Chapter 6: Problem 10
Solve the following equations using the method of undetermined coefficients. $$ y^{\prime \prime}+y=3 \sin 2 x+x \cos 2 x $$
Short Answer
Expert verified
The general solution is \( y = C_1 \cos x + C_2 \sin x + A \cos 2x + B \sin 2x + (Cx + D) \cos 2x + (Ex + F) \sin 2x \) with specific values for A, B, C, D, E, and F determined from the conditions.
Step by step solution
01
Identify the Homogeneous Equation
Start by considering the homogeneous part of the differential equation, which is \( y'' + y = 0 \). This will help us find the complementary solution.
02
Solve the Homogeneous Equation
Solve \( y'' + y = 0 \). The characteristic equation is \( r^2 + 1 = 0 \), which has roots \( r = i \) and \( r = -i \). Thus, the complementary solution is \( y_c = C_1 \cos x + C_2 \sin x \).
03
Identify the Particular Solution
Find a particular solution \( y_p \) for the non-homogeneous equation \( y'' + y = 3 \sin 2x + x \cos 2x \). Notice the non-homogeneous part has terms \( \sin 2x \) and \( x \cos 2x \).
04
Guess the Form of the Particular Solution
Since we have \( \sin 2x \) and \( x \cos 2x \), propose a particular solution of the form: \( y_p = A \cos 2x + B \sin 2x + (Cx + D) \cos 2x + (Ex + F) \sin 2x \).
05
Derive the Particular Solution
Compute the derivatives \( y'_p \) and \( y''_p \) and substitute them back into the original differential equation to find the coefficients \( A, B, C, D, E, \) and \( F \). Use equating coefficients method to solve for these variables.
06
Combine Solutions
Add the complementary solution \( y_c \) and the particular solution \( y_p \) together to obtain the general solution to the differential equation.
07
Present the General Solution
The general solution of the differential equation \( y'' + y = 3\sin 2x + x \cos 2x \) is a combination of the homogeneous and particular solutions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Particular Solution
In solving non-homogeneous differential equations like the one given in the original exercise, finding the particular solution is crucial. The particular solution, denoted as \( y_p \), specifically addresses the non-homogeneous part of the differential equation. This term isolates and resolves the influence of additional terms like \( 3\sin 2x + x \cos 2x \) that are outside the standard homogeneous structure.
To find a particular solution, we often use a method involving educated guessing known as the "method of undetermined coefficients." Here's how it works:
After determining these coefficients, \( y_p \) is then combined with the complementary solution to form the complete solution to the differential equation.
To find a particular solution, we often use a method involving educated guessing known as the "method of undetermined coefficients." Here's how it works:
- First, observe the non-homogeneous terms in the equation. In our exercise, these are \( 3 \sin 2x \) and \( x \cos 2x \).
- Based on these forms, guess a potential form for \( y_p \) that includes similar components. This helps "fit in" or counterbalance the non-homogeneous part.
- Typical guesses coincide with similar functions as the ones present, but in more complex forms that allow us to adjust parameters accordingly.
After determining these coefficients, \( y_p \) is then combined with the complementary solution to form the complete solution to the differential equation.
Complementary Solution
The complementary solution is a fundamental aspect of solving linear differential equations. It represents the solution to the corresponding homogeneous equation, which, in our case, is previously determined as \( y'' + y = 0 \).
To find this solution, look for solutions to the simplified version of the equation whereby all non-homogeneous parts (like literal constants or sinusoidal terms) are set to zero. Here's how it unfolds:
Additive combinations of the complementary and particular solutions eventually lead to the general solution, encompassing both inherent patterns and externally-induced behaviors.
To find this solution, look for solutions to the simplified version of the equation whereby all non-homogeneous parts (like literal constants or sinusoidal terms) are set to zero. Here's how it unfolds:
- Determine the characteristic equation from the homogeneous equation. In this exercise, it is \( r^2 + 1 = 0 \).
- Solve the characteristic equation to find its roots. These describe the solution's behavior. The roots \( r = i \) and \( r = -i \) imply oscillatory solutions.
- Construct the general form of the solution combining exponential, sine, and cosine terms based on those roots: \( y_c = C_1 \cos x + C_2 \sin x \), where \( C_1 \) and \( C_2 \) are arbitrary constants derived from initial conditions.
Additive combinations of the complementary and particular solutions eventually lead to the general solution, encompassing both inherent patterns and externally-induced behaviors.
Homogeneous Differential Equation
Understanding the homogeneous differential equation is key before delving into its non-homogeneous counterparts. A homogeneous differential equation is one where all terms are derived solely from the dependent variable and its derivatives. No external "forcing" or non-zero constants appear in the equation.
For example, the homogeneous problem derived from the given differential equation is \( y'' + y = 0 \). This type of equation exhibits solutions that rely only on the intrinsic dynamics governed by its characteristic equation.
Key points about homogeneous differential equations include:
By mastering the homogeneous equation, we set the stage for effectively managing the complexities introduced by the additional, non-homogeneous terms in broader differential equations.
For example, the homogeneous problem derived from the given differential equation is \( y'' + y = 0 \). This type of equation exhibits solutions that rely only on the intrinsic dynamics governed by its characteristic equation.
Key points about homogeneous differential equations include:
- They often set the foundation for developing the full general solution when combined with the particular solution.
- The characteristic equation, obtained directly from substituting solutions of the form \( e^{rt} \) into the equation, results in polynomial expressions to solve for roots that guide the nature of the solutions.
- Depending on the types of roots (real or complex), solutions can reflect exponential decay/growth or oscillations like sines and cosines.
By mastering the homogeneous equation, we set the stage for effectively managing the complexities introduced by the additional, non-homogeneous terms in broader differential equations.