Problem 64 - Calculus Volume 3

Question

Solve the following equations using the method of undetermined coefficients. $$ y^{\prime \prime}+3 y^{\prime}-28 y=10 e^{4 x} $$

Step-by-Step Solution

Answer

The general solution is \( y(x) = C_1 e^{4x} + C_2 e^{-7x} + \frac{10}{11}xe^{4x} \).

Step 1: Identify the Homogeneous Equation

First, solve the homogeneous equation: \[ y'' + 3y' - 28y = 0 \] This will give us the complementary solution, \( y_c(x) \).

Step 2: Find the Characteristic Equation

The characteristic equation for the homogeneous differential equation is:\[ r^2 + 3r - 28 = 0 \]

Step 3: Solve the Characteristic Equation

Solve for \(r\) using the quadratic formula, \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 1, b = 3, c = -28 \):\[ r = \frac{-3 \pm \sqrt{3^2 - 4 \times 1 \times (-28)}}{2} \]\[ r = \frac{-3 \pm \sqrt{9 + 112}}{2} \]\[ r = \frac{-3 \pm 11}{2} \]This results in \( r_1 = 4 \) and \( r_2 = -7 \).

Step 4: Write the Complementary Solution

With roots \( r_1 = 4 \) and \( r_2 = -7 \), the complementary solution is:\[ y_c(x) = C_1 e^{4x} + C_2 e^{-7x} \]

Step 5: Guess the Particular Solution

For the non-homogeneous term \( 10e^{4x} \), guess a particular solution of the form:\[ y_p(x) = Ae^{4x} \]However, since \( e^{4x} \) is a part of the complementary solution, multiply by \( x \) to avoid duplication:\[ y_p(x) = Axe^{4x} \]

Step 6: Substitute and Solve for Coefficient

Substitute \( y_p(x) = Axe^{4x} \) into the original equation. First, find derivatives:\[ y_p' = Ae^{4x} + 4Axe^{4x} \]\[ y_p'' = 8Ae^{4x} + 16Axe^{4x} \]Substitute these into the differential equation:\[ (8Ae^{4x} + 16Axe^{4x}) + 3(Ae^{4x} + 4Axe^{4x}) - 28Axe^{4x} = 10e^{4x} \]Combine like terms:\[ 11Ae^{4x} = 10e^{4x} \]This gives \( A = \frac{10}{11} \).

Step 7: Write the General Solution

Combine the complementary and particular solutions to get the general solution:\[ y(x) = C_1 e^{4x} + C_2 e^{-7x} + \frac{10}{11}xe^{4x} \]

Key Concepts

Homogeneous Equation

A homogeneous equation in the context of differential equations refers to an equation where all terms depend on the function and its derivatives only. This means there are no external forces or inputs affecting the system. In the exercise we’re exploring, the homogeneous equation is:
\[y'' + 3y' - 28y = 0\]
The reason we first consider the homogeneous equation is that it helps us find the complementary solution, which is a crucial part of solving the complete equation.

• By solving the homogeneous equation, we determine the natural behavior of the system without external influences.
• This step lays the groundwork for understanding how non-homogeneous solutions affect the overall system's output.

Complementary Solution

The complementary solution, denoted as \( y_c(x) \), is derived from the homogeneous equation. It represents the general solution of the system without any external forces or reservations. After identifying the homogeneous equation, the next step involves finding its characteristic equation:

• For \( y'' + 3y' - 28y = 0 \), the characteristic equation is \( r^2 + 3r - 28 = 0 \).

Solving this characteristic equation gives us the roots \( r_1 = 4 \) and \( r_2 = -7 \). These roots help us form the complementary solution:
\[ y_c(x) = C_1 e^{4x} + C_2 e^{-7x} \]

• The constants \( C_1 \) and \( C_2 \) are determined by initial conditions or boundary values of the actual problem.
• This solution captures the behavior of the system when no external forces are applied.

Quadratic Formula

The quadratic formula is a reliable method used to find the roots of any quadratic equation. In differential equations, the characteristic equation often comes in the quadratic form, and using the quadratic formula provides a straightforward approach to finding the roots.
The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \). The quadratic formula is:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Applying this to the characteristic equation \( r^2 + 3r - 28 = 0 \):

• Here, \( a = 1 \), \( b = 3 \), and \( c = -28 \).
• Substituting these values gives the roots \( r_1 = 4 \) and \( r_2 = -7 \). These roots are crucial for forming the complementary solution.

The quadratic formula not only provides the roots but also reveals the nature of the solutions based on the discriminant \( b^2 - 4ac \). In this scenario, the positive discriminant suggests two distinct real roots.

Particular Solution

The particular solution \( y_p(x) \) of a differential equation addresses the non-homogeneous part of the equation. In the exercise, this part is described by the term \( 10e^{4x} \). Finding \( y_p(x) \) involves a method called undetermined coefficients, where we guess a solution form and determine unknown coefficients.

• Initially, guess a solution of the form \( y_p(x) = Ae^{4x} \). Since \( e^{4x} \) is already part of the complementary solution, multiply by \( x \) to adjust the guess: \( y_p(x) = Axe^{4x} \).
• Substituting \( y_p(x) \) into the differential equation helps us solve for \( A \).
• Through substitution and simplification, we find that \( A = \frac{10}{11} \).

The particular solution now is \( y_p(x) = \frac{10}{11}xe^{4x} \). This particular solution complements the complementary solution:

• It combines with the homogeneous solution to provide a complete picture of the system’s behavior with external forces included.
• Helps in crafting the general solution \( y(x) \), which is a blend of \( y_c(x) \) and \( y_p(x) \).