Consider nth order linear differential equation, y^((n) ) (x)+a_1 y^((n-1) ) (x)+⋯+a_(n-1) y'(x)+a_n y(x)=r(x) General Solution of above equation is YGeneral Solution = Y Complementary Function + YParticular Integral Ordinary Linear Differential Equation of the Second Order (d^2 y)/(dx^2 )+p(x) dy/dx+q(x)y=r(x) or in alternative notation, y"+p(x)y^'+q(x)y=r(x). Types of equation (1- x2) y'' - 2 x y' + 6 y = 0  y'' – 2 x 1 - x2y' + 61 - x2y = 0 : Homogeneous linear equation with variable coefficient y'' + 4 y' + 3 y = ex : Homogeneous linear equation with constant coefficient y'' y + y' = 0: Non Linear Differential Equation y'' + (sin x) y' + y = 0: Homogeneous linear equation with variable coefficient. Homogeneous Linear Ordinary Differential Equation with Constant Coefficients The linear homogeneous differential equation of the nth order if r(x)=0 y^((n)) (x)+a_1 y^((n-1)) (x)+⋯+a_(n-1) y'(x)+a_n y(x)=0, where a_1,a_2,…,a_n are constants which may be real or complex or functions of x. Using the linear differential operator L(D), this equation can be represented as L(D)y(x)=0, where, D=d/dx and L(D)=D^n+a_1 D^(n-1)+⋯+a_(n-1) D+a_n For each differential operator with constant coefficients, we can introduce the characteristic polynomial. L(m)=m^n+a_1 m^(n-1)+⋯+a_(n-1) m+a_n. The algebraic equation L(m)=m^n+a_1 m^(n-1)+⋯+a_(n-1) m+a_n. is called the characteristic equation (Auxiliary equation) of the differential equation. According to the fundamental theorem of algebra, a polynomial of degree n has exactly n roots, counting multiplicity. Let us consider in more detail the different cases of the roots of the characteristic equation and the corresponding formulas for the general solution of differential equations. Case 1. All Roots of the Characteristic Equation are Real and Distinct Assume that the characteristic equation L(m)=0 has n roots m_1,m_2,…,m_n. In this case the general solution of the differential equation is written in a simple form: y(x)=c_1 e^(m_1 x)+c_2 e^(m_2 x)+⋯+c_n e^(m_n x), where c_1,c_2,…,c_n are constants depending on initial conditions. Case 2. The Roots of the Characteristic Equation are Real and Multiple Let the characteristic equation L(m)=0 of degree n have m roots m_1,m_2,…,m_n., the multiplicity of which, respectively, is equal to m_1,m_2,…,m_nIt is clear that the following condition holds: m_1,m_2,…,m_nThen the general solution of the homogeneous differential equations with constant coefficients has the form y(x)=c_1 e^(m_1 x)+c_2 xe^(m_2 x)+⋯+c_k1 x^(k_1-1) e^(m_1 x)+⋯+c_n x^(km-1) e^(m_m x). It is seen that the formula of the general solution has exactly ki terms corresponding to each root m_iof multiplicity k_i. Case 3. The Roots of the Characteristic Equation are Complex and Distinct If the coefficients of the differential equation are real numbers, the complex roots of the characteristic equation will be presented in the form of conjugate pairs of complex numbers:

1. Unit 3 Lecture Notes
Ordinary Differential Equations -
Higher Order

2. Ordinary Differential equation with
Higher Order
• Consider nth order linear differential equation,
General Solution of above equation is
YGeneral Solution = Y Complementary Function + YParticular Integral
Ordinary Linear Differential Equation of the Second Order
or in alternative notation,
.
Types of equation
(1- x2) y'' - 2 x y' + 6 y = 0 y'' – y' + y = 0 : Homogeneous linear equation with
variable coefficient
y'' + 4 y' + 3 y = ex : Homogeneous linear equation with constant coefficient
y'' y + y' = 0: Non Linear Differential Equation
y'' + (sin x) y' + y = 0: Homogeneous linear equation with variable coefficient.

3. Homogeneous Linear Ordinary
Differential Equation with Constant
Coefficients
• The linear homogeneous differential equation of the nth order if
where are constants which may be real or complex or functions of .
Using the L(D), this equation can be represented as
where, and
For each differential operator with constant coefficients, we can introduce the characteristic
polynomial.
The algebraic equation
is called the characteristic equation (Auxiliary equation) of the differential equation.
According to the fundamental theorem of algebra, a polynomial of degree n has exactly n
roots, counting multiplicity. Let us consider in more detail the different cases of the roots of
the characteristic equation and the corresponding formulas for the general solution of
differential equations.

4. Case 1. All Roots of the
Characteristic Equation are Real
and Distinct
• Assume that the characteristic equation has n roots . In this case the general solution of the
differential equation is written in a simple form:
where are constants depending on initial conditions.

5. Case 2. The Roots of the
Characteristic Equation are Real
and Multiple
• Let the characteristic equation of degree n have m roots ., the multiplicity of which,
respectively, is equal to It is clear that the following condition holds:
Then the general solution of the homogeneous differential equations with constant
coefficients has the form
It is seen that the formula of the general solution has exactly ki terms corresponding to each
rootof multiplicity .

