Chapter 13: Problem 2
In Exercises \(2-7\) solve the boundary value problem. $$ y^{\prime \prime}-y=x, \quad y(0)=-2, \quad y(1)=1 $$
Short Answer
Expert verified
Answer: The solution to the boundary value problem is \(y(x) = -\frac{1}{e+1}e^x - \frac{2e}{e+1}e^{-x} - \frac{1}{2}x + \frac{3}{2}\).
Step by step solution
01
Identify the given equation and boundary conditions
We are given the boundary-value problem:
$$
y^{\prime\prime} - y = x; \quad y(0)=-2, \quad y(1)=1.
$$
Our task is to find a function y(x) that satisfies this equation and the given boundary conditions.
02
Understand the structure of the solution and find the complementary function
The general solution of the given equation has the form:
$$
y(x) = y_c(x) + y_p(x),
$$
where \(y_c(x)\) is the complementary function (solution of the homogeneous equation), and \(y_p(x)\) is the particular solution (related to the non-homogeneous term "\(x\) ").
The homogeneous equation is:
$$
y^{\prime\prime} - y = 0.
$$
We will assume a solution of the form \(y_c(x) = Ae^{rx}\). Substituting this into the homogeneous equation, we get:
$$
r^2e^{rx} - e^{rx} = 0.
$$
Factoring out \(e^{rx}\), we obtain:
$$
e^{rx} (r^2 - 1) = 0.
$$
Which yields r=1 and r=-1. Therefore the complementary function is:
$$
y_c(x) = C_1e^x + C_2e^{-x}.
$$
03
Find the particular solution
For the particular solution, we will assume a solution in the form \(y_p(x) = ax + b\). Taking the derivatives and substituting them into the given equation, we get:
$$
(ax+b)'' - (ax+b) = x.
$$
After calculating the derivatives, we get:
$$
-2a - (ax+b) = x.
$$
Comparing the coefficients of x and the constant terms, we find that:
$$
-2a = 1, \quad -a-b = 2.
$$
Solving this system of equations, we get \(a=-\frac{1}{2}\) and \(b=\frac{3}{2}\). Therefore, the particular solution is:
$$
y_p(x) = -\frac{1}{2}x + \frac{3}{2}.
$$
04
Combine the complementary and particular solutions
Now that we have both the complementary and particular solutions, let's combine them to form the general solution:
$$
y(x) = C_1e^x + C_2e^{-x} - \frac{1}{2}x + \frac{3}{2}.
$$
05
Apply the boundary conditions to find the constants
We are given the boundary conditions \(y(0)=-2\) and \(y(1)=1\). Let's apply these conditions to the general solution we found in the previous step.
For \(y(0)=-2\):
$$
-2 = C_1e^0 + C_2e^0 - \frac{1}{2}\cdot0 + \frac{3}{2}.
$$
Simplifying the equation, we get:
$$
C_1 + C_2 = -3.
$$
For \(y(1)=1\):
$$
1 = C_1e^1 + C_2e^{-1} - \frac{1}{2}\cdot1 + \frac{3}{2}.
$$
Simplifying the equation, we get:
$$
C_1 + \frac{1}{e}C_2 = 1.
$$
06
Solve for the constants C1 and C2
Now we need to solve the linear system:
$$
\begin{cases}
C_1 + C_2 = -3 \\
C_1 + \frac{1}{e}C_2 = 1
\end{cases}
$$
Solving this system, we find \(C_1 = -\frac{1}{e+1}\) and \(C_2 = -\frac{2e}{e+1}\).
07
Write the final solution
Substituting the values of the constants C1 and C2 into the general solution, we obtain the solution to the boundary value problem:
$$
y(x) = -\frac{1}{e+1}e^x - \frac{2e}{e+1}e^{-x} - \frac{1}{2}x + \frac{3}{2}.
$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
Understanding differential equations is essential in various fields of science and engineering, as they describe relationships involving rates of change. A differential equation is a mathematical equation that relates some function of one or more variables with its derivatives. In simpler terms, it tells us how a quantity changes as it is influenced by other quantities. For instance, in physics, these equations can model the way the velocity of an object changes over time, reflecting its acceleration.
Differential equations can be categorized based on their order, linearity, and whether they are ordinary (if there's only one independent variable) or partial (when there are multiple independent variables). The order of a differential equation is determined by the highest derivative present in the equation. For example, the equation from the exercise given, \( y'' - y = x \), is an example of a second-order linear ordinary differential equation, because the highest derivative is the second derivative \( y'' \) and it only contains one independent variable.
When solving differential equations, it's often beneficial to look at simplified forms of the equation to find solutions. This involves finding the general solution, which encompasses all possible solutions, often including arbitrary constants. These constants are typically determined by additional information, such as initial or boundary conditions, leading to a specific solution applicable to the problem at hand. The exercise given where you are to solve \( y'' - y = x \) with certain boundary conditions is an example of how we apply these techniques in practice.
