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Chapter 4: Problem 18

Verify that the differential operator defined by $$ L[y]=y^{(n)}+p_{1}(t) y^{(n-1)}+\cdots+p_{n}(t) y $$ is a linear differential operator. That is, show that $$ L\left[c_{1} y_{1}+c_{2} y_{2}\right]=c_{1} L\left[y_{1}\right]+c_{2} L\left[y_{2}\right] $$ where \(y_{1}\) and \(y_{2}\) are \(n\) times differentiable functions and \(c_{1}\) and \(c_{2}\) are arbitrary constants. Hence, show that if \(y_{1}, y_{2}, \ldots, y_{n}\) are solutions of \(L[y]=0,\) then the linear combination \(c_{1} y_{1}+\cdots+c_{n} y_{n}\) is also a solution of \(L[y]=0 .\)

Short Answer

Expert verified
To prove the linearity of the given differential operator, we showed that the operator preserves linear combinations, i.e., \(L[c_1 y_1 + c_2 y_2] = c_1 L[y_1] + c_2 L[y_2]\) for any functions \(y_1\) and \(y_2\) and constants \(c_1\) and \(c_2\). Moreover, we demonstrated that if \(y_1, y_2, \ldots, y_n\) are solutions of \(L[y]=0\), then any linear combination of them, \(c_1 y_1 + \cdots + c_n y_n\), is also a solution of \(L[y]=0\). This establishes the linearity of the given differential operator.

Step by step solution

01

Apply operator to a linear combination

Using a linear combination of the functions \(y_1\) and \(y_2\), apply the given operator to \(c_1 y_1 + c_2 y_2\): $$ L[c_1 y_1 + c_2 y_2] = (c_1 y_1 + c_2 y_2)^{(n)} + p_1(t) (c_1 y_1 + c_2 y_2)^\bs{(n-1)} + \cdots + p_n(t) (c_1 y_1 + c_2 y_2). $$
02

Separate the derivatives and functions

Using the linearity properties of differentiation and constant multiplication, separate the derivatives and functions: $$ L[c_1 y_1 + c_2 y_2] = c_1 (y_1^\bs{(n)} + p_1(t) y_1^\bs{(n-1)} + \cdots + p_n(t) y_1) + c_2 (y_2^\bs{(n)} + p_1(t) y_2^\bs{(n-1)} + \cdots + p_n(t) y_2). $$
03

Display operator applied to each function

Now, observe that the expression can be written as the sum of the operator applied to each function, multiplied by their corresponding constants: $$ L[c_1 y_1 + c_2 y_2] = c_1 L[y_1] + c_2 L[y_2]. $$
04

Conclude linearity of the operator

Since we have shown that \(L[c_1 y_1 + c_2 y_2] = c_1 L[y_1] + c_2 L[y_2]\) for any functions \(y_1\) and \(y_2\) and constants \(c_1\) and \(c_2\), the differential operator defined by \(L[y]\) is indeed linear.
05

Analyze given solutions

We are given that \(y_1, y_2, \ldots, y_n\) are solutions of \(L[y]=0\). That is, $$ L[y_i] = 0, \text{ for } i = 1, 2, \ldots, n. $$
06

Apply the operator to the linear combination

Now, consider the linear combination \(c_1 y_1 + \cdots + c_n y_n\). Applying the differential operator to this linear combination, we get: $$ L[c_1 y_1 + \cdots + c_n y_n] = c_1 L[y_1] + \cdots + c_n L[y_n]. $$
07

Show that the linear combination is a solution of L[y]=0

Using the fact that each \(L[y_i] = 0\) for \(i = 1, 2, \ldots, n\), compute: $$ L[c_1 y_1 + \cdots + c_n y_n] = c_1 L[y_1] + \cdots + c_n L[y_n] = c_1 (0) + \cdots + c_n (0) = 0. $$ Thus, the linear combination \(c_1 y_1 + \cdots + c_n y_n\) is also a solution of \(L[y]=0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
Understanding the basics of differential equations is crucial for students in various scientific and engineering disciplines. A differential equation is a mathematical equation that relates some function with its derivatives. It provides a way of describing the change in one variable in relation to another, making it essential for modeling a wide array of dynamic systems, from the decay of radioactive substances to the motion of celestial bodies.

Differential equations can be broadly classified into two types: ordinary differential equations (ODEs), which involve functions of a single variable and their derivatives, and partial differential equations (PDEs), which involve functions of several variables and their partial derivatives. The key idea behind solving a differential equation is finding the unknown function, often called the 'solution' to the equation, which satisfies the relationship described by the differential equation. This exercise tackles ODEs, focusing on a specific kind called linear differential equations, characterized by the linearity in the unknown function and its derivatives.
Linearity of Differential Operators
The concept of linearity in the context of differential operators is tied to two primary properties: additivity and homogeneity. An operator is considered linear if applying it to a sum of functions or multiplying a function by a scalar yields the same result as performing these operations after applying the operator.

