Problem 10 - Advanced Engineering Mathematics

Question

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

Step 1: Identify the Type of Equation

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

Step 2: Solve the Homogeneous Equation

First, solve the homogeneous equation, which is \(y^{\prime \prime} + 2y^{\prime} = 0\). Its characteristic equation is \(r^2 + 2r = 0\), giving roots \(r = 0\) and \(r = -2\). Thus, the complementary solution is \(y_c = C_1 + C_2 e^{-2x}\).

Step 3: Formulate the Particular Solution

To solve the non-homogeneous part, propose a particular solution. Since the right-hand side is \(2x + 5 - e^{-2x}\), we use \(y_p = Ax + B\) for the polynomial \(2x + 5\) and \(Ce^{-2x}\) for the exponential part. Idea is to take derivatives of \(y_p = Ax + B + Ce^{-2x}\).

Step 4: Differentiate the Particular Solution

Calculate the first and second derivatives of \(y_p\). We get \(y_p' = A - 2Ce^{-2x}\) and \(y_p'' = 4Ce^{-2x}\).

Step 5: Insert Particular Solution into the Equation

Substitute \(y_p\), \(y_p'\), and \(y_p''\) into the original differential equation: \(4Ce^{-2x} + 2(A - 2Ce^{-2x}) = 2x + 5 - e^{-2x}\). Simplify the equation to find coefficients.

Step 6: Solve for Coefficients

Collect like terms: Solve \(2A = 2\) for \(A\), which gives \(A = 1\). Solve \(2B = 5\) for \(B\), giving \(B = \frac{5}{2}\). For the exponential part, solve \(2C - 3C = -1\) to find \(C = 1\).

Step 7: Combine Solutions

The general solution is the sum of the complementary and particular solutions: \(y(x) = C_1 + C_2 e^{-2x} + x + \frac{5}{2} + e^{-2x}\).

Key Concepts

Undetermined Coefficients

The method of undetermined coefficients is a technique used to find a particular solution to certain types of non-homogeneous differential equations. It is especially useful when the non-homogeneous part is a linear combination of exponentials, polynomials, sine, or cosine functions.

The basic idea is to assume a form for the particular solution, with undetermined coefficients.

• For a polynomial non-homogeneous part like "2x + 5", assume a polynomial form.
• For an exponential term like "e^{-2x}", assume an exponential form.

In the exercise, the assumed particular solution was the linear polynomial plus an exponential:
\[ y_p = Ax + B + Ce^{-2x} \]
Given this form, the next step is to adjust it by calculating derivatives and substituting them back into the original equation to find the values of the coefficients \(A\), \(B\), and \(C\).

Non-Homogeneous Equations

A non-homogeneous differential equation contains a term that is not dependent on the function or its derivatives. This term is often referred to as the **non-homogeneous part** or the *forcing function*.
In the given exercise, the differential equation is non-homogeneous and can be written as:

• \( y^{\prime \prime} + 2y^{\prime} = 2x + 5 - e^{-2x} \)

The expression "2x + 5 - e^{-2x}" is what makes the equation non-homogeneous.

The equation does not equal zero but rather a function of \(x\). Solving such equations involves finding both a complementary solution (solves the homogeneous equation) and a particular solution (accounts for the non-homogeneous part).

Complementary Solution

The complementary solution is derived by solving the homogeneous version of the differential equation.
Homogeneous equations set the differential equation to zero, removing the non-homogeneous part:

• \( y^{\prime \prime} + 2y^{\prime} = 0 \)

The solution to this is found by forming a characteristic equation:
\[ r^2 + 2r = 0 \]
Solving for \(r\) gives roots \(r=0\) and \(r=-2\), leading to the complementary solution:

• \( y_c = C_1 + C_2 e^{-2x} \)

This solution represents the homogeneous behavior of the equation without the influence of the non-homogeneous part. It consists of arbitrary constants \(C_1\) and \(C_2\) which are determined by initial or boundary conditions. These conditions were not part of the given exercise.

Particular Solution

The particular solution addresses the non-homogeneous aspect of the differential equation. For a chosen form that matches the function type on the right side of the equation, coefficients are determined to satisfy the entire equation.
Assuming a particular form:

• \( y_p = Ax + B + Ce^{-2x} \)

This includes a polynomial part \(Ax + B\) to counter the linear portion "2x + 5", and an exponential term \(Ce^{-2x}\) to account for "-e^{-2x}".

Once assumed, differentiate this trial solution, substitute back into the differential equation, and resolve to find the values of \(A\), \(B\), and \(C\).

• From the exercise, \( A = 1 \), \( B = \frac{5}{2} \), and \( C = 1 \).

Adding this particular solution to the complementary solution provides the general solution to the differential equation.