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Unit 3 - Notes

MTH166 7 min read

Unit 3: Linear Differential Equations

1. Introduction

A linear differential equation of order $n$ with constant coefficients is of the form:

$\frac{d^n y}{d x^n} + a_1 \frac{d^{n-1} y}{d x^{n-1}} + \dots + a_n y = Q(x)$

Where $a_1, a_2, \dots, a_n$ are constants and $Q(x)$ is a function of $x$ .

General Solution Structure:
The complete solution consists of two parts:

Complementary Function (C.F.): The solution to the homogeneous equation (where $Q(x) = 0$ ).
Particular Integral (P.I.): A specific solution satisfying the non-homogeneous equation.

$y_{general} = y_{c} (C.F.) + y_{p} (P.I.)$

2. Solution by Operator Method (D-Operator)

We introduce the differential operator $D = \frac{d}{dx}$ . Consequently, $D^2 = \frac{d^2}{dx^2}$ , and so on. The differential equation can be written as:
$f(D)y = Q(x)$

2.1 Finding the Complementary Function (C.F.)

To find the C.F., we solve the Auxiliary Equation (A.E.) obtained by replacing $D$ with $m$ and equating $f(m) = 0$ .

Nature of Roots ( $m_1, m_2, \dots$ )	Form of C.F.
Real and Distinct ( $m_1 \neq m_2$ )	$y_c = c_1 e^{m_1 x} + c_2 e^{m_2 x}$
Real and Equal ( $m_1 = m_2 = m$ )	$y_c = (c_1 + c_2 x) e^{mx}$
Complex Conjugates ( $\alpha \pm i\beta$ )	$y_c = e^{\alpha x} (c_1 \cos \beta x + c_2 \sin \beta x)$
Repeated Complex ( $\alpha \pm i\beta, \alpha \pm i\beta$ )	$y_c = e^{\alpha x} [(c_1 + c_2 x)\cos \beta x + (c_3 + c_4 x)\sin \beta x]$

2.2 Finding the Particular Integral (P.I.)

The Particular Integral is given by $y_p = \frac{1}{f(D)} Q(x)$ . The method depends on the form of $Q(x)$ .

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^-1 using Binomial Theorem (1 +/- D)^-1". Branch 4: "Q(x) = e^{ax}V(x)". Box: "Apply Shift Rule: e^{ax}[1/f(D+a)]V(x)". Use distinct colors for each branch (e.g., green, blue, orange, purple). Professional diagrammatic style with clear arrows.]

Case I: $Q(x) = e^{ax}$

Rule: Replace $D$ with $a$ .
$P.I. = \frac{1}{f(D)} e^{ax} = \frac{1}{f(a)} e^{ax}$
Exception: If $f(a) = 0$ , then $P.I. = x \frac{1}{f'(a)} e^{ax}$ .

Case II: $Q(x) = \sin(ax)$ or $\cos(ax)$

Rule: Replace $D^2$ with $-a^2$ .
$P.I. = \frac{1}{f(D^2)} \sin(ax) = \frac{1}{f(-a^2)} \sin(ax)$
Exception: If the denominator becomes zero, multiply by $x$ and differentiate the operator with respect to $D$ .

Case III: $Q(x) = x^m$ (Algebraic Polynomial)

Rule: Convert the operator into the form $[1 \pm \phi(D)]^{-1}$ and expand using the Binomial Theorem.
Expansions:
- $(1-D)^{-1} = 1 + D + D^2 + D^3 + \dots$
- $(1+D)^{-1} = 1 - D + D^2 - D^3 + \dots$

Case IV: $Q(x) = e^{ax} V(x)$

Rule (Exponential Shift):
$\frac{1}{f(D)} (e^{ax} V) = e^{ax} \frac{1}{f(D+a)} V$
After shifting $e^{ax}$ to the left, operate $\frac{1}{f(D+a)}$ on $V(x)$ .

Case V: $Q(x) = x V(x)$

Rule:
$\frac{1}{f(D)} (xV) = \left[ x - \frac{f'(D)}{f(D)} \right] \frac{1}{f(D)} V$

3. Method of Undetermined Coefficients

This method is used when $Q(x)$ is a standard function (exponential, polynomial, sinusoidal, or a sum/product of these) and the operator method is either difficult to apply or strictly restricted by the problem statement.

Procedure:

Find the Complementary Function ( $y_c$ ).
Assume a trial solution $y_p$ based on the form of $Q(x)$ .
Differentiate $y_p$ and substitute it into the original DE.
Equate coefficients of like terms to solve for the unknown constants.

