Chapter 3 – High-Order Differential Equations

3.1 Theory of Linear Equations

3.1.1 What is an IVP?

For a linear differential equation, an $n^{th}$ -order Initial-Value Problem (IVP) is

(1) $\begin{equation*}a_n (x) \frac{d^n y}{dx^n} + a_{n-1} (x) \frac{d^{n-1}y}{dx^{n-1}} + … + a_1 (x) \frac{dy}{dx} = g(x),\end{equation*}$

which is solved using the given Initial-Value Conditions (IVC) as

(2) $\begin{equation*}y(x_0) = y_0, \quad y^\prime (x_0) = y_1, \quad …, \quad y^{(n-1)} (x_0) = y_{n-1} .\end{equation*}$

3.1.2 What is a BVP?

For a linear differential equation, a second-order Boundary-Value Problem (BVP) is

(3) $\begin{equation*}a_2(x) \frac{d^2y}{dx^2} + a_1 (x) \frac{dy}{dx} + a_0 (x) y = g(x) ,\end{equation*}$

which is solved using the Boundary Condition (BC) given at some points as

(4) $\begin{equation*}y(a) = y_0, \quad y(b) = y_1 .\end{equation*}$

3.1.3 Linearly Dependent Functions

A set of functions are linearly dependent in an interval $I$ , if constants of $c_1, c_2, …, c_n$ exist in such a way that

(5) $\begin{equation*}c_1 f_1 (x) + c_2 f_2 (x) + … + c_n f_n (x) = 0 ,\end{equation*}$

for every $x$ in the interval $I$ . Note that all the constants cannot be zero.

3.1.4 Criterion for Linearly Independent Solutions

A set of solutions of the homogeneous linear $n^{th}$ -order DE on an interval $I$ are linearly independent if and only if the Wronskian of the solutions is non-zero.

3.1.5 Wronskian of a Set of Functions

The Wronskian of a set of functions is a determinant of a matrix and defined as follows, provided that all the functions have at least $n-1$ derivatives,

(6) $\begin{equation*}W(f_1, f_2, \dots, f_n) = \begin{vmatrix} f_1 & f_2 & \dots & f_n \\ f^\prime_1 & f^\prime_2 & \dots & f^\prime_n \\ \vdots & \vdots & \vdots & \vdots \\ f^{(n)}_1 & f^{(n)}_2 & \dots & f^{(n)}_n \end{vmatrix}\end{equation*}$

3.2 Reduction of Order

3.2.1 What is Reduction of Order method?

Suppose $y_1 (x)$ is a solution to the following homogeneous linear second-order differential equation.

(7) $\begin{equation*}a_2(x) y^{\prime \prime} + a_1 (x) y^\prime + a_0 (x) y = 0.\end{equation*}$

Recall that the solutions, $y_1(x)$ and $y_2(x)$ , are linearly independent on an interval $I$ . Thus, their ratio $\frac{y_2(x)}{y_1(x)}$ is non-constant on the same interval, meaning that $\frac{y_2(x)}{y_1(x)} = u(x)$ .
Now, by solving the first-order differential equation of $y_2(x) = u(x) y_1(x)$ , we can find the second solution as well.

3.2.2 How does Reduction of Order work?

By dividing the homogeneous linear second-order differential equation by $a_2(x)$ , we will recover the standard form as

(8) $\begin{equation*}y^{\prime \prime} + P(x) y^\prime + Q(x) y = 0\end{equation*}$

Suppose that $y_1(x)$ is a known solution of the DE and $y_1(x) \neq 0$ , then

(9) $\begin{equation*}y_2(x) = y_1(x) \int \frac{e^{- \int P(x) dx}}{y_1^2 (x)} dx .\end{equation*}$

3.3 Homogeneous Linear Differential Equations with Constant Coefficients

3.3.1 Form an Auxiliary Equation

Consider the following second-order homogeneous linear differential equation with constant coefficients

(10) $\begin{equation*}a y^{\prime \prime} + b y^\prime + c y = 0.\end{equation*}$

Assume that the solution has the general form of $y = e^{mx}$ , then after substituting $y^\prime$ and $y^{\prime \prime}$ into the equation, the auxiliary equation will be found as

(11) $\begin{equation*}a m^2 + bm + c = 0\end{equation*}$

This equation will have the following two roots

(12) $\begin{equation*}m_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\end{equation*}$

3.3.2 Different Scenarios for the Auxiliary Equation

$\sqrt{b^2 - 4ac} > 0 ~$ : $\quad m_1$ and $m_2$ are real and distinct $\quad \rightarrow \quad y = c_1 e^{m_1 x} + c_2 e^{m_2 x}$ .
$\sqrt{b^2 - 4ac} = 0 ~$ : $\quad m_1$ and $m_2$ are real and equal $\quad \rightarrow \quad y = c_1 e^{m_1 x} + c_2 x e^{m_1 x}$ .
$\sqrt{b^2 - 4ac} < 0 ~$ : $\quad m_1$ and $m_2$ are conjugate complex numbers, and $m_1 = \alpha + i \beta$ and $m_2 = \alpha - i \beta$ . Then, $y = e^{\alpha x} \left( c_1 \cos{\beta x} + c_2 \sin{\beta x} \right)$ .

