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Interactive Audio Lesson
Listen to a student-teacher conversation explaining the topic in a relatable way.
Exponential Forcing Function
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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!
Polynomial Forcing Function
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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!
Trigonometric Forcing Function
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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.
Introduction & Overview
Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.
Quick Overview
Standard
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.
Detailed
Detailed Summary
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.
Example Breakdown:
-
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.
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Audio Book
Dive deep into the subject with an immersive audiobook experience.
Example 1: Exponential Forcing Function
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Solve:
y′′−3y′+2y =ex
Step 1: Solve the homogeneous part
Auxiliary equation: r2−3r+2=0⇒(r−1)(r−2)=0⇒r =1,2
So,
y =C e^x+C e^(2x)
c 1 2
Step 2: Guess particular solution
Since e^x is already part of y , we guess:
c
y =Ax e^x
Step 3: Substitute into original equation
Compute:
• y =A e^x+Ax e^x
p′
• y =A e^x+A e^x+Ax e^x =2A e^x+Ax e^x
p′′
Substitute into LHS:
(2A e^x+Ax e^x)−3(A e^x+Ax e^x)+2Ax e^x =e^x
Simplify:
2A e^x+Ax e^x−3A e^x−3Ax e^x+2Ax e^x =e^x ⇒(−A e^x)=e^x ⇒A=−1
So,
y =−x e^x
p
Final Solution:
y(x)=C e^x+C e^(2x)−x e^x
1 2
Detailed Explanation
In this example, we are solving a differential equation with an exponential forcing function. The first step involves finding the homogeneous solution by solving the auxiliary equation. Here, we find two roots which allow us to write the complementary function. Next, we assume a form for the particular solution, knowing that e^x is already part of the complementary function. By modifying the guess to Ax e^x, we substitute back into the original equation to find the constants. After simplification, we determine A = -1, giving us the particular solution. Finally, we combine this with the complementary function to arrive at the complete solution.
Examples & Analogies
Imagine a bridge that is swaying due to the wind (exponential forces). The natural vibrations of the bridge represent the complementary function, which is part of its inherent structure. However, the additional sway caused by the wind requires us to adjust our calculations to account for the added force. Just as we adjust our equations to include this effect, civil engineers adjust models to account for real-world forces acting on structures.
Example 2: Polynomial Forcing Function
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Solve:
y′′+y =x2
Step 1: Homogeneous part: Auxiliary equation: r2+1=0⇒r =±i
y =C cos(x)+C sin(x)
c 1 2
Step 2: Guess particular solution
Since RHS is x2, try:
y =Ax2+Bx+C
p
Compute derivatives:
• y =2Ax+B
p′
• y =2A
p′′
Substitute:
2A+(Ax2+Bx+C)=x2 ⇒Ax2+Bx+(2A+C)=x2
Comparing both sides:
• A=1
• B =0
• 2A+C =0⇒2+C =0⇒C =−2
So,
y =x2−2
p
Final Solution:
y(x)=C cos(x)+C sin(x)+x2−2
Detailed Explanation
In this case, we solve a polynomial force applied to a dynamic system. We start by determining the homogeneous solution using the auxiliary equation, which yields complex roots. This leads us to the complementary function involving sine and cosine. For the particular solution, we guess a polynomial form due to the nature of the forcing function. After substituting our guess back into the differential equation, we equate the coefficients of similar powers of x to solve for A, B, and C. Combining these results gives us the final solution.
Examples & Analogies
Consider a child riding a bike down a gentle hill and occasionally pushing off the ground with their feet. The natural motion of the bike represents the system's inherent behavior, while the pushes from the child act like the polynomial forces applied to the bike’s motion. By calculating both the pedal (complementary response) and the child’s pushes (particular integral), we can accurately predict the bike's overall movement.
Example 3: Trigonometric Forcing Function
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Solve:
y′′+4y =cos(2x)
Step 1: Homogeneous part: Auxiliary equation: r2+4=0⇒r =±2i
y =C cos(2x)+C sin(2x)
c 1 2
Since cos(2x) is part of the complementary function, try:
y =x(Acos(2x)+Bsin(2x))
p
Calculate derivatives and substitute into the equation, then compare coefficients to find A,B.
Detailed Explanation
In this example, we are dealing with a trigonometric function as the forcing term. First, we find the homogeneous solution using the auxiliary equation, leading to a complementary function formed by sine and cosine with a frequency of 2. Since one of our terms is already part of the complementary function, we modify our guess for the particular solution by multiplying by x to avoid repetition. By substituting this new form into the differential equation and comparing coefficients, we can solve for the unknown constants A and B, completing the solution process.
Examples & Analogies
Think of waves in the ocean interacting with a buoy. The natural oscillations of the buoy represent its inherent response to the water (the homogeneous solution). When additional waves hit the buoy, similar to the forcing function, we need to adjust our model to account for these new forces. Just as we analyze both the buoy's natural movements and the added oceanic waves, engineers analyze both complementary and particular responses in their calculations.
Definitions & Key Concepts
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
-
Method of Undetermined Coefficients: A technique for finding particular solutions to non-homogeneous linear differential equations.
-
Complementary Function (y_c): The solution of the corresponding homogeneous equation.
-
Particular Integral (y_p): A form guessed based on the forcing function to solve the non-homogeneous equation.
Examples & Real-Life Applications
See how the concepts apply in real-world scenarios to understand their practical implications.
Examples
-
Example 1: Solve
y′′−3y′+2y = e^xleading to final solutiony(x) = C₁ e^x + C₂ e^(2x) - x e^x. -
Example 2: Solve
y′′ + y = x²leading to the final solutiony(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.
Memory Aids
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
🎵 Rhymes Time
-
When solving ODEs with a function so mean, Don't forget the complement in your routine.
📖 Fascinating Stories
-
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.
🧠 Other Memory Gems
-
To Remember the steps: H-G-S-M-R (Homogeneous, Guess, Substitute, Modify, Resolve)
🎯 Super Acronyms
Remember P.S.G.** for the order
- P**articular
- **S**ubstitute
- **G**eneral.
Flash Cards
Review key concepts with flashcards.
Glossary of Terms
Review the Definitions for terms.
-
Term: Complementary Function (y_c)
Definition:
The solution to the homogeneous part of a differential equation.
-
Term: Particular Integral (y_p)
Definition:
The specific solution to the non-homogeneous part of a differential equation, found using guesses based on the forcing function.
-
Term: Auxiliary Equation
Definition:
A quadratic equation that helps in finding the complementary function from a linear differential equation.
-
Term: Forcing Function
Definition:
The non-homogeneous part of a differential equation, often denoted as f(x).