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A short class with Sarah and her students. The other ways are under More.
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Let's start with our first example which is y′′−3y′+2y = e^x. The first step is to solve the homogeneous part of the equation. Can anyone tell me how to find the complementary function?
We need to solve the auxiliary equation, right?
Exactly! The auxiliary equation is r²−3r+2=0. Now, can anyone factor this equation for me?
(r−1)(r−2)=0, so the roots are r=1 and r=2.
Well done! Thus, the complementary function is y_c = C₁ e^x + C₂ e^(2x). Now, for the particular solution, what form do you think we should guess?
Since we have e^x, we should try y_p = A x e^x to avoid duplication.
Right! Now, let’s substitute y_p into the original equation and gather like terms. How do we do that?
We need to compute the derivatives and substitute them into the left-hand side, then simplify.
Exactly! After substitution, we find that -Ae^x = e^x, leading to A = -1. Thus, our particular solution is y_p = -x e^x.
To summarize, our final solution is y(x) = C₁ e^x + C₂ e^(2x) - x e^x. Well done everyone!
Now, let’s look at the second example, y′′ + y = x². What do we do first again?
We find the complementary function by solving the homogeneous part.
Correct! The auxiliary equation is r² + 1 = 0. Who can tell me the roots?
The roots are complex: r = ±i.
Right! So, the complementary function becomes y_c = C₁ cos(x) + C₂ sin(x). Now, for our particular solution, which form will we try?
We should guess y_p = Ax² + Bx + C because it’s a polynomial of degree 2.
Exactly! After substituting and simplifying, what do we find when we compare coefficients?
We find A = 1, B = 0, and C = -2. So y_p = x² - 2.
Exactly! Therefore, our final solution is y(x) = C₁ cos(x) + C₂ sin(x) + x² - 2. Great job!
Finally, let’s tackle y′′ + 4y = cos(2x). What’s our first step?
We start with the complementary function by solving the homogeneous part.
Correct again! The auxiliary equation is r² + 4 = 0. What does that give us?
The roots are r = ±2i.
Right! So our complementary function is y_c = C₁ cos(2x) + C₂ sin(2x). Now for the particular solution, what should we try?
Since cos(2x) is already in y_c, we can try y_p = x(A cos(2x) + B sin(2x)).
Exactly! Now substitute this into the original equation. What will we have on the left side?
We’ll differentiate our guess and substitute both derivatives into the equation.
Right again! After substitution, we will compare coefficients and solve for A and B. This is key to finding our final solution.
In summary, we need to account for overlaps with y_c and adjust accordingly to solve this trigonometrically infused equation.
A technique for finding particular solutions to non-homogeneous linear differential equations.
The solution of the corresponding homogeneous equation.
A form guessed based on the forcing function to solve the non-homogeneous equation.
Example 1: Solve y′′−3y′+2y = e^x leading to final solution y(x) = C₁ e^x + C₂ e^(2x) - x e^x.
Example 2: Solve y′′ + y = x² leading to the final solution y(x) = C₁ cos(x) + C₂ sin(x) + x² - 2.
Example 3: Solve y′′ + 4y = cos(2x), addressing duplication techniques and obtaining the solution with adjustments.
When solving ODEs with a function so mean, Don't forget the complement in your routine.
Imagine you're a detective finding unknown coefficients in a case against a tricky function. Each derived term leads you closer to the truth in the form of solutions.
To Remember the steps: H-G-S-M-R (Homogeneous, Guess, Substitute, Modify, Resolve)
Remember P.S.G. for the order: Particular, Substitute, General.
Term
Complementary Function
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Term
Particular Integral
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The solution to the homogeneous part of a differential equation.
The specific solution to the non-homogeneous part of a differential equation, found using guesses based on the forcing function.
A quadratic equation that helps in finding the complementary function from a linear differential equation.
The non-homogeneous part of a differential equation, often denoted as f(x).
This section presents illustrative examples demonstrating the application of the method of undetermined coefficients to solve non-homogeneous linear differential equations.
The section features three specific examples that showcase how to solve differential equations with exponential, polynomial, and trigonometric forcing functions. Each example walks through the steps of determining the complementary function and the particular integral using the method of undetermined coefficients.
In this section, we explore illustrative examples that demonstrate the application of the method of undetermined coefficients for solving non-homogeneous linear differential equations. The focus is on presenting a step-by-step approach to tackling each type of forcing function, including exponential, polynomial, and trigonometric.
Exponential Forcing Function: We solve the equation y′′−3y′+2y = e^x. This involves finding the complementary function from the auxiliary equation and guessing the form of the particular integral, leading to the solution.
Polynomial Forcing Function: The second example addresses the equation y′′ + y = x². It illustrates how to find the complementary function and choose an appropriate trial solution to find the particular integral.
Trigonometric Forcing Function: The final example involves solving y′′ + 4y = cos(2x), which requires careful selection of the trial solution due to the overlap with the complementary function.
These examples not only provide practical applications of the method but also illustrate the reasoning behind each step, highlighting significant techniques such as substitution and comparison of coefficients.