Problem 35 - A First Course in Differential Equations with Modeling Applications

Question

(a) Verify that \(y_{p_{1}}=3 e^{2 x}\) and \(y_{p_{2}}=x^{2}+3 x\) are, respectively, particular solutions of \(y^{\prime \prime}-6 y^{\prime}+5 y=-9 e^{2 x}\) and \(y^{\prime \prime}-6 y^{\prime}+5 y=5 x^{2}+3 x-16\). (b) Use part (a) to find particular solutions of \(y^{\prime \prime}-6 y^{\prime}+5 y=5 x^{2}+3 x-16-9 e^{2 x}\) and \(\quad y^{\prime \prime}-6 y^{\prime}+5 y=-10 x^{2}-6 x+32+e^{2 x}\).

Step-by-Step Solution

Answer

Sum particular solutions: \(3e^{2x} + x^{2} + 3x\) and \(e^{2x} - 2x^{2} - 6x\).

Step 1: Verify Particular Solution for First Equation

We first substitute \(y_{p_{1}}=3 e^{2x}\) into the equation \(y'' - 6y' + 5y = -9e^{2x}\). Begin by computing the derivatives:- First derivative: \(y_{p_{1}}' = 6e^{2x}\)- Second derivative: \(y_{p_{1}}'' = 12e^{2x}\)Substitute into the differential equation:\[ 12e^{2x} - 6(6e^{2x}) + 5(3e^{2x}) = 12e^{2x} - 36e^{2x} + 15e^{2x} = -9e^{2x} \]Thus, \(y_{p_{1}}=3 e^{2x}\) is indeed a particular solution of the first equation.

Step 2: Verify Particular Solution for Second Equation

Substitute \(y_{p_{2}}=x^{2}+3x\) into the equation \(y'' - 6y' + 5y = 5x^2 + 3x - 16\). Calculate the derivatives:- First derivative: \(y_{p_{2}}' = 2x + 3\)- Second derivative: \(y_{p_{2}}'' = 2\)Substitute into the differential equation:\[ 2 - 6(2x + 3) + 5(x^2 + 3x) = 2 - 12x - 18 + 5x^2 + 15x = 5x^2 + 3x - 16 \]Thus, \(y_{p_{2}}=x^{2}+3x\) is a particular solution of the second equation.

Step 3: Find Particular Solution for Combined Equation

Now consider \(y'' - 6y' + 5y = 5x^2 + 3x - 16 - 9e^{2x}\). A particular solution would be the sum of the solutions from parts (a), i.e., \(y_{p} = y_{p_{1}} + y_{p_{2}} = 3e^{2x} + x^{2} + 3x\). This is because the right side of the equation is a combination of the two parts we verified as particular solutions.

Step 4: Find Particular Solution for Altered Equation

For \(y'' - 6y' + 5y = -10x^2 - 6x + 32 + e^{2x}\), the idea is to adjust the previously found particular solutions. Notice:- Change the polynomial component so that it becomes: \(-2(x^{2} + 3x) = -2x^{2} - 6x\)- For the exponential part, simply take \(y_{p_{1}} = e^{2x}\)Thus, the particular solution can be computed as:\[y_{p_{3}} = e^{2x} - 2(x^{2} + 3x) = e^{2x} - 2x^{2} - 6x\]

Key Concepts

Particular Solutions

In the context of differential equations, a particular solution is a specific solution that satisfies a given non-homogeneous differential equation. It incorporates the particular non-homogeneous parts of the equation's right-hand side. In simpler terms, when you have a differential equation that looks something like this:

• \( y'' - 6y' + 5y = f(x) \)

The goal is to find a function \( y_p \) which, when substituted into the equation, yields exactly \( f(x) \) on the right-hand side. Particular solutions are essential when dealing with linear differential equations, as they help us solve for the specific form that includes the effects of external forces or inputs to the system. These solutions differ from general solutions, which include constants of integration and represent the family of all possible solutions for the homogeneous part of the equation.

Second Order Linear Equations

A second order linear differential equation is an equation that involves the second derivative of a function. These equations often appear in both physics and engineering, modeling systems that involve acceleration or curvature. The general form of such an equation is:

• \( a(x)y'' + b(x)y' + c(x)y = f(x) \)

In this equation, \( a(x), b(x), \) and \( c(x) \) are functions of \( x \), while \( f(x) \) is the non-homogeneous part or the input function. The term "linear" indicates that the unknown function and its derivatives appear to the first power (there are no products of the function or its derivatives). Second order linear equations are significant because they can describe a wide variety of physical phenomena, such as oscillations, waves, and electrical circuits.

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique used to find particular solutions to linear differential equations with constant coefficients. This method is particularly effective for equations where the non-homogeneous term \( f(x) \) is of a particular type, such as exponential, polynomial, or sinusoidal functions. The key idea is simple:

• Guess a form for the particular solution \( y_p \), with coefficients that are initially undetermined.
• Substitute this guess into the differential equation.
• Solve for the coefficients so that the equation holds true.

Using this approach, you create an educated guess based on the form of \( f(x) \). For example, if \( f(x) = e^{2x} \), then a suitable assumption for \( y_p \) might be \( Ae^{2x} \). You then calculate the derivatives of this assumption, plug them back into the differential equation, and solve for the unknown coefficients \( A \). This process adjusts the solution specifically to satisfy the particular input provided by \( f(x) \).

Verification of Solutions

Verifying a solution to a differential equation involves demonstrating that it actually satisfies the given equation. This is a crucial step to ensure that the particular solution you've worked out is accurate. Verification typically involves:

• Computing the derivatives of the proposed particular solution.
• Substituting these derivatives, along with the solution itself, back into the original differential equation.
• Checking if the left-hand side equals the right-hand side of the equation.

For example, if you have already determined that \( y_{p_{1}} = 3e^{2x} \) is a solution for
\( y'' - 6y' + 5y = -9e^{2x} \), you would
substitute back, compute the required derivatives, and simplify to confirm that the equality holds.
This ensures that all terms balance as expected, confirming the validity of the proposed particular solution. Verification helps in not only proving your solution works but also deepening your understanding of how differential equations function.