Problem 57 - Calculus 3

Question

Solve the following equations using the method of undetermined coefficients. $$y^{\prime \prime}+16 y=e^{-2 x}$$

Step-by-Step Solution

Answer

The general solution is \(y = C_1 \cos(4x) + C_2 \sin(4x) + \frac{1}{20}e^{-2x}\).

Step 1: Solve the Homogeneous Equation

To solve the homogeneous equation, set the right-hand side to zero:\[ y'' + 16y = 0 \]The characteristic equation is obtained by replacing \(y\) with \(e^{rx}\), which gives us:\[ r^2 + 16 = 0 \]Solve the characteristic equation for \(r\):\[ r^2 = -16 \]\[ r = \, \pm 4i \]The general solution for the homogeneous part (complementary solution \(y_c\)) is then:\[ y_c = C_1 \cos(4x) + C_2 \sin(4x) \] where \(C_1\) and \(C_2\) are constants.

Step 2: Determine the Form of Particular Solution

Use the method of undetermined coefficients to guess the form of a particular solution \(y_p\). The non-homogeneous term is \(e^{-2x}\), which suggests a guess of the form \(y_p = Ae^{-2x}\), where \(A\) is a constant to be determined.

Step 3: Calculate the Derivatives of Particular Solution

Differentiate the guessed particular solution:\[ y_p = Ae^{-2x} \]First derivative:\[ y_p' = -2Ae^{-2x} \]Second derivative:\[ y_p'' = 4Ae^{-2x} \]

Step 4: Substitute and Solve for Coefficients

Substitute the derivatives into the original non-homogeneous equation:\[ y'' + 16y = e^{-2x} \]Substituting, we have:\[ 4Ae^{-2x} + 16(Ae^{-2x}) = e^{-2x} \]Combine like terms:\[ (4A + 16A)e^{-2x} = e^{-2x} \]\[ 20Ae^{-2x} = e^{-2x} \]Equate coefficients and solve for \(A\):\[ 20A = 1 \]\[ A = \frac{1}{20} \]

Step 5: Write the General Solution

The general solution to the differential equation is the sum of the complementary and particular solutions:\[ y = y_c + y_p = C_1 \cos(4x) + C_2 \sin(4x) + \frac{1}{20}e^{-2x} \]

Step 6: Final Check

Verify the solution by differentiating the particular solution parts and substituting back into the original equation to ensure everything satisfies the initial differential equation. This confirms correctness.

Key Concepts

Homogeneous Equations

A homogeneous equation is a type of differential equation where, upon setting the non-homogeneous part to zero, the equation remains balanced. For example, in the given problem, setting the equation to zero leads us to:

• \( y'' + 16y = 0 \)

Solving this involves finding the characteristic equation and its roots. The characteristic equation arises when we assume a solution of the form \( y = e^{rx} \). Substituting this form into the homogeneous equation allows us to solve for \( r \), finding the equation's inherent structure. The roots of the characteristic equation determine the form of the general solution or complementary solution, \( y_c \). For second-order linear homogeneous equations with constant coefficients, the solutions often involve combinations of exponential, sine, and cosine functions.

Non-Homogeneous Equations

Non-homogeneous equations include an additional term that is not present in homogeneous equations. This term means the equation does not equal zero, introducing external influence or a driving force to the system. Take the initial equation:

• \( y'' + 16y = e^{-2x} \)

The term \( e^{-2x} \) makes it non-homogeneous. Solving such equations involves finding solutions to both the homogeneous part (\( y_c \)) and a particular solution (\( y_p \)), which responds directly to the non-homogeneous part. Together, these solutions form the general solution for the entire equation, which accounts for both natural and forced behaviors of the system described by the differential equation.

Particular Solution

A particular solution, \( y_p \), directly addresses the non-homogeneous part of the differential equation. The method of undetermined coefficients is a common approach to determining this solution. With undetermined coefficients, you begin by guessing a form that mirrors the non-homogeneous term. In our example, since the non-homogeneous term is \( e^{-2x} \), we guess:

• \( y_p = Ae^{-2x} \)

Next, differentiate and substitute back into the original equation to solve for the constant coefficient \( A \). This gives a solution that specifically counteracts the effect of the non-homogeneous term, like calibrating a response to an external force. The perfected guesswork leads us to fine-tune the particulars of \( y_p \), thus completing the profile of the system's behavior.

Characteristic Equation

The characteristic equation is crucial in solving homogeneous differential equations. It emerges by switching the differential equation into polynomials through assuming the solution form \( y = e^{rx} \). Replacing derivatives with polynomial functions of \( r \), it turns the problem from differential into algebraic. For the given problem:

• \( y'' + 16y = 0 \) leads to the characteristic equation \( r^2 + 16 = 0 \).

Factoring and solving for \( r \) determines the nature of the roots:

• Complex roots like \( \pm 4i \) indicate oscillatory behavior. Solutions are then expressed in terms of sine and cosine.

This step is pivotal as it prescribes the baseline solution for the homogeneous part—laying the groundwork for further solving both homogeneous and non-homogeneous differential equations.