mx+Ny} \) is an integrating factor.
1. Solve \( (x^3 + y^3) dx - x y^2 dy = 0 \).
Solution: Here \( M = x^3 + y^3 \) and \( N = -xy^2 \)
\[ \frac{\partial M}{\partial y} = 3y^2, \quad \frac{\partial N}{\partial x} = -y^2 \]Since \( \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} \), the given equation is not exact and homogenous
Now \( Mx + Ny = (x^3 + y^3)x + (-xy^2)y = x^4 \neq 0 \).
\( \therefore \) Integrating factor (I.F.) = \( \frac{1}{Mx+Ny} = \frac{1}{x^4} \)
Multiplying the given equation with \( \frac{1}{x^4} \), we get
\[ \left( \frac{x^3}{x^4} + \frac{y^3}{x^4} \right) dx + \left( \frac{-xy^2}{x^4} \right) dy = 0 \]or
\[ \left( \frac{1}{x} + \frac{y^3}{x^4} \right) dx + \left( \frac{y^2}{x^3} \right) dy = 0, \]which is an exact equation.
The general solution is
\[ \int_{y \hspace{0.1cm} is \hspace{0.1cm} \text{constant}} \left( \frac{1}{x} + \frac{y^3}{x^4} \right) dx + \int 0 dy = c \]i.e.,
\[ \int_{y \hspace{0.1cm} is \hspace{0.1cm} \text{constant}} \frac{dx}{x} + y^3\int \frac{1}{x^4} dx = c \]or
\[ \log x - \frac{y^3}{3x^3} = c \]Method III: If \( Mdx + Ndy = 0 \) is of the form \( yf(xy)dx + xg(xy)dy = 0 \) and \( Mx - Ny \neq 0 \), then \( \frac{1}{Mx - Ny} \) is an I.F.
1. Solve \( (xy \sin xy + \cos xy) y dx + (xy \sin xy - \cos xy) x dy = 0 \)
Solution: The given equation is of the form \( y f(xy) dx + x g(xy) dy = 0 \) and is not exact.
Here, \( M = (xy \sin xy + \cos xy) y \) and \( N = (xy \sin xy - \cos xy) x \).
Now, \( M x - N y = x^2 y^2 \sin xy + xy \cos xy - x^2 y^2 \sin xy + xy \cos xy = 2xy \cos xy \neq 0 \).
Thus, the integrating factor (I.F.) is:
\[ \text{I.F.} = \frac{1}{M x - N y} = \frac{1}{2xy \cos xy} \]
Multiplying the given equation with \( \frac{1}{2xy \cos xy} \), we get:
\[ \left( \frac{xy \sin xy}{2xy \cos xy} + \frac{\cos xy}{2xy \cos xy} \right) y dx + \left( \frac{xy \sin xy}{2xy \cos xy} - \frac{\cos xy}{2xy \cos xy} \right) x dy = 0 \]
or
\[ \left( \frac{y}{2} \tan xy + \frac{1}{2x} \right) dx + \left( \frac{x}{2} \tan xy - \frac{1}{2y} \right) dy = 0 \]
which is an exact equation.
The general solution is:
\[ \int_{y \hspace{0.1cm} is \hspace{0.1cm} \text{constant}} \left( \frac{y}{2} \tan xy + \frac{1}{2x} \right) dx - \int \frac{1}{2y} dy = c \]
i.e.,
\[ \frac{y}{2} \int_{y \hspace{0.1cm} is \hspace{0.1cm} \text{constant}} \tan xy \, dx + \frac{1}{2} \int \frac{1}{x} \, dx - \frac{1}{2} \int \frac{1}{y} \, dy = c \]
or
\[ \left( \frac{y}{2} \right) \frac{\log \sec xy}{y} + \frac{1}{2} \log x - \frac{1}{2} \log y = c \]
or
\[ \log \left( \frac{x \sec xy}{y} \right) = 2c \]
or
\[ x \sec xy = y c_1, \quad \text{where } c_1 = e^{2c} \]
Method IV: If \( Mdx + Ndy = 0 \) be a non-exact D.E. and
(a) If \( \frac{1}{N} \left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \right)= f(x) \) or \( k (\text{constant}) \), then \( e^{\int f(x) dx} \) or \( e^{\int k dx} \) is an I.F.
(b) If \( \frac{1}{M} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) = g(y) \) or \( k (\text{constant}) \), then \( e^{\int g(y) dy} \) or \( e^{\int k dx} \) is an I.F.