For instance, the differential operator from the exercise, represented by \( L[y] \), is defined to operate on a function \( y \) and can be expressed as a sum of derivatives weighted by functions \( p_i(t) \). To demonstrate that \( L[y] \) is linear, we must show that it distributes over addition and scalar multiplication, which aligns with the superposition principle. The step-by-step solution provided beautifully illustrates these properties by decomposing the application of \( L \) onto a linear combination of functions \( y_1 \) and \( y_2 \), ultimately offering students a clear pathway to grasp the definition of linearity for differential operators.
Superposition Principle
The superposition principle is a fundamental concept in linear systems, stating that the response caused by two or more stimuli is the sum of the responses that would have been caused by each stimulus individually. In terms of differential equations, this signifies that if we have several solutions to a linear homogeneous differential equation, any linear combination of these solutions is also a solution.

Applied to our exercise, the principle enables us to understand why, given \( n \) solutions \( y_1, y_2, \ldots, y_n \) to the homogeneous equation \( L[y] = 0 \), a mix of these solutions, as in \( c_1 y_1 + c_2 y_2 + \ldots + c_n y_n \), remains a solution. The steps of the solution methodically unfold the logic behind the superposition principle, showing its power in simplifying and solving complex differential equations and highlighting an essential strategy for students tackling linear differential equations.

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Most popular questions from this chapter

Consider the spring-mass system, shown in Figure \(4.2 .4,\) consisting of two unit masses suspended from springs with spring constants 3 and \(2,\) respectively. Assume that there is no damping in the system. (a) Show that the displacements \(u_{1}\) and \(u_{2}\) of the masses from their respective equilibrium positions satisfy the equations $$ u_{1}^{\prime \prime}+5 u_{1}=2 u_{2}, \quad u_{2}^{\prime \prime}+2 u_{2}=2 u_{1} $$ (b) Solve the first of Eqs. (i) for \(u_{2}\) and substitute into the second equation, thereby obtaining the following fourth order equation for \(u_{1}:\) $$ u_{1}^{\mathrm{iv}}+7 u_{1}^{\prime \prime}+6 u_{1}=0 $$ Find the general solution of Eq. (ii). (c) Suppose that the initial conditions are $$ u_{1}(0)=1, \quad u_{1}^{\prime}(0)=0, \quad u_{2}(0)=2, \quad u_{2}^{\prime}(0)=0 $$ Use the first of Eqs. (i) and the initial conditions (iii) to obtain values for \(u_{1}^{\prime \prime}(0)\) and \(u_{1}^{\prime \prime \prime}(0)\) Then show that the solution of Eq. (ii) that satisfies the four initial conditions on \(u_{1}\) is \(u_{1}(t)=\cos t .\) Show that the corresponding solution \(u_{2}\) is \(u_{2}(t)=2 \cos t .\) (d) Now suppose that the initial conditions are $$ u_{1}(0)=-2, \quad u_{1}^{\prime}(0)=0, \quad u_{2}(0)=1, \quad u_{2}^{\prime}(0)=0 $$ Proceed as in part (c) to show that the corresponding solutions are \(u_{1}(t)=-2 \cos \sqrt{6} t\) and \(u_{2}(t)=\cos \sqrt{6} t\) (e) Observe that the solutions obtained in parts (c) and (d) describe two distinct modes of vibration. In the first, the frequency of the motion is \(1,\) and the two masses move in phase, both moving up or down together. The second motion has frequency \(\sqrt{6}\), and the masses move out of phase with each other, one moving down while the other is moving up and vice versa. For other initial conditions, the motion of the masses is a combination of these two modes.

Consider the nonhomogeneous \(n\) th order linear differential equation $$ a_{0} y^{(n)}+a_{1} y^{(n-1)}+\cdots+a_{n} y=g(t) $$ where \(a_{0}, \ldots, a_{n}\) are constants. Verify that if \(g(t)\) is of the form $$ e^{\alpha t}\left(b_{0} t^{m}+\cdots+b_{m}\right) $$ then the substitution \(y=e^{\alpha t} u(t)\) reduces the preceding equation to the form $$ k_{0} u^{(n)}+k_{1} u^{(n-1)}+\cdots+k_{n} u=b_{0} t^{m}+\cdots+b_{m} $$ where \(k_{0}, \ldots, k_{n}\) are constants. Determine \(k_{0}\) and \(k_{n}\) in terms of the \(a^{\prime}\) 's and \(\alpha .\) Thus the problem of determining a particular solution of the original equation is reduced to the simpler problem of determining a particular solution of an equation with constant coefficients and a polynomial for the nonhomogeneous term.

Find the general solution of the given differential equation. $$ y^{\prime \prime \prime}-y^{\prime \prime}-y^{\prime}+y=0 $$

Find the general solution of the given differential equation. $$ y^{\prime \prime \prime}-3 y^{\prime \prime}+3 y^{\prime}-y=0 $$

Find the general solution of the given differential equation. Leave your answer in terms of one or more integrals. $$ y^{\prime \prime \prime}-y^{\prime}=\csc t, \quad 0
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