Problem 39 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given differential equation by undetermined coefficients. $$y^{\prime \prime}+4 y^{\prime}+4 y=2 x+6$$

Step-by-Step Solution

Answer

The solution is \( y = C_1 e^{-2x} + C_2 x e^{-2x} + \frac{1}{2}x + \frac{5}{4} \).

Step 1: Identify the Form of the Differential Equation

The given differential equation is of the form \( y'' + 4y' + 4y = 2x + 6 \). This is a linear non-homogeneous differential equation with constant coefficients.

Step 2: Solve the Complementary Equation

To find the complementary solution \(y_c\), solve the associated homogeneous equation: \( y'' + 4y' + 4y = 0 \). The characteristic equation is \( r^2 + 4r + 4 = 0 \). This can be factored to \( (r+2)^2 = 0 \), giving a repeated root \( r = -2 \). Thus, the complementary solution is \( y_c = C_1 e^{-2x} + C_2 x e^{-2x} \).

Step 3: Choose a Form for the Particular Solution

Since the non-homogeneous term on the right side is \( 2x + 6 \), propose a particular solution of the form \( y_p = Ax + B \) where \(A\) and \(B\) are constants to be determined.

Step 4: Differentiate the Particular Solution

Find the first and second derivatives of the proposed particular solution: \( y_p = Ax + B \) gives \( y_p' = A \) and \( y_p'' = 0 \).

Step 5: Substitute into the Differential Equation

Substitute \( y_p \), \( y_p' \), and \( y_p'' \) into the original differential equation: \( 0 + 4A + 4(Ax + B) = 2x + 6 \). Simplifying gives \( 4Ax + 4A + 4B = 2x + 6 \).

Step 6: Solve for Constants A and B

Match the coefficients on both sides of the equation: from \(4A = 2 \), solve for \(A\) to get \( A = \frac{1}{2} \). From \(4A + 4B = 6\), substitute \(A = \frac{1}{2}\) to find \( B = \frac{5}{4} \).

Step 7: Write the General Solution

Combine the complementary solution and the particular solution to get the general solution: \( y = y_c + y_p = C_1 e^{-2x} + C_2 x e^{-2x} + \frac{1}{2}x + \frac{5}{4} \).

Key Concepts

Undetermined Coefficients

The method of Undetermined Coefficients is a technique used to find particular solutions of linear non-homogeneous differential equations. It is especially useful when the non-homogeneous term in the equation is a simple polynomial, exponential, sine, or cosine function. The essence of this method lies in assuming a possible form of the particular solution. We then determine the unknown coefficients by substituting back into the original equation. This process involves:

• Identifying the nature of the non-homogeneous term.
• Proposing an appropriate form for the particular solution with undetermined coefficients.
• Calculating the derivatives of this assumed solution.
• Substituting these into the given differential equation to find the values of the coefficients.

A key advantage of this method is its simplicity and applicability to a wide range of problems with straightforward non-homogeneous terms. Remember: it only works well for certain types of functions.

Non-homogeneous Equation

A non-homogeneous differential equation is one that involves a non-zero term that is not a function of the dependent variable and its derivatives. In the equation presented, this term is "2x + 6", which makes the equation "non-homogeneous". Compared to a homogeneous equation, where the right-hand side is zero, non-homogeneous equations require solving an additional part called the "particular solution".
A non-homogeneous equation generally has the form:
\[ ay^{(n)} + by^{(n-1)} + \, \ldots \, + ny = g(x) \]

• The term \( g(x) \) is the non-homogeneous part.
• The goal in solving such equations is to find a solution to the full equation, combining solutions to the homogeneous part and this non-homogeneous term.

The combination of these two solutions gives the general solution, which expresses all possible solutions of the equation.

Complementary Solution

The Complementary Solution, often symbolized as \( y_c \), addresses the homogeneous aspect of the differential equation. It represents the component of the general solution that accounts for the left-hand part \( y'' + 4y' + 4y = 0 \). To find this, typically a characteristic equation is used, arising from setting the differential equation equal to zero.
For our exercise:

• The characteristic equation is obtained by treating the derivatives as polynomial terms: \( r^2 + 4r + 4 \).
• Solving this gives repeated roots, in this case, \( r = -2 \).
• The complementary solution for a repeated root \( r \) is \( y_c = C_1e^{-2x} + C_2xe^{-2x} \).

This shows how the solutions superpose and adapt when dealing with repeated roots, ensuring the completeness of the solution.

Particular Solution

The Particular Solution \( y_p \) is crafted to address the specific nature of the non-homogeneous term \( 2x + 6 \) in our equation. Its role is to find one specific solution to the differential equation that includes the non-homogeneous part.
Here's how we determine it:

• We assume a technical form that aligns with the form of the non-homogeneous term. Here, \( y_p = Ax + B \) was chosen, as it mirrors the nature of \( 2x + 6 \).
• Calculate the derivatives: \( y_p' = A \) and \( y_p'' = 0 \).
• Substitute these into the differential equation to establish parameters for \( A \) and \( B \).
• Solve the resulting system of equations, deriving specific values: \( A = \frac{1}{2} \) and \( B = \frac{5}{4} \).

Combining the particular solution with the complementary solution results in the full description of the equation's solution.