6. Case 3. The Roots of the
Characteristic Equation are
Complex and Distinct
If the coefficients of the differential equation are real numbers, the complex roots of the
characteristic equation will be presented in the form of conjugate pairs of complex numbers:
In this case the general solution is written as
Second order Homogeneous Differential Equations with Constant Coefficients
Consider an equation
y'' + a y' + b y = 0
where a and b are real constants.
Auxiliary equation of homogeneous differential equation is given by
m2 + a m + b = 0 --Auxiliary equation
Since the Auxiliary equation is quadratic, we have two roots:
m1 = and m2 =
Thus, there are three possible situations for the roots of m1 and m2 of the characteristic
equation:
Case I a2 – 4b > 0 m1 and m2 are distinct real roots
Case II a2 – 4b = 0 m1 = m2 , a real double root
Case III a2 – 4b < 0 m1 and m2 are two complex conjugate roots
We have following cases:
Example 1. y” – y’ – 6y = 0
Solution: Characteristic equation of the given differential equation is
m2 – m – 6 = 0 (m + 2) (m – 3) = 0
we have two distinct roots

7. 1. SI Model (Susceptible →
Infected)
• Assumptions:
- Population is divided into Susceptible (S) and Infected (I).
- No recovery; once infected, individuals stay infected.
Equations: Total population N = S + I is constant.
Using S(t) = N(t) – I(t), substitute into the equation:
Solution: Separate variables and integrate:
This leads to a logistic-type solution:
2. SIS Model (Susceptible → Infected → Susceptible)
Assumptions: Individuals get infected and then recover but do not gain immunity (they return
to the susceptible class).
Equations: Let S(t) + I(t) = N(t)
Solution: General form:
Steady state I* = N(1 - γ/β), if β > γ

8. 3. SIR Model (Susceptible →
Infected → Recovered)
• Assumptions:
- Individuals who recover gain immunity.
- Population is divided into Susceptible (S), Infected (I), and Recovered (R).
Equations:
Use total population S(t) + I(t) + R(t) = N(t).
Key Relationship:
=> ln(S(t)) = -(β / γN(t)) * R(t) + C
This gives an implicit solution showing the trade-off between susceptible and recovered
populations.
3.9 Newton’s law of cooling
Newton’s Law of Cooling describes the rate at which an object changes temperature through
radiation or conduction to its surrounding medium. The law is governed by a first-order linear
differential equation.
Let:
- T(t): Temperature of the object at time t
- Ts: Constant surrounding (ambient) temperature
- k: Positive constant depending on the nature of the object
According to Newton’s Law of Cooling:
dT/dt = - k(T - Ts)

9. 2. Solution of the Differential
Equation
• This is a first-order separable differential equation:
Step 1: Separate variables
dT / (T - Ts) = - k dt
Step 2: Integrate both sides
∫ 1 / (T - Ts) dT = ∫ -k dt
ln|T - Ts| = -kt + C
Step 3: Solve for T
T - Ts = Ce- kt
T(t) = Ts + Ce- kt
Using initial condition T(0) = T0:
C = T0 - Ts

10. 3. Final Solution
• T(t) = Ts + (T0 - Ts) * e-kt

11. T(t) = 20 + (90 - 20)e-kt = 20 + 70e-
kt
At t = 10, T = 60:
60 = 20 + 70e-10k → 40 = 70e-10k
→ e-10k = 4/7
-10k = ln(4/7) → k = -1/10 * ln(4/7)
T(20) = 20 + 70 * (4/7)2 = 20 + 70 *
• 3.10 Single Compartmental Modelling, Bending of beam Models
Single Compartmental Modelling:
Single-compartment modeling is a simplified approach in pharmacokinetics that views the
body as a single, well-mixed compartment. It assumes that drugs distribute rapidly and
uniformly throughout this compartment, with no barriers to movement. This model is often
used to describe the distribution and elimination of drugs that quickly equilibrate with tissues
and fluids.
Let:
- C(t): Drug concentration in the compartment at time t
- k: Elimination rate constant (k > 0)
- D: Initial dose of the drug
- V: Volume of distribution
The rate of elimination is proportional to the concentration:
dC/dt = -kC

12. 2. Solution of the Differential
Equation
• This is a first-order linear differential equation:
Step 1: Separate variables
dC / C = -k dt
Step 2: Integrate both sides
∫1/C dC = ∫-k dt
ln|C| = -kt + C₁
Step 3: Solve for C
C(t) = C₀ * e-kt
Where C₀ is the initial concentration (C₀ = D/V).

13. Governing Differential Equation
• The bending of slender beams is described by the Euler–Bernoulli beam equation:
EI d⁴y(x)/dx⁴ = q(x)
here:
E = Modulus of Elasticity (Young's Modulus)
I = Moment of Inertia of the beam's cross-section
y(x) = Deflection of the beam at position x
q(x) = Distributed load per unit length (can be a function of x)

14. Case 1: Simply Supported Beam
with Uniform Load
• Let q(x) = q₀ (constant), and the beam is simply supported at both ends x = 0 and x = L.
Boundary Conditions:
• At x = 0: y = 0, M = 0 d²y/dx² = 0
⇒
• At x = L: y = 0, M = 0 d²y/dx² = 0
⇒
Solution:
The deflection curve is given by:
y(x) = (q₀ / 24EI) * (x⁴ - 2Lx³ + L²x²)
Maximum deflection occurs at the midpoint x = L/2:
y_max = (5q₀L⁴) / (384EI)
Free Oscillation
Consider a spring OA suspended vertically from a fixed support at O. let a body of mass being
large in comparison with mass of the spring, that the later may be neglected. Let be the
elongation produced by the mass hanging in equilibrium, and then B is called the position of
static equilibrium.
Let be the restoring force per unit stretch of the spring due to elasticity.
For the equilibrium at
Let the mass be displaced through a further distance y from equilibrium position.
The acceleration of the mass m at this position is and the forces acting upon it are weight mg
down wards and restoring force upward.
Writing
Damped Free Oscillation