3.3.3 Higher-Order Equations

Consider the following high-order homogeneous linear differential equation with constant coefficients

(13) $\begin{equation*}a_n y^{(n)} + a_{n-1} y^{(n-1)} + \dots + a_1 y^{prime} + a_0 y = 0,\end{equation*}$

the auxiliary equation will be an $n$ -th degree polynomial

(14) $\begin{equation*}a_n m^n + a_{n-1} m^{n-1} + \dots + a_1 m + a_0 = 0.\end{equation*}$

The solution to the differential equation depends on the roots of the auxiliary equation and is determined similarly to the previous section.

3.4 Non-Homogeneous Linear Differential Equations

A non-homogeneous linear differential equation has a general form of

(15) $\begin{equation*}a_n y^{(n)} + a_{n-1} y^{(n-1)} + ... + a_1 y^\prime + a_0 y = g(x).\end{equation*}$

We need to find the complementary function, $y_c$ , and also any particular solution, $y_p$ , of the non-homogeneous equation. Then, the solution to the given differential equation will be $y=y_c + y_p$ . The complementary function, $y_c$ , is the solution to the associated homogenous differential equation, which is

(16) $\begin{equation*}a_n y^{(n)} + a_{n-1} y^{(n-1)} + ... + a_1 y^\prime + a_0 y = 0.\end{equation*}$

The complementary function is found using the methods that we learnt so far; however, there are different methods to find the particular solutions.

3.4.1 Method of Undetermined Coefficients

The idea of this method is an educated guess about the form of $y_p$ based on the input function $g(x)$ . This method is only suitable if the coefficients, $a_i$ , are constants, and $g(x)$ is a constant, a polynomial function, an exponential function $e^{\alpha x}$ , sine or cosine functions $\sin \beta x$ or $\cos \beta x$ , or finite sums and products of these functions.
The following functions are some examples of the types of inputs $g(x)$ that are appropriate:

$\begin{equation*}g(x) = 10, \end{equation*}$

$\begin{equation*}g(x) = x^2 - 5x,\end{equation*}$

$\begin{equation*}g(x) = 15x - 6 + 8e^{-x},\end{equation*}$

$\begin{equation*}g(x) = \sin 3x - 5 x \cos 2x,\end{equation*}$

$\begin{equation*}g(x) = x e^x \sin x + (3 x^2 - 1) e^{-4x}.\end{equation*}$

And here are some examples of the types of inputs $g(x)$ that are not appropriate:

$\begin{equation*}g(x) = ln x\end{equation*}$

$\begin{equation*}g(x) = \frac{1}{x}\end{equation*}$

$\begin{equation*}g(x) = \tan x\end{equation*}$

$\begin{equation*}g(x) = \sin ^{-1} x\end{equation*}$

To find a particular solution to a non-homogeneous linear differential equation using the method of undetermined coefficients, we first need to assume a form of $y_p$ based on $g(x)$ and then substitute $y_p$ in the differential equation to find the coefficients.

Note that if any $y_{p_{i}}$ contains terms that duplicate terms in $y_c$ , then that $y_{p_{i}}$ must be multiplied by $x^n$ , where $n$ is the smallest positive integer that eliminates that duplication.

3.4.2 Method of Variation of Parameters

First, we need to write the linear second-order differential equation in the standard form,

(17) $\begin{equation*}y^{\prime \prime} + P(x) y^{\prime} + Q(x) y = f(x).\end{equation*}$

Then, the particular solution is $y_p = u_1 (x) y_1 (x) + u_2 (x) y_2 (x)$ , where $y_1$ and $y_2$ form a fundamental set of solutions of the associated homogenous form of the differential equation, and

(18) $\begin{equation*}u^{\prime}_1 = \frac{W_1}{W} = - \frac{y_2 f(x)}{W} \rightarrow u_1(x) = - \int_{x_0}^x \frac{y_2(t) f(t)}{W(t)}\end{equation*}$

