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Non Exact DE MADE EXACT USING INTEGRATING FACTOR - Differential Equations
Yu Jei Abat
Overview
This video explains how to solve non-exact differential equations (DEs) by transforming them into exact DEs using an integrating factor. It details the two main formulas for finding integrating factors based on whether the resulting function is in terms of 'x' or 'y'. The process involves calculating partial derivatives to check for exactness, applying the appropriate integrating factor formula, multiplying the original DE by the integrating factor, re-checking for exactness, and finally solving the resulting exact DE. Two examples are worked through to illustrate the method.
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Chapters
- Non-exact DEs cannot be solved directly using the standard method for exact DEs.
- The goal is to find an 'integrating factor' that, when multiplied by the entire DE, makes it exact.
- It's crucial to have watched a video on solving exact DEs first, as that method is used after transforming a non-exact DE.
Understanding the nature of differential equations as exact or non-exact is the first step in choosing the correct solution method. Non-exact DEs require an extra step to be solvable.
- Formula 1: If (∂M/∂y - ∂N/∂x) / N results in a function of only 'x' (f(x)), the integrating factor is e^(∫f(x)dx).
- Formula 2: If (∂N/∂x - ∂M/∂y) / M results in a function of only 'y' (f(y)), the integrating factor is e^(-∫f(y)dy).
- The choice between formulas depends on whether the resulting expression can be simplified to a function of a single variable (x or y) and a constant.
- An integrating factor must simplify to a constant or a function of a single variable; expressions with both 'x' and 'y' cannot be used.
These formulas provide a systematic way to find the multiplier that converts a non-exact DE into a solvable exact DE, guiding the choice based on the structure of the equation.
If (∂M/∂y - ∂N/∂x) / N simplifies to -1, then f(x) = -1, and the integrating factor is e^(∫-1 dx) = e^(-x).
- The initial DE (y - xy)dx + x dy = 0 is tested for exactness and found to be non-exact.
- Using Formula 1, (∂M/∂y - ∂N/∂x) / N = -1, which is a function of x (f(x) = -1).
- The integrating factor is calculated as e^(∫-1 dx) = e^(-x).
- Multiplying the original DE by e^(-x) transforms it into an exact DE.
- The resulting exact DE is then solved using the standard method, yielding the solution C = xy e^(-x).
This example demonstrates the practical application of the integrating factor method, showing how a non-exact DE can be systematically solved by transforming it into an exact one.
The original DE (y - xy)dx + x dy = 0 is multiplied by the integrating factor e^(-x) to become (y - xy)e^(-x)dx + xe^(-x)dy = 0, which is then solved.
- The DE (2xy² - 2y)dx + (3x²y - 4x)dy = 0 is tested and found to be non-exact.
- Attempting Formula 1 yields a function of both x and y, which is unusable.
- Using Formula 2, (∂N/∂x - ∂M/∂y) / M results in -1/y, a function of y (f(y) = -1/y).
- The integrating factor is calculated as e^(-∫(-1/y)dy) = e^(∫(1/y)dy) = e^(ln|y|) = y.
- Multiplying the original DE by 'y' makes it exact.
- The resulting exact DE is solved, yielding the solution x²y³ - 2xy² = C.
This example highlights the importance of choosing the correct integrating factor formula and demonstrates how to handle cases where one formula leads to an unusable expression, while the other provides a solvable path.
The DE (2xy² - 2y)dx + (3x²y - 4x)dy = 0 is multiplied by the integrating factor 'y' to become (2xy³ - 2y²)dx + (3x²y² - 4xy)dy = 0, which is then solved.
Key takeaways
- Non-exact differential equations require an integrating factor to be transformed into exact differential equations before they can be solved.
- The choice of integrating factor formula depends on whether the derived function is solely dependent on 'x' or 'y'.
- An integrating factor must simplify to a function of a single variable (or a constant) to be valid.
- After finding and applying the integrating factor, the transformed DE must be re-checked for exactness.
- The final step in solving a non-exact DE is to solve the now-exact DE using the standard method.
- The solution to a differential equation represents a family of curves, often expressed with an arbitrary constant 'C'.
Key terms
Non-exact Differential EquationExact Differential EquationIntegrating FactorPartial DerivativeM(x, y)dx + N(x, y)dy = 0f(x)f(y)e^(∫f(x)dx)e^(-∫f(y)dy)Integration by Parts
Test your understanding
- What is the primary purpose of an integrating factor in solving differential equations?
- How do you determine which of the two integrating factor formulas to use for a given non-exact DE?
- What condition must be met by the expression derived from (∂M/∂y - ∂N/∂x) / N or (∂N/∂x - ∂M/∂y) / M for it to be a valid basis for an integrating factor?
- Why is it necessary to re-check for exactness after multiplying a non-exact DE by its integrating factor?
- Describe the steps involved in solving an exact differential equation after it has been obtained from a non-exact one.