Abstract:

This lecture, the second on exact and non-exact differential equations, focuses on solving ordinary differential equations (ODEs) that do not initially satisfy the condition for exactness. The core method introduced is the use of an "integrating factor," Lambda (λ), which is a function that, when multiplied by the non-exact ODE, transforms it into an exact differential equation that can then be solved using previously learned techniques. The lecture derives the general partial differential equation that the integrating factor must satisfy and demonstrates how finding a specific solution for Lambda, often by assuming it depends on only one variable (x or y), simplifies the problem significantly. Two detailed examples are presented: a specific non-exact equation solved by finding an integrating factor dependent on y, and the general linear first-order ODE, where the integrating factor dependent on x is shown to be the known exponential factor. A third example illustrates how different ways of writing a separable ODE can result in either an exact or a non-exact form, with the integrating factor linking the two. The lecture concludes the series on first-order ODEs and announces the transition to second-order equations.

Solving Non-Exact Differential Equations Using Integrating Factors

• 0:00: Introduction to non-exact differential equations, following the previous lecture on exact ones, and the concept of an "integrating factor" (Lambda, λ) to make them exact.
• 0:56: Presents Example 1: The equation x dy/dx - xy² - y = 0, rewritten in differential form as (xy² + y) dx - x dy = 0.
• 2:09: Identifies A(x,y) = xy² + y and B(x,y) = -x for Example 1.
• 3:00: Performs the integrability test for Example 1, showing dA/dy = 2xy + 1 and dB/dx = -1. Since they are not equal, the equation is not exact (4:02).
• 4:20: Introduces the idea of multiplying the non-exact equation by an integrating factor Lambda(x,y) to make it exact (LambdaA dx + LambdaB dy = 0).
• 6:13: States the condition for the new equation to be exact: d(LambdaA)/dy = d(LambdaB)/dx.
• 7:20: Derives the general partial differential equation (PDE) that Lambda must satisfy: A * dLambda/dy - B * dLambda/dx + Lambda * (dA/dy - dB/dx) = 0. Acknowledges this PDE is generally difficult but only a specific solution for Lambda is needed (9:35).
• 11:07: Applies the Lambda PDE to Example 1. Makes the simplifying assumption that Lambda only depends on y (Lambda(y)), eliminating the dLambda/dx term.
• 13:55: Solves the resulting ordinary differential equation for Lambda(y): y * dLambda/dy + 2Lambda = 0, finding the integrating factor Lambda = 1/y² (16:21).
• 16:27: Verifies that multiplying the original equation by 1/y² (giving (x + 1/y) dx - x/y² dy = 0) results in an exact equation (19:10) by showing the new A' and B' satisfy the integrability test.
• 19:26: Solves the now exact equation using the method from the previous lecture, finding the function U(x,y) = x²/2 + x/y + C.
• 24:52: Derives the implicit solution x²/2 + x/y = K (by setting U=constant) and then the explicit general solution y = 2x / (C_new - x²) (26:35). Encourages viewers to verify this solution in the original non-exact equation.
• 27:49: Presents Example 2: The general linear first-order inhomogeneous ODE dy/dx + P(x)y = Q(x). Rewrites in differential form (P(x)y - Q(x)) dx + 1 dy = 0.
• 29:52: Identifies A(x,y) = P(x)y - Q(x) and B(x,y) = 1. Shows it's not exact (31:09) as dA/dy = P(x) and dB/dx = 0.
• 31:58: Applies the Lambda PDE to Example 2. Makes the simplifying assumption that Lambda only depends on x (Lambda(x)).
• 34:03: Solves the resulting ODE for Lambda(x): dLambda/dx = P(x) * Lambda, finding the integrating factor Lambda(x) = exp(∫P(x) dx) (35:30), which matches the known factor for this type of equation.
• 36:36: Uses this Lambda to solve the now exact linear equation, demonstrating how this method reproduces the standard solution formula y = exp(-∫P(x) dx) * (∫exp(∫P(x) dx)Q(x) dx + C_final) (43:00 - 44:40).
• 46:41: Presents Example 3: Separable differential equations dy/dx = g(x)h(y).
• 47:07: Shows one differential form (g(x) dx - 1/h(y) dy = 0) which is exact (47:53).
• 48:08: Shows another differential form (g(x)h(y) dx - 1 dy = 0) which is generally not exact (50:16).
• 50:24: Explains that the exactness depends on how the equation is written.
• 51:25: Identifies that multiplying the non-exact form by Lambda = 1/h(y) (51:32) transforms it into the exact form, confirming 1/h(y) as the integrating factor for this presentation.
• 52:38: Concludes the series of lectures on first-order differential equations and announces that future lectures will cover second-order differential equations.