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Find a General Solution to the Following Higher-order Equations

Higher Order Equations with Constant Coefficients

This section is devoted to an introductory study of higher order linear equations with constant coefficients. This is an extension of the study of $ 2^{\mbox{nd}}$ order linear equations with constant coefficients (see, Section 8.3).

The standard form of a linear $ n^{\mbox{th}}$ order differential equation with constant coefficients is given by

$\displaystyle L_n(y) = f(x) \;\; {\mbox{ on }} \;\; I,$ (8.6.1)

where

$\displaystyle L_n \equiv \frac{d^n}{dx^n} + a_1 \frac{d^{n-1}}{dx^{n-1}} + \cdots + a_{n-1} \frac{d}{dx} + a_n $

is a linear differential operator of order $ n$ with constant coefficients, $ a_1, a_2, \ldots, a_n$ being real constants (called the coefficients of the linear equation) and the function $ f(x)$ is a piecewise continuous function defined on the interval $ I.$ We will be using the notation $ y^{(n)}$ for the $ n^{\mbox{th}}$ derivative of $ y.$ If $ f(x) \equiv 0,$ then (8.6.1) which reduces to
$\displaystyle L_n(y) = 0 \;\; {\mbox{ on }} \;\; I,$ (8.6.2)

is called a homogeneous linear equation, otherwise (8.6.1) is called a non-homogeneous linear equation. The function $ f$ is also known as the non-homogeneous term or a forcing term.

DEFINITION 8.6.1 A function $ y$ defined on $ I$ is called a solution of (8.6.1) if $ y$ is $ n$ times differentiable and $ y$ along with its derivatives satisfy (8.6.1).

As in Section 8.3, we first take up the study of (8.6.2). It is easy to note (as in Section 8.3) that for a constant $ {\lambda},$

$\displaystyle L_n(e^{{\lambda}x}) = p({\lambda}) e^{{\lambda}x}$

where,
$\displaystyle p({\lambda}) = {\lambda}^n + a_1 {\lambda}^{n-1} + \cdots + a_n$ (8.6.3)

DEFINITION 8.6.3 (Characteristic Equation) The equation $ p({\lambda}) = 0,$ where $ p({\lambda}) $ is defined in (8.6.3), is called the CHARACTERISTIC EQUATION of (8.6.2).

Note that $ p({\lambda}) $ is of polynomial of degree $ n$ with real coefficients. Thus, it has $ n$ zeros (counting with multiplicities). Also, in case of complex roots, they will occur in conjugate pairs. In view of this, we have the following theorem. The proof of the theorem is omitted.

THEOREM 8.6.4 $ e^{{\lambda}x}$ is a solution of (8.6.2) on any interval $ I \subset {\mathbb{R}}$ if and only if $ {\lambda}$ is a root of (8.6.3)

  1. If $ {\lambda}_1, {\lambda}_2, \ldots, {\lambda}_n$ are distinct roots of $ p({\lambda}) = 0,$ then

    $\displaystyle e^{{\lambda}_1 x}, e^{{\lambda}_2 x}, \ldots, e^{{\lambda}_n x}$

    are the $ n$ linearly independent solutions of (8.6.2).
  2. If $ {\lambda}_1$ is a repeated root of $ p({\lambda}) = 0$ of multiplicity $ k,$ i.e., $ {\lambda}_1$ is a zero of (8.6.3) repeated $ k$ times, then

    $\displaystyle e^{{\lambda}_1 x}, x e^{{\lambda}_1 x}, \ldots, x^{k-1}e^{{\lambda}_1 x}$

    are linearly independent solutions of (8.6.2), corresponding to the root $ {\lambda}_1$ of $ p({\lambda}) = 0.$
  3. If $ {\lambda}_1 = {\alpha}+ i \beta$ is a complex root of $ p({\lambda}) = 0,$ then so is the complex conjugate $ \overline{{\lambda}_1} = {\alpha}- i \beta.$ Then the corresponding linearly independent solutions of (8.6.2) are

    $\displaystyle y_1 = e^{{\alpha}x} \bigl( \cos (\beta x) + i \sin (\beta x) \big... ...d }} \;\; y_2 = e^{{\alpha}x} \bigl( \cos (\beta x) - i \sin (\beta x) \bigr).$

    These are complex valued functions of $ x.$ However, using super-position principle, we note that

    $\displaystyle \frac{y_1 + y_2}{2} = e^{{\alpha}x} \cos (\beta x) \;\; {\mbox{ and }} \;\; \frac{y_1 - y_2}{2i} = e^{{\alpha}x} \sin (\beta x)$

    are also solutions of (8.6.2). Thus, in the case of $ {\lambda}_1 = {\alpha}+ i \beta$ being a complex root of $ p({\lambda}) = 0,$ we have the linearly independent solutions

