Unit 2: Linear differential equation-I

I. Foundations — Nature and standard form

A linear differential equation relates an unknown function and its derivatives so that each appears only to the first power and no products of the function or its derivatives occur. Linearity permits superposition, separates homogeneous and forced responses, and supports systematic solution through characteristic roots and differential operators.

A. Introduction to linear differential equation

The defining feature of a linear differential equation is that the dependent variable and all its derivatives occur linearly.

• General form: An (n)th-order linear ordinary differential equation is

a_n(x)y^(n) + a_(n-1)(x)y^(n-1) + ... + a_1(x)y' + a_0(x)y = r(x)

where (x) is the independent variable, (y=y(x)) is the dependent variable, (y^{(k)}) is its (k)th derivative, (a_0,\ldots,a_n) are coefficient functions, (a_n(x)\neq0), and (r(x)) is the forcing function.

• Order: The order is the highest derivative present; for example,

y''' - 2y' + 5y = 0

is a third-order equation because (y''') is the highest derivative.

• Linearity conditions: A differential equation is linear in (y) when:

  • (y,y',\ldots,y^{(n)}) have exponent one.
  • Derivatives are not multiplied together.
  • Coefficients depend only on the independent variable (x), not on (y).

• Linear examples:

y' + x²y = sin x
y'' + 3xy' - e^x y = x

In each equation, coefficients such as (x^2), (3x), and (-e^x) depend only on (x).

• Nonlinear examples:

yy' + 1 = 0
(y')² + y = x
y'' + sin y = 0

The terms (yy'), ((y')^2), and (\sin y) violate linearity.

• Homogeneous equation: If (r(x)=0), the equation is homogeneous:

a_n(x)y^(n) + ... + a_0(x)y = 0

• Non-homogeneous equation: If (r(x)\neq0), the equation is non-homogeneous or forced. Its solution combines the complementary solution of the homogeneous equation with a particular integral.

• Constant coefficients: If every (a_k) is constant, characteristic-polynomial and operator methods apply directly; for example, (y''-5y'+6y=0).

II. General Solution — Complementary and particular components

A solution is a sufficiently differentiable function that satisfies the differential equation throughout a stated interval. An (n)th-order equation normally requires (n) independent arbitrary constants in its general solution.

A. Solution of linear differential equation

The complete solution of a linear equation consists of the homogeneous response plus one particular response to the forcing term.

• Solution principle: For the operator (L),

L[y] = r(x)

if (y_c) satisfies (L[y_c]=0) and (y_p) satisfies (L[y_p]=r(x)), then

y = y_c + y_p

Here, (L) is a linear differential operator, (y_c) is the complementary function, and (y_p) is a particular integral.

• Superposition: Linearity gives

L[c₁u + c₂v] = c₁L[u] + c₂L[v]

where (u,v) are differentiable functions and (c_1,c_2) are constants. Thus, any linear combination of homogeneous solutions is also a homogeneous solution.

• General versus particular solution:

  1. General solution: Contains the full number of arbitrary constants; an (n)th-order equation normally gives (c_1,\ldots,c_n).
  2. Particular solution: Results after initial or boundary conditions determine those constants.

• Initial conditions: Conditions at one point, such as

y(x₀)=A,   y'(x₀)=B

determine constants for a second-order equation. Here, (x_0) is the initial point and (A,B) are prescribed values.

• Worked example: Solve (y'-2y=e^{3x}).
  • Homogeneous part: (y'-2y=0) gives (y_c=Ce^{2x}).
  • Assume (y_p=Ae^{3x}). Substitution gives (3Ae^{3x}-2Ae^{3x}=e^{3x}), hence (A=1).
  • Therefore,

y = Ce^(2x) + e^(3x)

where (C) is an arbitrary constant.

• Existence and uniqueness: For a normalized equation with continuous coefficients near (x_0), prescribed values of (y,y',\ldots,y^{(n-1)}) at (x_0) determine one local solution.

III. Independence of Solutions — Building a fundamental set

A homogeneous (n)th-order linear equation needs (n) linearly independent solutions to form its complete complementary function. Independence is tested on the interval where the equation is considered.

A. Linear dependence and linear independence of solution

Solutions are linearly independent when no nontrivial constant linear combination of them is identically zero.

• Linear dependence: Functions (y_1,\ldots,y_n) are dependent if constants (c_1,\ldots,c_n), not all zero, exist such that

c₁y₁(x) + c₂y₂(x) + ... + c_n y_n(x) = 0

for every (x) in the interval.

• Linear independence: They are independent if the identity above implies

c₁ = c₂ = ... = c_n = 0

• Simple comparison:

  1. Dependent pair: (e^x) and (3e^x), because (3e^x-(3e^x)=0), using coefficients (3) and (-1).
  2. Independent pair: (e^x) and (e^{-x}); neither is a constant multiple of the other.

• Wronskian test: For (n) functions, define

W(y₁,...,y_n)(x) =
| y₁        y₂        ... y_n       |
| y₁'       y₂'       ... y_n'      |
| ...       ...       ... ...        |
| y₁^(n-1)  y₂^(n-1)  ... y_n^(n-1) |

If (W(x_0)\neq0) at some point (x_0), the functions are linearly independent.

• Example of the test: For (y_1=e^x) and (y_2=e^{-x}),

W = | e^x    e^(-x)  |
    | e^x   -e^(-x)  | = -2

Since (W=-2\neq0), the solutions are independent.

