Problem 57 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given differential equation by undetermined coefficients. $$y^{\prime \prime}+y^{\prime}+y=x \sin x$$

Step-by-Step Solution

Answer

The differential equation solution is \(y(x) = e^{-x/2}(C_1 \cos(\frac{\sqrt{3}}{2}x) + C_2 \sin(\frac{\sqrt{3}}{2}x)) + x \sin x\).

Step 1: Identify the Homogeneous Equation

The given differential equation is \(y'' + y' + y = x \sin x\). The corresponding homogeneous equation is \(y'' + y' + y = 0\).

Step 2: Solve the Homogeneous Equation

For the equation \(y'' + y' + y = 0\), assume a solution of the form \(y = e^{rx}\). Substituting this into the equation gives the characteristic equation \(r^2 + r + 1 = 0\). Solve this quadratic using the quadratic formula: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1, b=1, c=1\). This yields complex roots \(r = \frac{-1 \pm i\sqrt{3}}{2}\). Thus, the complementary solution is \(y_c = e^{-x/2}(C_1 \cos(\frac{\sqrt{3}}{2}x) + C_2 \sin(\frac{\sqrt{3}}{2}x))\).

Step 3: Formulate the Particular Solution Assumption

To find a particular solution \(y_p\) for \(x \sin x\), use undetermined coefficients. Assume \(y_p(x) = x(A \cos x + B \sin x)\). Include \(x\) to account for the non-homogeneous term, as the forcing function \(x\sin x\) suggests it.

Step 4: Differentiate the Particular Solution Assumption

Differentiate \(y_p(x) = x(A \cos x + B \sin x)\) to obtain \(y_p'(x)\) and \(y_p''(x)\). Using the product rule, we get:1. \(y_p'(x) = A (\cos x - x \sin x) + B (\sin x + x \cos x)\)2. \(y_p''(x) = A (-2\sin x - x \cos x) + B (2\cos x - x \sin x)\).

Step 5: Substitute into the Original Equation

Substitute \(y_p(x)\), \(y_p'(x)\), and \(y_p''(x)\) into the original differential equation. Simplify the equation to find coefficients \(A\) and \(B\):\[(-xA \cos x - xB \sin x) + (A \cos x - xB \sin x + B \sin x + xA \cos x) + (A \cos x - xA \sin x + B \sin x + xB \cos x) = x \sin x\]

Step 6: Solve for Coefficients

After simplifying, equate coefficients of \(\cos x\) and \(\sin x\) with those on the right side:- For \(\sin x\): \(-xA + xA + B = 1\), hence \(B = 1\).- For \(\cos x\): \(-xB + xB + A = 0\), hence \(A = 0\).

Step 7: Write the Particular Solution

With \(A = 0\) and \(B = 1\), the particular solution \(y_p(x)\) is \(x \sin x\).

Step 8: Combine Solutions for the General Solution

The general solution to the differential equation is the sum of the homogeneous and particular solutions:\[y(x) = y_c(x) + y_p(x) = e^{-x/2}(C_1 \cos(\frac{\sqrt{3}}{2}x) + C_2 \sin(\frac{\sqrt{3}}{2}x)) + x \sin x\]

Key Concepts

Homogeneous Equation

When solving differential equations, it is important to first address the homogeneous part of the equation. This simplification helps us understand how solutions behave when the non-homogeneous or external inputs are removed. In our exercise, the original equation is \( y'' + y' + y = x \sin x \). By setting the right side to zero, we get the corresponding homogeneous equation: \( y'' + y' + y = 0 \). This step isolates the equation to focus on the intrinsic characteristics of the system.To solve the homogeneous equation, we assume a solution of the form \( y = e^{rx} \). This approach often reveals key information about the system because the exponential function is fundamental to understanding growth and decay processes. Thus, the homogeneous equation is essential for building the complementary solution, which forms part of the overall solution.

Particular Solution

In differential equations, a particular solution addresses the specific effects of a non-homogeneous component. Here, the term \( x \sin x \) represents an external excitation or input to the system. To find a particular solution (\( y_p \)), we propose a form that mirrors the non-homogeneous term.In our problem, we hypothesize \( y_p(x) = x(A \cos x + B \sin x) \). The introduction of \( x \) accounts for the presence of a linear factor in the non-homogeneous term. Finding the correct form is crucial as it ensures that our particular solution matches the complexity of the external input.By differentiating \( y_p(x) \) and substituting back into the original differential equation, we systematically solve for the constants \( A \) and \( B \). In this exercise, calculations reveal that \( A = 0 \) and \( B = 1 \), leading to \( y_p(x) = x \sin x \) as our particular solution.

Characteristic Equation

A fundamental step in solving homogeneous linear differential equations is determining the characteristic equation. This equation is derived from substituting a trial solution \( y = e^{rx} \) into the homogeneous equation. For our case, \( y'' + y' + y = 0 \), this substitution yields the characteristic equation \( r^2 + r + 1 = 0 \).The characteristic equation is solved to find the roots \( r \), which are crucial for determining the complementary solution. Solving \( r^2 + r + 1 = 0 \) using the quadratic formula, we obtain complex roots: \( r = \frac{-1 \pm i\sqrt{3}}{2} \). This indicates that the solutions are oscillatory due to the imaginary component.The complex roots lead to a complementary solution of the form: \( y_c = e^{-x/2}(C_1 \cos(\frac{\sqrt{3}}{2}x) + C_2 \sin(\frac{\sqrt{3}}{2}x)) \). Understanding the characteristic equation helps us see how the system responds naturally, either by oscillating or simply decaying.

Undetermined Coefficients

The method of undetermined coefficients offers a straightforward way to find particular solutions for certain types of non-homogeneous linear differential equations. This technique involves guessing a form of the solution based on the non-homogeneous term's nature and structure. For our example, the term is \( x \sin x \).The approach requires making an educated guess about the form of the particular solution, \( y_p(x) \), including the functions that appear in the non-homogeneous term. Here, we chose \( y_p(x) = x(A \cos x + B \sin x) \). This hypothesis is informed by the form of \( x \sin x \), and the requirement to accommodate the linear \( x \).After forming the guess, we differentiate this assumed function and substitute it back into the original equation. The key is solving for the coefficients (\( A \) and \( B \)), which are often possible due to clever algebraic simplifications. Ultimately, the method of undetermined coefficients simplifies complex differential equation problems into manageable algebraic ones, aiding us in arriving at the correct particular solution.