First Order Differential Equations
2.4 Integrating Factors
When faced with a non-exact first-order differential equation, the method of integrating factors provides a systematic way to transform it into an exact equation that can be solved. This section explores the techniques of utilizing integrating factors for solving differential equations.
Sometimes a differential equation that is not initially exact can be transformed into an exact one by multiplying through by an appropriate function, [asciimath]mu(x, y)[/asciimath]. Consider the equation
[asciimath](3x+2y^2)dx+2xydy=0[/asciimath].
It is not exact because [asciimath]M_y = 4y[/asciimath] and [asciimath]N_x = 2y[/asciimath] do not match. However, if we multiply the entire equation by a function [asciimath]mu(x) = x[/asciimath], it becomes
[asciimath](3x^2+2xy^2)dx+2x^2ydy=0[/asciimath].
This equation is now exact as [asciimath]M_y = N_x = 4xy[/asciimath]. This modified equation can then be solved using the exact equation methods discussed in Section 2.3.
The function [asciimath]mu(x, y)[/asciimath] is known as an integrating factor for the equation if, when multiplied by the equation, it results in an exact equation. In formal terms, if multiplying the differential equation by [asciimath]mu(x, y)[/asciimath] as in
[asciimath]mu(x, y)M(x, y)dx + mu(x, y)N(x, y)dy = 0[/asciimath]
makes it exact, then [asciimath]mu(x, y)[/asciimath] is the integrating factor.
Method for Finding the Special Integrating Factor
When you encounter a first-order differential equation in the form [asciimath]Mdx + Ndy = 0[/asciimath] that is neither separable nor linear, you can still potentially solve it by finding a special integrating factor. Follow these steps:
1. Compute partial derivatives: Compute [asciimath]M_y[/asciimath] and [asciimath]N_x[/asciimath] .
2. Check for exactness:
- If [asciimath]M_y=N_x[/asciimath], then the equation is already exact, and no integrating factor is needed.
- If [asciimath]M_y!=N_x[/asciimath], the equation is not exact, and you may proceed to find an integrating factor.
3. Find a special integrating factor:
- Compute the expression [asciimath](M_y-N_x)/N[/asciimath] (i). If (i) is a function of [asciimath]x[/asciimath] only, then an integrating factor is given by [asciimath]mu(x)=e^(int (M_y-N_x)/N dx[/asciimath].
- If (i) is not a function of [asciimath]x[/asciimath] only, compute the expression [asciimath](N_x-M_y)/M[/asciimath] (ii). If (ii) is a function of [asciimath]y[/asciimath] only, then an integrating factor is given by [asciimath]mu(y)=e^(int (N_x-M_y)/M dy[/asciimath] .
4. Apply the integrating factor: Multiply the entire differential equation by the integrating factor [asciimath]mu[/asciimath] to transform it to an exact equation.
5. Solve the exact equation: Once the equation is made exact, solve it using the method outlined in Section 2.3 for exact equations.
Solve [asciimath](2x^2+y)dx+(x^2y-x)dy=0[/asciimath]
Show/Hide Solution
A quick inspection shows that the equation is neither separable nor linear nor exact. Therefore, we check if a special integrating factor exists:
[asciimath](M_y-N_x)/N=(1-(2xy-1))/(x^2y-x)=(2(1-xy))/(-x(1-xy))=-2/x[/asciimath]
Since (i) is the function of only [asciimath]x[/asciimath], an integrating factor is given by
[asciimath]mu(x)=e^(int (M_y-N_x)/N dx[/asciimath]
[asciimath]=e^(int -2/x dx[/asciimath]
[asciimath]=x^(-2)[/asciimath]
Multiplying [asciimath]mu(x)=x^(-2)[/asciimath] by the original differential equation, we obtain the exact equation
[asciimath](2+yx^(-2))dx+(y-x^(-1))dy=0[/asciimath]
Solving the equation using the exact method, we get the implicit solution
[asciimath]2x-yx^(-1)+(y^2)/2=C[/asciimath]
Try an Example
Section 2.4 Exercises
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- Find an integrating factor for the following equation: [asciimath](xy + x + 2y +1)dx + (x+1)dy = 0[/asciimath]
Show/Hide Answer
[asciimath]mu(x)=e^x[/asciimath]
- For the given differential equation, a) Determine the integrating factor. b) Find a general solution.
[asciimath](4x^2+y)dx+(x^2y-x)dy = 0[/asciimath]
Show/Hide Answer
a) [asciimath]mu(x)=x^(-2)[/asciimath]
b) [asciimath]4x-y x^-1+1/2y^2=C[/asciimath]
- Solve the differential equation: [asciimath](4x^2-y)dx+(-4x^2y+x)dy = 0[/asciimath]
Show/Hide Answer
[asciimath]4x+y x^-1-2y^2=C[/asciimath]
- For the given differential equation, a) Determine the integrating factor. b) Find a general solution.
[asciimath]ysin(y)dx+x(sin(y)-ycos(y))dy = 0[/asciimath]
Show/Hide Answer
a) [asciimath]mu(y)=y/(sin(y)[/asciimath]
b) [asciimath](xy)/(sin(y))=C[/asciimath]
- Find an integrating factor for the following equation: [asciimath](xy + x + 2y +1)dx + (x+1)dy = 0[/asciimath]
