Problem 54 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given differential equation by undetermined coefficients. $$y^{\prime \prime}+y^{\prime}+\frac{1}{4} y=e^{x}(\sin 3 x-\cos 3 x)$$

Step-by-Step Solution

Answer

The general solution is \( y = C_1 e^{-\frac{1}{2}x} + C_2 xe^{-\frac{1}{2}x} + e^x \left( \frac{1}{19} \sin 3x + \frac{7}{57} \cos 3x \right) \).

Step 1: Identify the Complementary Function

To solve the differential equation, first find the complementary function (CF) by solving the associated homogeneous equation: \( y'' + y' + \frac{1}{4} y = 0 \). The characteristic equation \( m^2 + m + \frac{1}{4} = 0 \) is solved using the quadratic formula:\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Substitute \( a = 1 \), \( b = 1 \), \( c = \frac{1}{4} \), giving\[ m = \frac{-1 \pm \sqrt{1 - 1}}{2} = -\frac{1}{2} \].Thus, the CF is \( y_c = C_1 e^{-\frac{1}{2}x} + C_2 xe^{-\frac{1}{2}x} \).

Step 2: Formulate the Particular Solution

Apply the method of undetermined coefficients to find the particular solution (PS) to the non-homogeneous equation.Assume the PS form:\( y_p = e^{x}(A \sin 3x + B \cos 3x) \), where \( A \) and \( B \) are coefficients to be determined.

Step 3: Differentiate and Substitute

Differentiate \( y_p \) to find \( y_p' \) and \( y_p'' \):\[ y_p' = e^x(A \cos 3x \cdot 3 + B \sin 3x \cdot 3) + e^x(A \sin 3x + B \cos 3x) \],\[ y_p'' = e^x(-A \sin 3x \cdot 9 + B \cos 3x \cdot 9) + 2e^x(3A \cos 3x + 3B \sin 3x) \].Substitute \( y_p \), \( y_p' \), and \( y_p'' \) into the original equation and equate coefficients of \( e^x \sin 3x \) and \( e^x \cos 3x \).

Step 4: Solve for Coefficients

After substitution, gather terms:- For \( e^x \sin 3x \), the coefficient is \( A \cdot 10 + 3B = 1 \).- For \( e^x \cos 3x \), the coefficient is \( B \cdot 10 - 3A = -1 \).Solve this system of equations:From \( 10A + 3B = 1 \) and \( 10B - 3A = -1 \).Solving gives \( A = \frac{1}{19} \) and \( B = \frac{7}{57} \).

Step 5: Write the Particular Solution

Substitute the values of \( A \) and \( B \) back into the assumed form:\[ y_p = e^x \left( \frac{1}{19} \sin 3x + \frac{7}{57} \cos 3x \right) \].

Step 6: Form the General Solution

Combine the complementary function and the particular solution to form the general solution:\[ y = y_c + y_p = C_1 e^{-\frac{1}{2}x} + C_2 xe^{-\frac{1}{2}x} + e^x \left( \frac{1}{19} \sin 3x + \frac{7}{57} \cos 3x \right) \].

Key Concepts

Complementary Function

In the realm of differential equations, the complementary function (CF) plays a crucial role. It describes the solution to the associated homogeneous differential equation. For a given non-homogeneous equation, the homogeneous part is obtained by setting the non-homogeneous term to zero. In our example, we started with the homogeneous equation \( y'' + y' + \frac{1}{4} y = 0 \).

The next step involves finding its characteristic equation, which is essential to derive the complementary function. The characteristic equation for our problem is derived as \( m^2 + m + \frac{1}{4} = 0 \). By applying the quadratic formula, we find the roots, \( m = -\frac{1}{2} \). With repeated roots, we form the CF using these roots which leads to:

• \( y_c = C_1 e^{-\frac{1}{2}x} + C_2 xe^{-\frac{1}{2}x} \)

Understanding the CF helps grasp how solutions behave when the system is left to its own devices without external input. Remember, this part of the solution represents the general behavior of the system under homogeneous conditions.

Particular Solution

The particular solution (PS) in the context of differential equations addresses the influence of non-homogeneous terms, such as external forces or inputs to the system. This requires a distinct method from the complementary function. Here, we apply the method of undetermined coefficients, which cleverly guesses the form of the particular solution based on the non-homogeneous part.

For our equation, we hypothesized a PS of the form:

• \( y_p = e^{x}(A \sin 3x + B \cos 3x) \)

By differentiating this assumed solution, we could substitute back into the given differential equation. This step is where we find the values of \( A \) and \( B \) that satisfy the equation. Substitution and equating coefficients allow us to solve a system of equations leading to exact values for the coefficients; here, \( A = \frac{1}{19} \) and \( B = \frac{7}{57} \). The PS complements the CF, capturing specific external influences.

Non-homogeneous Differential Equation

Differential equations come in two flavors: homogeneous and non-homogeneous. The key distinction is the presence of a non-zero term independent of the dependent variable, often referred to as the forcing function or input. Our original equation:

• \( y'' + y' + \frac{1}{4} y = e^{x}(\sin 3x - \cos 3x) \)

is non-homogeneous due to the \( e^{x}(\sin 3x - \cos 3x) \) term.

Solving non-homogeneous differential equations requires combining solutions of the homogeneous equation (i.e., the complementary function) with a particular solution. This reflects both natural system responses and the impact of external forces. Identifying this non-homogeneous component is crucial in determining the appropriate approach, like undetermined coefficients or variation of parameters, to find the entire solution.

Characteristic Equation

The characteristic equation is a pivotal concept when dealing with linear differential equations, particularly of homogeneous type. It transitions us from the original differential equation to an algebraic equation that's typically easier to solve. In our example, by focusing on the homogeneous part \( y'' + y' + \frac{1}{4} y = 0 \), we developed the characteristic equation as follows:

• \( m^2 + m + \frac{1}{4} = 0 \)

The quadratic form of the characteristic equation allows us to use the quadratic formula to find the roots. These roots are crucial because they dictate the form of the complementary function. Here, solving the quadratic equation yields \( m = -\frac{1}{2} \) (a repeated root), which is indicative of solutions in the form \( e^{-\frac{1}{2}x} \) involving multiples of \( x \) due to the repeat. Comprehending and crafting the characteristic equation bridges the initial problem setup to obtaining valid and meaningful solutions.