(19) $\begin{equation*}u^{\prime}_2 = \frac{W_2}{W} = \frac{y_1 f(x)}{W} \rightarrow u_2(x) = \int_{x_0}^x \frac{y_1(t) f(t)}{W(t)}\end{equation*}$

And, for higher-order non-homogeneous differential equations of the form

(20) $\begin{equation*}y^{(n)} + P_{n-1} (x) y^{(n-1)} + ... + P_1(x) y^\prime + P_0(x) y = f(x),\end{equation*}$

we have

(21) $\begin{equation*}u^{\prime}_k = \frac{W_k}{W}, \quad k=1,2,...,n.\end{equation*}$

3.5 Cauchy-Euler Differential Equations

The Cauchy-Euler equation, Euler-Cauchy equation, Euler equation, or equidimensional equation, is any linear differential equation of the form

(22) $\begin{equation*}a_n x^n \frac{d^n y}{dx^n} + a_{n-1} x^{n-1} \frac{d^{n-1}y}{dx^{n-1}} + ... + a_1 x \frac{dy}{dx} + a_0 y = g(x), \end{equation*}$

where the coefficients $a_n, a_{n-1}, ..., a_0$ are constants.
To find the complementary function, i.e. the solution to the associated homogeneous differential equation, we assume that the solution has a general form of $y = x^m$ , and find its derivatives. Then, we substitute the obtained derivates inside the differential equation to find the auxiliary equation.

3.5.1 Second-Order Cauchy-Euler Differential Equation

The homogeneous second-order linear Cauchy-Euler differential equation has a general form of

(23) $\begin{equation*}ax^2 \frac{d^2 y}{dx^2} + bx \frac{dy}{dx} + cy = 0.\end{equation*}$

The auxiliary equation is

(24) $\begin{equation*}am^2 + (b-a) m + c = 0.\end{equation*}$

Based on the roots of the auxiliary equation, there will be three different solutions to the given ordinary differential equation.

Distinct Real Roots. $y = c_1 x^{m_1} + c_2 x^{m_2}$ .
Repeated Real Roots. $y = c_1 x^{m_1} + c_2 x^{m_1} ln x$ .
Conjugate Complex Roots. $y = x^{\alpha} \left( c_1 \cos \left( \beta ln x \right) + c_2 \sin \left( \beta ln x \right) \right)$ .

3.6 High-Order Linear Differential Equations with Variable Coefficients

3.6.1 Power Series Solution

Whenever the coefficients of a high-order differential equation are variable, we assume that the solution has a form of power series. The solution and its derivatives will have the following form

(25) $\begin{equation*}y = \sum_{n=0}^{\infty} c_n x^n,\end{equation*}$

(26) $\begin{equation*}y^{\prime} = \sum_{n=1}^{\infty} n c_n x^{n-1},\end{equation*}$

(27) $\begin{equation*}y^{\prime \prime} = \sum_{n=2}^{\infty} n (n-1) c_n x^{n-2},\end{equation*}$

$\begin{equation*}\vdots\end{equation*}$

Then, we will substitute the above derivatives in the original differential equation and shift the summation index so that all of the power series are of degree $n$ . By finding the constants of $c_k$ , the solutions to the given differential equations are found.

3.7 Problems

Reference

Dennis G. Zill. Advanced Engineering Mathematics, $6^{th}$ edition. Jones $&$ Bartlett Learning. 2016.

$g(x) = 1$	$y_p = A$
$g(x) = 5x + 7$	$y_p = Ax+B$
$g(x) = 10x^2 - 8x$	$y_p = Ax^2 + Bx + C$
$g(x) = x^3 - 2x^2$	$y_p = Ax^3 + Bx^2 + Cx + D$
$g(x) = \sin 4x$	$y_p = A \cos 4x + B \sin 4x$
$g(x) = \cos 2x$	$y_p = A \cos 4x + B \sin 4x$
$g(x) = e^{5x}$	$y_p = A e^{5x}$
$g(x) = (x-1) e^{5x}$	$y_p = (Ax+B) e^{5x}$
$g(x) = 2x^2 e^{5x}$	$y_p = (Ax^2 +Bx + C) e^{5x}$
$g(x) = e^{3x} \sin 4x$	$y_p = Ae^{3x} \cos 4x + B e^{3x} \sin 4x$
$g(x) = 5x^2 \sin 4x$	$y_p = (Ax^2+Bx+C) \cos 4x + (Ex^2+Fx+G) \sin 4x$
$g(x) = xe^{3x} \cos 4x$	$y_p = (Ax+B) \cos 4x + (Cx+D) e^{3x} \sin 4x$