Choice of Trial Solution:

Term in $Q(x)$	Assumed Trial Solution $y_p$
$k e^{ax}$	$A e^{ax}$
$k x^n$ ( $n=1$ )	$Ax + B$
$k x^2$	$Ax^2 + Bx + C$
$k \sin(ax)$ or $k \cos(ax)$	$A \cos(ax) + B \sin(ax)$

Modification Rule: If any term in the assumed trial solution is already present in the C.F., multiply the trial solution by $x$ (or $x^2$ if the root is repeated).

4. Method of Variation of Parameters

This is a general method used to find the particular integral for linear DEs of the form $y'' + P(x)y' + Q(x)y = R(x)$ , especially when $R(x)$ is not a standard form (e.g., $\sec x, \tan x$ ).

A step-by-step block diagram explaining the "Method of Variation of Parameters" for a second-order D... — AI-generated image — may contain inaccuracies

, [y_1', y_2']]. Note: "If W ≠ 0, proceed".
Step 3 Block: "Compute Parameters u and v". Show formulas: u = -Integral[(y_2 R(x)) / W] dx and v = Integral[(y_1 R(x)) / W] dx.
Step 4 Block: "Final Solution". Formula: y_p = u(x)y_1 + v(x)y_2.
Connect blocks with downward arrows. Use a clean, academic layout with math symbols clearly rendered.]

Algorithm:

Find the C.F.: $y_c = c_1 y_1 + c_2 y_2$ .
Calculate the Wronskian ( $W$ ):
$W = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = y_1 y_2' - y_2 y_1'$
The Particular Integral is given by:
$y_p = u(x) y_1 + v(x) y_2$
Where:
$u(x) = - \int \frac{y_2 R(x)}{W} dx$
$v(x) = \int \frac{y_1 R(x)}{W} dx$
General Solution: $y = y_c + y_p$ .

5. Euler-Cauchy Equation (Linear Equation with Variable Coefficients)

Also known as the Homogeneous Linear Equation. It has the form:
$x^n \frac{d^n y}{d x^n} + a_1 x^{n-1} \frac{d^{n-1} y}{d x^{n-1}} + \dots + a_n y = Q(x)$

Note that the power of $x$ matches the order of the derivative.

Transformation Procedure:
To solve, we transform it into a linear DE with constant coefficients.

Substitution: Let $x = e^z$ or $z = \ln x$ .
Operator Transformation:
Define .
- $x \frac{d}{dx} \equiv \theta$
- $x^2 \frac{d^2}{dx^2} \equiv \theta(\theta - 1)$
- $x^3 \frac{d^3}{dx^3} \equiv \theta(\theta - 1)(\theta - 2)$

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Bottom Caption: "Converting Variable Coefficients to Constant Coefficients".]

Steps to Solve:

Substitute the $\theta$ operators into the equation.
Replace $Q(x)$ with the function in terms of $z$ (replace $x$ with $e^z$ ).
Solve the resulting constant coefficient DE for $y(z)$ .
Back-substitute $z = \ln x$ to get the final solution $y(x)$ .

6. Simultaneous Differential Equations

This involves solving a system of linear differential equations with constant coefficients involving two or more dependent variables (e.g., $x(t), y(t)$ ) and one independent variable ( $t$ ).

Method of Operators (Elimination Method):

Consider the system:

$f_1(D)x + g_1(D)y = T_1(t)$
$f_2(D)x + g_2(D)y = T_2(t)$

Steps:

Write the equations in operator form ( $D = \frac{d}{dt}$ ).
Treat the operators $f(D)$ and $g(D)$ like algebraic coefficients.
Eliminate one variable (say $y$ ) by multiplying equation (1) by $g_2(D)$ and equation (2) by $g_1(D)$ , then subtracting.
This yields a single DE in terms of $x$ and $t$ :
$L(D)x = R(t)$
Solve this DE to find $x(t)$ .
Substitute back into one of the original equations (or eliminate similarly) to find .
- Note: Solving separately introduces extra constants. Substitute $x(t)$ into the equations to relate the constants of $y(t)$ to $x(t)$ .

7. Summary of Key Formulae

Auxiliary Equation: $f(m) = 0$ .
P.I. for $e^{ax}$ : $\frac{1}{f(a)}e^{ax}$ ( $f(a) \neq 0$ ).
P.I. for $\sin(ax)$ : Replace $D^2 \to -a^2$ .
Wronskian: $W = y_1 y_2' - y_2 y_1'$ .
Variation of Parameters: $y_p = -y_1 \int \frac{y_2 Q}{W} dx + y_2 \int \frac{y_1 Q}{W} dx$ .
Euler Substitution: $x = e^z, \quad x^2 D^2 = \theta(\theta-1)$ .

Unit 2

Unit 4