Unit 2: Ordinary Differential Equation

Order and Degree of a Differential Equation
First Order ODEs
Second Order ODEs (Homogeneous)
Second Order ODEs (Non-Homogeneous)

Order and Degree of a Differential Equation

Order

The order of an ordinary differential equation (ODE) is the order of the highest derivative present in the equation.

Example 1: dy/dx + 2y = x². The highest derivative is dy/dx (1st derivative). So, the order is 1.
Example 2: d²y/dx² + 3(dy/dx)⁴ + 5y = 0. The highest derivative is d²y/dx² (2nd derivative). So, the order is 2.

Degree

The degree of an ODE is the highest power (exponent) of the highest order derivative, after the equation has been made free from radicals and fractional powers of the derivatives.

Example 1: (d²y/dx²)³ + dy/dx = 0. The highest order derivative is d²y/dx². Its power is 3. So, the degree is 3.
Example 2: d²y/dx² + 3(dy/dx)⁴ + 5y = 0. The highest order derivative is d²y/dx². Its power is 1. So, the degree is 1. (The power of the lower-order derivative doesn't matter).
Example 3 (Tricky): y'' = (1 + (y')²)^3/2. To find the degree, we must remove the fractional power.
Square both sides: (y'')² = (1 + (y')²)³.
The highest derivative is y'', and its power is 2. So, Order = 2, Degree = 2.

First Order ODEs

A first-order ODE has the general form dy/dx = f(x, y) or M(x,y)dx + N(x,y)dy = 0.

Variable Separable

This is the simplest form. The equation can be rearranged so that all terms involving 'y' are on one side (with 'dy') and all terms involving 'x' are on the other side (with 'dx').

Form: f(y) dy = g(x) dx
Solution: Integrate both sides.
∫ f(y) dy = ∫ g(x) dx + C
(where C is the constant of integration)
Example: dy/dx = x/y.
Rearrange: y dy = x dx.
Integrate: ∫y dy = ∫x dx + C
Solution: y²/2 = x²/2 + C or y² - x² = C'

Exact Equation

An equation of the form M(x,y)dx + N(x,y)dy = 0 is called "exact" if it is the total differential (du) of some function u(x,y).

Condition for Exactness:
∂M/∂y = ∂N/∂x
(The partial derivative of M with respect to y equals the partial derivative of N with respect to x).
Solution Method:
1. Check if ∂M/∂y = ∂N/∂x. If yes, proceed.
2. The solution is given by:
  ∫ M dx (treating y as constant) + ∫ (terms in N not containing x) dy = C

Integrating Factors (I.F.)

If an equation M(x,y)dx + N(x,y)dy = 0 is not exact (i.e., ∂M/∂y ≠ ∂N/∂x), we can sometimes find a function μ(x,y), called an Integrating Factor (I.F.), to multiply the entire equation by. The new equation `(μM)dx + (μN)dy = 0` will then be exact.

Finding the I.F. can be hard. Two common cases are:

If `(1/N) * (∂M/∂y - ∂N/∂x)` is a function of x only, say f(x), then:
I.F. = e^∫f(x)dx
If `(1/M) * (∂N/∂x - ∂M/∂y)` is a function of y only, say g(y), then:
I.F. = e^∫g(y)dy

Linear Equations

A first-order linear ODE has a standard form:

dy/dx + P(x)y = Q(x)

where P(x) and Q(x) are functions of x (or constants).

Solution Method:
1. Identify P(x) and Q(x) from the standard form.
2. Calculate the Integrating Factor (I.F.):
  I.F. = e^∫P(x)dx
3. The general solution is given by the formula:
  y * (I.F.) = ∫ [ Q(x) * (I.F.) ] dx + C
Example: dy/dx + (1/x)y = x.
Here, P(x) = 1/x and Q(x) = x.
I.F. = e^∫(1/x)dx = e^ln(x) = x.
Solution: y * (x) = ∫ [ x * (x) ] dx + C
xy = ∫x² dx + C
xy = x³/3 + C or y = x²/3 + C/x

Second Order ODEs (Homogeneous)

We will focus on linear second-order ODEs with constant coefficients.

The general solution (y) is the sum of two parts: the Complementary Function (y_c) and the Particular Integral (y_p).

y = y_c + y_p

Homogeneous Form: a y'' + b y' + c y = 0

The homogeneous form is when the right-hand side is zero. The solution to this part is the Complementary Function (y_c).

Step 1: The Auxiliary Equation (A.E.)

We assume a trial solution of the form y = e^mx. Substituting this into the ODE gives:

a(m²e^mx) + b(me^mx) + c(e^mx) = 0

e^mx (am² + bm + c) = 0

Since e^mx cannot be zero, we must solve the quadratic equation:

am² + bm + c = 0

This is the Auxiliary Equation (A.E.). Let the roots be m₁ and m₂.

Step 2: Find y_c based on the roots

There are three possible cases for the roots of the A.E.:

Case	Roots (m₁, m₂)	Complementary Function (y_c)
Case 1: Real and Distinct Roots	m₁ and m₂ are real, m₁ ≠ m₂	y_c = C₁e^m₁x + C₂e^m₂x
Case 2: Real and Equal Roots	m₁ = m₂ = m (real)	y_c = (C₁ + C₂x)e^mx
Case 3: Complex Roots	m = α ± iβ (complex conjugate pair)	y_c = e^αx (C₁cos(βx) + C₂sin(βx))

where C₁ and C₂ are arbitrary constants, determined by initial/boundary conditions.

Second Order ODEs (Non-Homogeneous)

The non-homogeneous form is a y'' + b y' + c y = f(x), where f(x) is not zero.

Method for Finding Particular Integrals (y_p)

The syllabus mentions the "Method for finding particular integrals," which typically refers to the Method of Undetermined Coefficients. This method involves making an educated guess for the form of y_p based on the form of f(x).

If f(x) is...	Trial Solution for y_p
A constant, k	A constant, A
A polynomial, e.g., ax² + bx + c	A polynomial of the same degree, e.g., Ax² + Bx + C
An exponential, ke^ax	Ae^ax
A sine or cosine, e.g., k sin(ax) or k cos(ax)	A sin(ax) + B cos(ax) (must include both)
A product, e.g., e^ax sin(bx)	e^ax (A sin(bx) + B cos(bx))

Step-by-step to find y_p:

CRITICAL Exam Pitfall (Resonance):
If your chosen trial solution for y_p (or any part of it) is already present in the Complementary Function (y_c), it will not work (it will just give 0 when plugged in).
Solution: Multiply your entire trial solution by x.
If that solution is *still* in y_c (which happens in Case 2, repeated roots), multiply by x².
Example: y'' + y = sin(x).
A.E: m² + 1 = 0 → m = ±i → y_c = C₁cos(x) + C₂sin(x).
f(x) = sin(x). Our normal trial would be y_p = A sin(x) + B cos(x).
Problem: This is the same form as y_c!
Corrected Trial: y_p = x (A sin(x) + B cos(x)).