Problem 44 - Differential Equations and Linear Algebra

Question

Use the annihilator method to solve the given differential equation. $$y^{\prime \prime}-8 y^{\prime}+16 y=5 e^{4 x}.$$

Step-by-Step Solution

Answer

The general solution for the given differential equation using the annihilator method is: \[ y(x) = C_1 e^{4x} + (C_2 - \frac{5}{28})xe^{4x} \] where \(C_1\) and \(C_2\) are constants to be determined based on initial conditions or boundary conditions, if given.

Step 1: Identify the complementary function (CF)

First, let's rewrite the given differential equation (DE) and solve its homogeneous equation: \[ y''(x) - 8y'(x) + 16y(x) = 0 \] For the homogeneous equation, we can let \(y = e^{rx}\), where r is a constant. Plug this into the equation: \[ (r^2 - 8r + 16)e^{rx} = 0 \] Since \(e^{rx} \neq 0\), we have a quadratic equation for the constant r: \[ r^2 - 8r + 16 = 0 \] By factoring the quadratic equation, we get: \[ (r - 4)^2 = 0 \] The root of this equation is \(r=4\), and since it is a repeated root, the complementary function (CF) will have the form: \[ y_c(x) = C_1 e^{4x} + C_2 xe^{4x} \] Where \(C_1\) and \(C_2\) are constants to be determined later.

Step 2: Find the particular solution using annihilator method

The given DE is: \[ y''(x) - 8y'(x) + 16y(x) = 5e^{4x} \] Since the right side (RHS) of the DE is of the form \(5e^{4x}\), we can apply the annihilator method. The annihilator for \(e^{4x}\), in this case, will be \((D-4)\), where \(D\) is the differential operator. So, we apply \((D-4)\) twice (as the RHS already contains the complementary function): \[ (D-4)^2 y_p(x) = 5e^{4x} \] Now, we find the general solution for the equation: \[ y_p''(x) - 8y_p'(x) + 16y_p(x) = 5e^{4x} \] Assume the particular solution \(y_p(x) = Axe^{4x}\), where \(A\) is a constant to be determined later. Then: \[ y_p'(x) = A(4xe^{4x} + 4e^{4x}) \] \[ y_p''(x) = A(4^2e^{4x} + 8e^{4x}) \] Substitute these expressions to the DE: \[ A(4^2e^{4x} + 8e^{4x}) - 8A(4xe^{4x} + 4e^{4x}) + 16Axe^{4x} = 5e^{4x} \] Simplify the equation: \[ (4A - 32A)e^{4x} + (64A - 64A)xe^{4x} = 5e^{4x} \] Comparing the coefficients, we obtain: \[ 4A - 32A = 5 \Rightarrow A = -\frac{5}{28} \] Thus, the particular solution is: \[ y_p(x) = -\frac{5}{28}xe^{4x} \]

Step 3: Find the general solution

Combine the complementary function and particular solution to obtain the general solution: \[ y(x) = y_c(x) + y_p(x) = C_1 e^{4x} + C_2 xe^{4x} -\frac{5}{28}xe^{4x} \] So, the general solution for the given DE is: \[ y(x) = C_1 e^{4x} + (C_2 - \frac{5}{28})xe^{4x} \] Where \(C_1\) and \(C_2\) are constants to be determined based on initial conditions or boundary conditions, if given.

Key Concepts

Complementary Function

When solving linear differential equations like the one given, a critical step involves finding the Complementary Function (CF). This function represents the solution to the associated homogeneous equation, which assumes the right-hand side (RHS) is zero. For our differential equation: \[ y''(x) - 8y'(x) + 16y(x) = 0 \]we replace it with a trial function \( y = e^{rx} \), where \( r \) is an unknown constant. This leads to a characteristic equation—a crucial quadratic equation—providing insights into the nature of the roots. In our example, the characteristic equation is:\[ r^2 - 8r + 16 = 0 \]Factoring gives \((r - 4)^2 = 0\), indicating a repeated root \( r = 4 \). A repeated root means our complementary function will comprise terms that increase linearly with \( x \). Specifically, the CF is:- \( y_c(x) = C_1 e^{4x} + C_2 xe^{4x} \)Here, \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial or boundary conditions. The inclusion of \( xe^{4x} \) accounts for the repetition of the root, ensuring a more comprehensive solution space.

Particular Solution

Finding the Particular Solution involves catering to the nonhomogeneous part of the differential equation—here, this is given by \( 5e^{4x} \). Using the annihilator method, we introduce a differential operator \((D-4)\) that would "annihilate" or zero out \( e^{4x} \). Hence, applying it twice on our equation is necessary as the RHS shares terms with our complementary function.Assume our particular solution \( y_p(x) = Axe^{4x} \)—a form adjusted due to the repeated root shared with the CF. By substituting \( y_p \), its derivatives, and simplifying, matching coefficients allow us to solve for \( A \). In this scenario, calculations reveal:- \( A = -\frac{5}{28} \)So, our particular solution becomes:- \( y_p(x) = -\frac{5}{28}xe^{4x} \)This matches specifically the extra term in the differential equation, satisfying the equation's nonhomogeneous part by introducing correct balancing terms.

Homogeneous Equation

The Homogeneous Equation is foundational in solving differential equations, arising when the equation equals zero—used to form the general solution's structural backbone.In simple terms, it extracts the natural modes of the system's behavior without external inputs or forces. For: \[ y''(x) - 8y'(x) + 16y(x) = 0 \]we address it by locating its characteristic equation through a substitution method, namely: \( y = e^{rx} \). The resulting quadratic equation: \[ r^2 - 8r + 16 = 0 \]leads us to analyze solutions based on root properties:- Repeated roots imply solutions that grow similarly yet differ by multiplying by \( x \).In essence, the familiarity with the homogeneous equation allows us to form the complementary function crucial for the DE's general solution structure. The fundamental solutions adopted here lay the groundwork upon which particular solutions are carefully constructed and compounded.