Problem 53 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given differential equation by undetermined coefficients. $$y^{\prime \prime}-2 y^{\prime}+5 y=e^{x} \sin x$$

Step-by-Step Solution

Answer

\( y = e^x (C_1 \cos(2x) + C_2 \sin(2x)) + e^x \left( \frac{1}{13} \cos x + \frac{4}{13} \sin x \right) \)

Step 1: Identify the Type of Differential Equation

The given differential equation is: \( y'' - 2y' + 5y = e^x \sin x \). This is a non-homogeneous linear differential equation with constant coefficients.

Step 2: Solve the Homogeneous Equation

First, solve the associated homogeneous equation \( y'' - 2y' + 5y = 0 \). This can be solved by finding the roots of the characteristic equation: \( r^2 - 2r + 5 = 0 \).

Step 3: Solve the Characteristic Equation

The characteristic equation is \( r^2 - 2r + 5 = 0 \). Solve for \( r \) using the quadratic formula: \( r = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} \).

Step 4: Calculate Roots of the Characteristic Equation

Calculate the discriminant: \( (-2)^2 - 4 \times 1 \times 5 = 4 - 20 = -16 \). Since the discriminant is negative, the roots are complex: \( r = 1 \pm 2i \).

Step 5: Write the General Solution of the Homogeneous Equation

The general solution of the homogeneous equation is \( y_h = e^x (C_1 \cos(2x) + C_2 \sin(2x)) \), where \( C_1 \) and \( C_2 \) are constants.

Step 6: Form the Particular Solution Guess

For the particular solution, use the method of undetermined coefficients. Guess the form \( y_p = e^x (A \cos x + B \sin x) \).

Step 7: Differentiate the Particular Solution Guess

Differentiate \( y_p = e^x (A \cos x + B \sin x) \) to find \( y'_p \) and \( y''_p \): \( y'_p = e^x ((A - B) \cos x + (A + B) \sin x) \).

Step 8: Compute the Second Derivative

Calculate the second derivative: \( y''_p = e^x ((A - 3B) \cos x + (3A + B) \sin x) \).

Step 9: Substitute into the Differential Equation

Substitute \( y_p, y'_p, \) and \( y''_p \) into \( y'' - 2y' + 5y = e^x \sin x \).

Step 10: Simplify to Find Coefficients

Set coefficients of \( \sin x \) and \( \cos x \) equal:1. \( A - 3B - 2(A - B) + 5A = 0 \) 2. \( 3A + B - 2(A + B) + 5B = 1 \).

Step 11: Solve the System of Equations

Solve the equations: \( (4A - B = 0) \) and \( (A + 3B = 1) \). First, from \( 4A = B \), substitute into the second equation: \( A + 3(4A) = 1 \).

Step 12: Calculate A and B

Solving \( A + 12A = 1 \) gives \( A = \frac{1}{13} \). Substitute back, \( B = 4 \times \frac{1}{13} = \frac{4}{13} \).

Step 13: Write the Particular Solution

The particular solution is \( y_p = e^x \left( \frac{1}{13} \cos x + \frac{4}{13} \sin x \right) \).

Step 14: Write the General Solution

Combine the homogeneous and particular solutions: \( y = e^x (C_1 \cos(2x) + C_2 \sin(2x)) + e^x \left( \frac{1}{13} \cos x + \frac{4}{13} \sin x \right) \).

Key Concepts

Non-homogeneous Differential Equation

A non-homogeneous differential equation stands apart because it includes a term without the dependent variable, making the solution a bit more interesting. In the provided problem, we have the differential equation \( y'' - 2y' + 5y = e^x \sin x \). This is non-homogeneous due to the \( e^x \sin x \).
Understanding this concept is key:

• The left side involves differential terms (\( y'' \), \( y' \), and \( y \)).
• The right side is a function of \( x \), which doesn't involve \( y \).
• The ultimate solution will be a combination of solving both its homogeneous counterpart \( y'' - 2y' + 5y = 0 \) and finding a particular solution for the non-homogeneous part.

Grasping these distinct sections of the equation helps bridge the knowledge for finding solutions through methods like undetermined coefficients.

Characteristic Equation

The characteristic equation is a pivotal tool when solving linear differential equations, especially homogeneous ones. For our equation \( y'' - 2y' + 5y = 0 \), the characteristic equation comes out as \( r^2 - 2r + 5 = 0 \). To derive it:

• Transform the differential equation into an algebraic one where each differentiation translates into a power of \( r \).
• Set the equation as \( r^2 - 2r + 5 = 0 \) to solve for roots.

The solutions of this equation give us the exponential functions, and potentially trigonometric functions, that form the structure of the equation's homogeneous solution.
This equation allows us to determine how the homogeneous part of the differential equation behaves over time. The nature of roots (real or complex) from this characteristic equation directs how the solutions are expressed.

Complex Roots

When dealing with differential equations, complex roots indicate oscillatory motion through sine and cosine functions in their solution. For the characteristic equation \( r^2 - 2r + 5 = 0 \), solving gives us complex roots \( r = 1 \pm 2i \). Here's what happens:

• The complex roots arise due to a negative discriminant in our quadratic formula setup.
• The form \( a \pm bi \) translates into the exponential \( e^{ax} \) damping (or growing) a wave function \( \cos(bx) \) and \( \sin(bx) \).

In this instance, our homogeneous solution becomes \( y_h = e^x (C_1 \cos(2x) + C_2 \sin(2x)) \).
The exponential part reflects any growth or decay based on the root's real part, while the imaginary part dictates oscillations, fittingly represented via sine and cosine. This intricate dance of numbers in complex solutions can elegantly map many real-world processes, such as electrical circuits and mechanical vibrations.

Particular Solution

The particular solution is concerned with addressing the non-homogeneous part of a differential equation. In the method of undetermined coefficients, we make an educated "guess" at the form of the particular solution based on the non-homogeneous term.In our problem, \( e^x \sin x \) suggests a form of \( y_p = e^x (A \cos x + B \sin x) \).
To find the specific values of \( A \) and \( B \):

• Derive the first and second derivatives of \( y_p \).
• Substitute back into the original differential equation.
• Equate coefficients of like terms to solve for \( A \) and \( B \).

Here, the values were solved as \( A = \frac{1}{13} \) and \( B = \frac{4}{13} \), giving the particular solution \( y_p = e^x \left( \frac{1}{13} \cos x + \frac{4}{13} \sin x \right) \).
Combining this particular solution with the homogeneous solution provides the general solution, encapsulating the full behavior of our original differential equation.