    $\displaystyle e^{{\alpha}x} \cos (\beta x) \;\; {\mbox{ and }} \;\; e^{{\alpha}x} \sin (\beta x).$

EXAMPLE 8.6.5

  1. Find the solution space of the differential equation

    $\displaystyle y^{\prime\prime\prime} - 6 y^{\prime\prime} + 11 y^\prime - 6 y = 0.$


    Solution: Its characteristic equation is

    $\displaystyle p({\lambda}) = {\lambda}^3 - 6 {\lambda}^2 + 11 {\lambda}- 6 = 0.$

    By inspection, the roots of $ p({\lambda}) = 0$ are $ {\lambda}= 1, 2, 3.$ So, the linearly independent solutions are $ e^{x}, e^{2 x}, e^{3 x}$ and the solution space is

    $\displaystyle \{ c_1 e^{x} + c_2 e^{2x} + c_3 e^{3x} \; : \; c_1, c_2, c_3 \in {\mathbb{R}}\}.$

  2. Find the solution space of the differential equation

    $\displaystyle y^{\prime\prime\prime} - 2 y^{\prime\prime} + y^\prime = 0.$


    Solution: Its characteristic equation is

    $\displaystyle p({\lambda}) = {\lambda}^3 - 2 {\lambda}^2 + {\lambda}= 0.$

    By inspection, the roots of $ p({\lambda}) = 0$ are $ {\lambda}= 0, 1, 1.$ So, the linearly independent solutions are $ 1, e^{x}, xe^{x}$ and the solution space is

    $\displaystyle \{ c_1 + c_2 e^{x} + c_3 x e^{x} \; : \; c_1, c_2, c_3 \in {\mathbb{R}}\}.$

  3. Find the solution space of the differential equation

    $\displaystyle y^{(4)} + 2 y^{\prime\prime} + y = 0.$


    Solution: Its characteristic equation is

    $\displaystyle p({\lambda}) = {\lambda}^4 + 2 {\lambda}^2 + 1 = 0.$

    By inspection, the roots of $ p({\lambda}) = 0$ are $ {\lambda}= i, i, -i, -i.$ So, the linearly independent solutions are $ \sin x, x \sin x, \cos x, x \cos x$ and the solution space is

    $\displaystyle \{c_1 \sin x + c_2 \cos x + c_3 x \sin x + c_4 x \cos x \; : \; c_1, c_2, c_3, c_4 \in {\mathbb{R}}\}.$

From the above discussion, it is clear that the linear homogeneous equation (8.6.2), admits $ n$ linearly independent solutions since the algebraic equation $ p({\lambda}) = 0$ has exactly $ n$ roots (counting with multiplicity).

DEFINITION 8.6.6 (General Solution) Let $ y_1, y_2, \ldots, y_n$ be any set of $ n$ linearly independent solution of (8.6.2). Then

$\displaystyle y = c_1 y_1 + c_2 y_2 + \cdots + c_n y_n$

is called a general solution of (8.6.2), where $ c_1, c_2, \ldots, c_n$ are arbitrary real constants.

EXAMPLE 8.6.7

  1. Find the general solution of $ y^{\prime\prime\prime} = 0.$
    Solution: Note that 0 is the repeated root of the characteristic equation $ {\lambda}^3 = 0.$ So, the general solution is

    $\displaystyle y = c_1 + c_2 x + c_3 x^2.$

  2. Find the general solution of

    $\displaystyle y^{\prime\prime\prime} + y^{\prime\prime}+ y^\prime + y = 0.$


    Solution: Note that the roots of the characteristic equation $ {\lambda}^3 + {\lambda}^2 + {\lambda}+ 1 = 0$ are $ -1, i, -i.$ So, the general solution is