Differential equations can be categorized based on their order, linearity, and whether they are ordinary (if there's only one independent variable) or partial (when there are multiple independent variables). The order of a differential equation is determined by the highest derivative present in the equation. For example, the equation from the exercise given, \( y'' - y = x \), is an example of a second-order linear ordinary differential equation, because the highest derivative is the second derivative \( y'' \) and it only contains one independent variable.
When solving differential equations, it's often beneficial to look at simplified forms of the equation to find solutions. This involves finding the general solution, which encompasses all possible solutions, often including arbitrary constants. These constants are typically determined by additional information, such as initial or boundary conditions, leading to a specific solution applicable to the problem at hand. The exercise given where you are to solve \( y'' - y = x \) with certain boundary conditions is an example of how we apply these techniques in practice.
Particular Solution
The particular solution is a single, specific solution to a non-homogeneous differential equation that not only satisfies the differential equation but also fits certain initial or boundary conditions given as part of the problem. Unlike the complementary function that addresses the associated homogeneous equation (an equation set to zero), the particular solution is directly affected by the non-homogeneous term, in this case, 'x'.
To identify a particular solution, we often use methods such as undetermined coefficients or variation of parameters, depending on the form of the non-homogeneous term. In our exercise, for instance, we see the method of undetermined coefficients in action. The choice of \( y_p(x) = ax + b \) as a trial solution is an educated guess based on the degree and type of the non-homogeneous term 'x'. After differentiating and substituting back into the original equation, we're able to determine the coefficients 'a' and 'b' that make the trial solution work. It's like fitting a key into a lock; only the correct shape will turn to unlock the answer, fulfilling the equation \( y'' - y = x \).
The accuracy of the particular solution, in any context, is the linchpin that ensures that the general solution is more than a theoretical construct. It makes the solution practical, applicable, and tailored to the given physical, engineering, or mathematical scenario present in the boundary value problem.
To identify a particular solution, we often use methods such as undetermined coefficients or variation of parameters, depending on the form of the non-homogeneous term. In our exercise, for instance, we see the method of undetermined coefficients in action. The choice of \( y_p(x) = ax + b \) as a trial solution is an educated guess based on the degree and type of the non-homogeneous term 'x'. After differentiating and substituting back into the original equation, we're able to determine the coefficients 'a' and 'b' that make the trial solution work. It's like fitting a key into a lock; only the correct shape will turn to unlock the answer, fulfilling the equation \( y'' - y = x \).
The accuracy of the particular solution, in any context, is the linchpin that ensures that the general solution is more than a theoretical construct. It makes the solution practical, applicable, and tailored to the given physical, engineering, or mathematical scenario present in the boundary value problem.
Complementary Function
The complementary function \( y_c(x) \) is the general solution to the corresponding homogeneous differential equation, which is obtained when the non-homogeneous term (any term not involving the function y or its derivatives) is set to zero. In our exercise, the homogeneous equation is \( y'' - y = 0 \) and it represents the equation without the external influence - in this case, without 'x'.
The purpose of finding the complementary function is to capture the behavior of the system's inherent properties, such as natural oscillations or decay rates, without any external forces acting on it. To find the complementary function, typically, a method known as the characteristic equation is used. By assuming a solution of the form \( e^{rx} \) and determining the values of 'r', we can find the general form of \( y_c(x) \). This part of the full solution reflects the natural response of the system described by the differential equation.
In the provided solution, we determined that \( y_c(x) = C_1e^x + C_2e^{-x} \). The constants \( C_1 \) and \( C_2 \) will later be determined using the boundary conditions, which, in this problem, allows us to explicitly mold the complementary function into the boundary value problem's context, ensuring the full solution adheres to the constraints given by the conditions \( y(0)=-2 \) and \( y(1)=1 \). The complementary function is as crucial as the particular solution; when combined, they form the complete answer to our differential equation.
The purpose of finding the complementary function is to capture the behavior of the system's inherent properties, such as natural oscillations or decay rates, without any external forces acting on it. To find the complementary function, typically, a method known as the characteristic equation is used. By assuming a solution of the form \( e^{rx} \) and determining the values of 'r', we can find the general form of \( y_c(x) \). This part of the full solution reflects the natural response of the system described by the differential equation.
In the provided solution, we determined that \( y_c(x) = C_1e^x + C_2e^{-x} \). The constants \( C_1 \) and \( C_2 \) will later be determined using the boundary conditions, which, in this problem, allows us to explicitly mold the complementary function into the boundary value problem's context, ensuring the full solution adheres to the constraints given by the conditions \( y(0)=-2 \) and \( y(1)=1 \). The complementary function is as crucial as the particular solution; when combined, they form the complete answer to our differential equation.