1. Solve \( (xy^2 - e^{1/x^3}) dx - x^2 y dy = 0 \)
Solution: Here, \( M = xy^2 - e^{1/x^3} \) and \( N = -x^2 y \).
\[ \frac{\partial M}{\partial y} = 2xy \quad \text{and} \quad \frac{\partial N}{\partial x} = -2xy \]
Since \( \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} \), the given equation is not exact.
Now, \( \frac{1}{N} \left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \right) = -\frac{1}{x^2 y} (2xy + 2xy) = -\frac{4}{x} = f(x) \).
Therefore,
\[ \text{I.F.} = e^{\int f(x) \, dx} = e^{-4 \int \frac{1}{x} \, dx} = e^{-4 \log x} = x^{-4} = \frac{1}{x^4} \]
Multiplying the given equation with \( \frac{1}{x^4} \), we get:
\[ \left( \frac{xy^2}{x^4} - \frac{e^{1/x^3}}{x^4} \right) dx - \frac{x^2 y}{x^4} dy = 0 \]
or
\[ \left( \frac{y^2}{x^3} - \frac{e^{1/x^3}}{x^4} \right) dx - \frac{y}{x^2} dy = 0 \]
which is an exact equation.
The general solution is:
\[ \int_{y=\text{constant}} \left( \frac{y^2}{x^3} - \frac{e^{1/x^3}}{x^4} \right) dx + \int(0) \, dy = c \]
That is,
\[ y^2 \int \frac{1}{x^3} \, dx - \int e^{1/x^3} \left( \frac{1}{x^4} \right) \, dx = c \]
or
\[ -\frac{y^2}{2x^2} + \frac{1}{3} e^{1/x^3} = c \]
2. Solve \( (3x^2 y^4 + 2xy) dx + (2x^3 y^3 - x^2) dy = 0 \)
Solution: Here, \( M = 3x^2 y^4 + 2xy \) and \( N = 2x^3 y^3 - x^2 \).
\[ \frac{\partial M}{\partial y} = 12x^2 y^3 + 2x \quad \text{and} \quad \frac{\partial N}{\partial x} = 6x^2 y^3 - 2x \]
Since \( \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} \), the given equation is not exact.
Now, \( \frac{1}{M} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) = \frac{1}{3x^2 y^4 + 2xy} (6x^2 y^3 - 2x - 12x^2 y^3 - 2x) = -\frac{2}{y} = g(y) \).
Therefore
\[ \text{I.F.} = e^{\int g(y) \, dy} = e^{-2 \int \frac{1}{y} \, dy} = e^{-2 \log y} = y^{-2} = \frac{1}{y^2} \]
Multiplying the given equation with \( \frac{1}{y^2} \), we get:
\[ \left( \frac{3x^2 y^4}{y^2} + \frac{2xy}{y^2} \right) dx + \left( \frac{2x^3 y^3}{y^2} - \frac{x^2}{y^2} \right) dy = 0 \]
or
\[ (3x^2 y^2 + \frac{2x}{y}) dx + (2x^3 y - \frac{x^2}{y^2}) dy = 0 \]
which is an exact equation.
The general solution is:
\[ \int_{y \hspace{0.1cm} is \hspace{0.1cm} \text{constant}} \left( 3x^2 y^2 + \frac{2x}{y} \right) dx + \int(0) \, dy = c \]
That is,
\[ 3y^2 \int x^2 \, dx + \frac{2}{y} \int x \, dx = c \]
or
\[ y^2 x^3 + \frac{x^2}{y} = c \]
Summary
A differential equation that is not exact can often be made exact by multiplying with an integrating factor (I.F.). Such equations are called equations reducible to exact differential equations. The I.F. transforms the equation so that it satisfies the condition ∂M/∂y = ∂N/∂x, allowing straightforward integration to obtain the solution.
FAQs on Equations Reducible to Exact Differential Equations
Q1: What does it mean when a differential equation is reducible to exact?
Ans: It means the given non-exact equation can be multiplied by an integrating factor (I.F.) so that it satisfies the exactness condition \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\).
Q2: What is an integrating factor?
Ans: An integrating factor is a function that, when multiplied with a non-exact differential equation, makes it exact and solvable.
Q3: What are common methods of finding integrating factors?
Ans: Methods include inspection, homogeneous functions, \(\frac{1}{Mx+Ny}\), \(\frac{1}{Mx-Ny}\), and exponential forms \(e^{\int f(x)dx}\) or \(e^{\int g(y)dy}\).