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Magazine Chapter 17: Problem 8
Solve the equations in Exercises by the method of undetermined coefficients. $$y^{\prime \prime}+y=2 x+3 e^{x}$$
Short Answer
Expert verified
The general solution is \( y = C_1 \cos(x) + C_2 \sin(x) + 2x + \frac{3}{2}e^x \).
Step by step solution
01
Identify Homogeneous Equation
Start by considering the homogeneous part of the differential equation: \( y'' + y = 0 \). The characteristic equation is \( r^2 + 1 = 0 \). Solving this, we get the roots \( r = i \) and \( r = -i \).
02
Solve Homogeneous Solution
Using the roots \( r = i \) and \( r = -i \), the general solution to the homogeneous equation is \( y_h = C_1 \cos(x) + C_2 \sin(x) \).
03
Determine Form of Particular Solution
For the non-homogeneous part \( 2x + 3e^x \), use the method of undetermined coefficients. Assume a particular solution \( y_p = Ax + B + Ce^x \).
04
Differentiate Particular Solution Assumption
The derivatives are required to substitute into the original equation. Calculate them: \( y_p' = A + Ce^x \) and \( y_p'' = Ce^x \).
05
Substitute and Solve for Coefficients
Substitute \( y_p \), \( y_p' \), and \( y_p'' \) into the original equation. This gives: \( Ce^x + Ax + B + Ce^x = 2x + 3e^x \). Simplify to \( 2Ce^x + Ax + B = 2x + 3e^x \).
06
Equate and Determine Coefficients
Compare coefficients on both sides: \( 2C = 3 \), so \( C = \frac{3}{2} \); \( A = 2 \); \( B = 0 \). Therefore, \( y_p = 2x + \frac{3}{2}e^x \).
07
Write General Solution
Combine the homogeneous and particular solutions to form the general solution: \( y = y_h + y_p = C_1 \cos(x) + C_2 \sin(x) + 2x + \frac{3}{2}e^x \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Homogeneous Equations
When dealing with differential equations, a critical step is to identify the homogeneous equation. These equations are characterized by setting the non-homogeneous (non-zero) terms of a differential equation to zero.
- For example, consider the equation: \( y'' + y = 2x + 3e^x \). The homogeneous part is what results when you remove the terms on the right: \( y'' + y = 0 \).
- Homogeneous equations are important because their solutions, known as the complementary solution, form the basis upon which particular solutions are built.
Undetermined Coefficients
The method of undetermined coefficients is a technique used to find particular solutions to non-homogeneous linear differential equations. This method is straightforward and efficient when the non-homogeneous term is of a specific form, such as polynomials, exponentials, or sines and cosines.
- The process involves guessing a form of the particular solution that resembles the non-homogeneous part of the equation.
- For instance, with an equation like \( y'' + y = 2x + 3e^x \), the terms "2x" and "3e^x" suggest a particular solution of the form \( y_p = Ax + B + Ce^x \).
Particular Solution
After determining the form for your particular solution, the next step is to find the specific coefficients that make it satisfy the non-homogeneous equation.
- From the earlier example equation \( y^{ ext{''}} + y = 2x + 3e^x \), with the proposed particular solution \( y_p = Ax + B + Ce^x \), we compute the derivatives: \( y_p' = A + Ce^x \) and \( y_p'' = Ce^x \).
- Substitute these forms back into the original equation, effectively placing them on the left-hand side to allow a term-by-term comparison to the right-hand side \(2x + 3e^x \).
