Problem 10 - Advanced Engineering Mathematics

Question

In Problems 1-26, solve the given differential equation by undetermined coefficients. $$ y^{\prime \prime}+2 y^{\prime}=2 x+5-e^{-2 x} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the differential equation type

The given equation is a second-order linear non-homogeneous differential equation: \[ y'' + 2y' = 2x + 5 - e^{-2x} \] This equation can be solved using the method of undetermined coefficients due to its specific non-homogeneous terms (polynomial and exponential).

Step 2: Solve the complementary equation (homogeneous part)

First, solve the complementary (homogeneous) equation:\[ y'' + 2y' = 0 \]To find the complementary solution, assume \( y_c = e^{rx} \) and substitute into the homogeneous equation:\[ r^2 e^{rx} + 2r e^{rx} = 0 \]Factor out \( e^{rx} \):\[ e^{rx}(r^2 + 2r) = 0 \]The characteristic equation is \( r^2 + 2r = 0 \).Solve for \( r \):\[ r(r + 2) = 0 \] \[ r = 0 \text{ or } r = -2 \]Thus, the complementary solution is:\[ y_c = C_1 + C_2 e^{-2x} \]

Step 3: Determine the form of the particular solution

The non-homogeneous part is \( 2x + 5 - e^{-2x} \).The particular solution \( y_p \) can be assumed to have the form \( y_p = Ax + B + Ce^{-2x} \).However, note that \( e^{-2x} \) is a solution of the complementary equation, so the form of \( y_p \) is adjusted to avoid duplication with \( y_c \). Let:\[ y_p = Ax + B + Cxe^{-2x} \]

Step 4: Substitute and determine coefficients for the particular solution

Compute the derivatives:\[ y_p' = A - 2Cx e^{-2x} + Ce^{-2x} \]\[ y_p'' = 4Cxe^{-2x} - 4Ce^{-2x} \]Substitute \( y_p, y_p', \text{ and } y_p'' \) into the original differential equation:\[ 4Cxe^{-2x} - 4Ce^{-2x} + 2(A - 2Cx e^{-2x} + Ce^{-2x}) = 2x + 5 - e^{-2x} \] Simplify and equate coefficients:1. From the polynomial terms \( Ax \) and constant terms: - \( 2A = 2 \Rightarrow A = 1 \) - \( 2B = 5 \Rightarrow B = rac{5}{2} \)2. From coefficients of \( e^{-2x} \) and terms: - \( 4C - 4C + 2C = -1 \Rightarrow 2C = -1 \Rightarrow C = -rac{1}{2} \)Thus, the particular solution is:\[ y_p = x + \frac{5}{2} - \frac{1}{2}xe^{-2x} \]

Step 5: Form the general solution

Combine the complementary and particular solutions to form the general solution:\[ y = y_c + y_p \]\[ y = C_1 + C_2 e^{-2x} + x + \frac{5}{2} - \frac{1}{2}xe^{-2x} \]

Key Concepts

Non-homogeneous Differential Equation

A non-homogeneous differential equation is a type of differential equation where the function is not set to zero. In simpler terms, it has terms that do not involve the dependent variable or its derivatives. Unlike homogeneous differential equations, their solutions consist of two parts. One part relates to the homogeneous equation, and the other addresses the non-homogeneous part.

To identify a non-homogeneous differential equation, look for the presence of an additional term — in this case, the term is on the right side of the equation, like the polynomial and exponential terms: \(2x + 5 - e^{-2x}\). These non-homogeneous terms mean the solution needs to account for both the structure of the equation and its driving force (or source) found in these terms.

• The given differential equation, \(y'' + 2y' = 2x + 5 - e^{-2x}\), is a classic example of a non-homogeneous differential equation.
• Since the equation is non-homogeneous, the solution must address the added complexity introduced by these extra terms.
  

Complementary Solution

The complementary solution refers to the solution of the homogeneous part of a differential equation. It is derived from the left side of the original equation, where the equation is set equal to zero.

To find the complementary solution, consider only the homogeneous portion of the equation: \(y'' + 2y' = 0\). Here, you assume a solution of the form \(y_c = e^{rx}\). To determine \(r\), substitute \(y_c\) back into the equation and solve the ensuing algebraic equation known as the characteristic equation:

• \(r^2 + 2r = 0\) is the homogeneous characteristic equation
• Solving this gives roots \(r = 0\) and \(r = -2\)
• These roots present the complementary solution as \(y_c = C_1 + C_2 e^{-2x}\)

The complementary solution represents the general behavior of the system without any external influence, responding generally to initial conditions or behavior of similar past systems.

Particular Solution

The particular solution is the part of the differential equation's solution that accounts for any non-homogeneous terms. These non-homogeneous terms should result in solutions that do not belong to the complementary solution. So, you need a strategy to account for these.

For the equation \(y'' + 2y' = 2x + 5 - e^{-2x}\), it is essential to hypothesize a form for the particular solution, \(y_p\), according to the non-homogeneous terms present:

• Initially assumed \(y_p = Ax + B + Ce^{-2x}\) needs to adjust due to overlap with the complementary solution.\

Modify this to avoid duplicating terms in the complementary solution:

• Use \(y_p = Ax + B + Cxe^{-2x}\)

After substitution into the original equation and solving for coefficients, this leads to:

• \(A = 1, B = \frac{5}{2}, C = -\frac{1}{2}\)
• Thus, \(y_p = x + \frac{5}{2} - \frac{1}{2}xe^{-2x}\)

Characteristic Equation

The characteristic equation comes into play when solving the complementary solution of a homogeneous differential equation. It is an algebraic equation whose solutions dictate the behavior and form of the complementary solution.

Obtained from assuming a solution of the form \(y_c = e^{rx}\), substituting into the differential equation gives rise to an equation in terms of \(r\). Specifically for the differential equation \(y'' + 2y' = 0\), substituting \(y_c = e^{rx}\) results in:

• \(r^2 + 2r = 0\)
• Solving gives \(r = 0\) and \(r = -2\)
• The characteristic equation is solved for these roots

These roots signify the exponents in the complementary solution. Real, distinct roots (as in this case) provide exponential functions in the solution, illustrating growth or decay tendencies in the system.