• Fundamental set: If (y_1,\ldots,y_n) are (n) independent solutions of an (n)th-order homogeneous equation, then

y_c = c₁y₁ + ... + c_n y_n

is its complete complementary function.

• Limitation of a zero Wronskian: For arbitrary differentiable functions, (W=0) alone may not always prove dependence. For solutions of the same linear homogeneous equation with continuous coefficients, however, the Wronskian is either everywhere zero or nowhere zero on the interval.

IV. Differential-Operator Method — Algebraic treatment of derivatives

The operator method rewrites a constant-coefficient differential equation as a polynomial in the derivative operator, turning the homogeneous problem into an algebraic root problem.

A. Method of solution of linear differential equation using differential operator

The differential operator (D) represents differentiation with respect to (x), allowing repeated derivatives to be manipulated symbolically.

• Operator notation:

D = d/dx,   D² = d²/dx²,   D^n = d^n/dx^n

Thus, (D[y]=y') and (D^2[y]=y'').

• Polynomial operator form: The equation

a_n y^(n) + ... + a₁y' + a₀y = r(x)

becomes

f(D)y = r(x)

where (f(D)=a_nD^n+\cdots+a_1D+a_0).

• Homogeneous procedure:

  • Replace (D) by an algebraic variable (m).
  • Solve the auxiliary equation (f(m)=0).
  • Construct the complementary function from the roots.

• Particular-integral notation:

y_p = [1/f(D)] r(x)

This inverse-operator notation means “find a function whose image under (f(D)) equals (r(x)).”

• Exponential forcing rule: Since (D(e^{ax})=ae^{ax}),

[1/f(D)]e^(ax) = e^(ax)/f(a),   provided f(a) ≠ 0

Here, (a) is a constant. If (f(a)=0), resonance occurs, and the trial form must be multiplied by a sufficient power of (x).

• Worked example: Solve ((D^2-3D+2)y=e^{4x}).
  • The auxiliary equation is (m^2-3m+2=(m-1)(m-2)=0).
  • Hence (y_c=C_1e^x+C_2e^{2x}).
  • Since (f(4)=16-12+2=6),

y_p = e^(4x)/6
y = C₁e^x + C₂e^(2x) + e^(4x)/6

• Applicability: The algebraic operator method is most direct for linear equations with constant coefficients; variable coefficients generally do not commute with (D), so ordinary polynomial manipulation may fail.

V. Second-Order Homogeneous Equations — Classification by roots

A second-order homogeneous equation with constant coefficients is solved by assuming an exponential solution and classifying the two roots of its auxiliary equation.

A. Solution of second order homogeneous linear differential equation with constant coefficient

For constants (a,b,c), with (a\neq0), consider

ay'' + by' + cy = 0

• Exponential trial: Setting (y=e^{mx}) gives (y'=me^{mx}) and (y''=m^2e^{mx}). Since (e^{mx}\neq0),

am² + bm + c = 0

is the auxiliary or characteristic equation.

• Distinct real roots: If (m_1\neq m_2) are real,

y = C₁e^(m₁x) + C₂e^(m₂x)

• Repeated real root: If (m_1=m_2=m), one exponential is insufficient; an independent second solution is (xe^{mx}):

y = (C₁ + C₂x)e^(mx)

• Complex conjugate roots: If roots are (m=\alpha\pm i\beta), with (\beta\neq0), Euler’s formula produces the real form

y = e^(αx)[C₁ cos(βx) + C₂ sin(βx)]

where (\alpha) controls exponential growth or decay and (\beta) controls oscillation.

• Worked example: For (y''-4y'+13y=0),

m² - 4m + 13 = 0
m = 2 ± 3i

Therefore,

y = e^(2x)[C₁ cos(3x) + C₂ sin(3x)]

• Initial data: Two independent constants agree with the second order of the equation and can normally be fixed by two conditions such as (y(0)) and (y'(0)).

VI. Higher-Order Homogeneous Equations — Multiplicity and complete solution

Higher-order constant-coefficient equations follow the same characteristic-root principle, but every root must contribute as many independent terms as its multiplicity requires.

A. Solution of higher order homogeneous linear differential equations with constant coefficient

An (n)th-order homogeneous equation is reduced to a degree-(n) characteristic polynomial.

• Standard equation:

a_n y^(n) + a_(n-1)y^(n-1) + ... + a₁y' + a₀y = 0

where all (a_k) are constants and (a_n\neq0).

• Characteristic equation:

a_n m^n + a_(n-1)m^(n-1) + ... + a₁m + a₀ = 0

Factoring this polynomial identifies all roots and multiplicities.

• Real root of multiplicity (s): If (m=r) occurs (s) times, its contribution is

e^(rx)[C₀ + C₁x + ... + C_(s-1)x^(s-1)]

where (r) is the root and (C0,\ldots,C{s-1}) are constants.

• Complex roots of multiplicity (s): For (\alpha\pm i\beta), multiply both sine and cosine terms successively by (1,x,\ldots,x^{s-1}):

e^(αx) Σ from k=0 to s-1 of x^k[A_k cos(βx) + B_k sin(βx)]

Here, (A_k,B_k) are arbitrary real constants.

• Worked example: Solve ((D-1)^2(D+2)y=0).
  • The root (m=1) has multiplicity two, while (m=-2) is simple.
  • The three independent solutions are (e^x), (xe^x), and (e^{-2x}).
  • Hence,

y = (C₁ + C₂x)e^x + C₃e^(-2x)

• Completeness check: The total number of arbitrary constants, counting root multiplicities, must equal the order (n). This confirms that the complementary function contains a full fundamental set of solutions.