    $\displaystyle y = c_1 e^{-x} + c_2 \sin x + c_3 \cos x.$

EXERCISE 8.6.8

  1. Find the general solution of the following differential equations:
    1. $ y^{\prime\prime\prime} + y^\prime = 0.$
    2. $ y^{\prime\prime\prime} + 5 y^\prime - 6 y= 0.$
    3. $ y^{iv} + 2 y^{\prime\prime} + y = 0.$
  2. Find a linear differential equation with constant coefficients and of order $ 3$ which admits the following solutions:
    1. $ \cos x, \sin x$ and $ e^{-3 x}.$
    2. $ e^x, e^{2x}$ and $ e^{3x}.$
    3. $ 1 , e^x$ and $ x.$
  3. Solve the following IVPs:
    1. $ y^{iv} - y = 0, \;\; y(0) = 0, y^{\prime}(0) = 0, y^{\prime\prime}(0) = 0, y^{\prime\prime\prime}(0) = 1.$
    2. $ 2 y^{\prime\prime\prime} + y^{\prime\prime} + 2 y^\prime + y = 0, \;\; y(0) = 0, y^{\prime}(0) = 1, y^{\prime\prime}(0) = 0. $
  4. Euler Cauchy Equations:
    Let $ a_0, a_1, \ldots, a_{n-1} \in {\mathbb{R}}$ be given constants. The equation
    $\displaystyle x^n \frac{d^ny}{d x^n} + a_{n-1} x^{n-1} \frac{d^{n-1}y}{d x^{n-1}} + \cdots + a_0 y = 0, \;\; x \in I$ (8.6.4)

    is called the homogeneous Euler-Cauchy Equation (or just Euler's Equation) of degree $ n.$ (8.6.4) is also called the standard form of the Euler equation. We define

    $\displaystyle L(y) = x^n \frac{d^ny}{d x^n} + a_{n-1} x^{n-1} \frac{d^{n-1}y}{d x^{n-1}} + \cdots + a_0 y.$

    Then substituting $ y = x^\lambda,$ we get

    $\displaystyle L(x^{\lambda}) = \bigl( \lambda (\lambda-1) \cdots (\lambda-n+1) ... ...-1} \lambda (\lambda-1) \cdots (\lambda-n+2) + \cdots + a_0 \bigr) x^\lambda.$

    So, $ x^\lambda$ is a solution of (8.6.4), if and only if
    $\displaystyle \lambda (\lambda-1) \cdots (\lambda-n+1) + a_{n-1} \lambda (\lambda-1) \cdots (\lambda-n+2) + \cdots + a_0 = 0.$ (8.6.5)

    Essentially, for finding the solutions of (8.6.4), we need to find the roots of (8.6.5), which is a polynomial in $ \lambda.$ With the above understanding, solve the following homogeneous Euler equations:
    1. $ x^3 y^{\prime\prime\prime} + 3 x^2 y^{\prime\prime} + 2 x y^\prime = 0.$
    2. $ x^3 y^{\prime\prime\prime} -6 x^2 y^{\prime\prime} + 11 x y^\prime - 6 y = 0.$
    3. $ x^3 y^{\prime\prime\prime} - x^2 y^{\prime\prime} + x y^\prime - y= 0.$
    For an alternative method of solving (8.6.4), see the next exercise.
  5. Consider the Euler equation (8.6.4) with $ x > 0$ and $ x \in I.$ Let $ x = e^t$ or equivalently $ t = \ln x.$ Let $ D = \frac{d}{dt}$ and $ d = \frac{d}{dx}.$ Then
    1. show that $ x d( y) = D y(t),$ or equivalently $ x \frac{d y}{dx} = \frac{ dy)}{dt}.$
    2. using mathematical induction, show that $ x^n d^n y = \bigl( D(D-1)\cdots (D-n+1)\bigr) y(t).$
    3. with the new (independent) variable $ t$ , the Euler equation (8.6.4) reduces to an equation with constant coefficients. So, the questions in the above part can be solved by the method just explained.

We turn our attention toward the non-homogeneous equation (8.6.1). If $ y_p$ is any solution of (8.6.1) and if $ y_h$ is the general solution of the corresponding homogeneous equation (8.6.2), then

$\displaystyle y = y_h + y_p$

is a solution of (8.6.1). The solution $ y$ involves $ n$ arbitrary constants. Such a solution is called the GENERAL SOLUTION of (8.6.1).

Solving an equation of the form (8.6.1) usually means to find a general solution of (8.6.1). The solution $ y_p$ is called a PARTICULAR SOLUTION which may not involve any arbitrary constants. Solving (8.6.1) essentially involves two steps (as we had seen in detail in Section 8.3).

Step 1: a) Calculation of the homogeneous solution $ y_h$ and
b) Calculation of the particular solution $ y_p.$

In the ensuing discussion, we describe the method of undetermined coefficients to determine $ y_p.$ Note that a particular solution is not unique. In fact, if $ y_p$ is a solution of (8.6.1) and $ u$ is any solution of (8.6.2), then $ y_p + u$ is also a solution of (8.6.1). The undetermined coefficients method is applicable for equations (8.6.1).

A K Lal 2007-09-12

Find a General Solution to the Following Higher-order Equations

Source: https://nptel.ac.in/content/storage2/courses/122104018/node77.html