Unit 4: Differential Equations

Table of Contents

1. Introduction to Differential Equations

A Differential Equation (DE) is an equation that relates a function with its derivatives.

This unit focuses on First Order, First Degree equations, which have the general form dy/dx = f(x, y) or M(x, y)dx + N(x, y)dy = 0.


2. Equations of First Order and First Degree

Standard types and their solution methods:

a) Variables Separable

The equation can be algebraically rearranged into the form f(y) dy = g(x) dx.

Solution: Integrate both sides: ∫ f(y) dy = ∫ g(x) dx + C

b) Homogeneous Equations

The equation can be written in the form dy/dx = F(y/x).

Solution: 1. Substitute y = vx. 2. This implies dy/dx = v + x(dv/dx). 3. The equation becomes v + x(dv/dx) = F(v). 4. This is now separable: x(dv/dx) = F(v) - v => dv / (F(v) - v) = dx / x. 5. Integrate and then substitute v = y/x back.

c) Linear Differential Equations

The equation has the standard form: dy/dx + P(x) · y = Q(x).

Solution: 1. Find the Integrating Factor (IF):

IF = e∫ P(x) dx
2. The solution is given by the formula:
y · (IF) = ∫ [ Q(x) · (IF) ] dx + C

d) Bernoulli's Equation (Reducible to Linear)

This equation has the form: dy/dx + P(x) · y = Q(x) · yn.

Solution: 1. Divide the entire equation by yn: y-n(dy/dx) + P(x)y(1-n) = Q(x). 2. Make the substitution: z = y(1-n). 3. Then dz/dx = (1-n)y-n(dy/dx). 4. Substitute these into the DE. It will become a linear DE in 'z' and 'x', which you can solve using the IF method. 5. Substitute z = y(1-n) back at the end.


3. Exact Differential Equations

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

i.e., M = ∂u/∂x and N = ∂u/∂y.

Condition for Exactness:
The equation is exact if and only if:
∂M/∂y = ∂N/∂x

Solution Method for Exact DEs:

If the equation is exact, the general solution is:

∫ M dx (treating y as constant)
+ ∫ (terms in N that do NOT contain x) dy
= C
Exam Tip: Always check for exactness first! It's often the simplest solution method if the condition ∂M/∂y = ∂N/∂x holds.

4. Integrating Factors

If a DE of the form Mdx + Ndy = 0 is not exact (i.e., ∂M/∂y ≠ ∂N/∂x), we can sometimes find an Integrating Factor (IF). This is a function we multiply the entire equation by to *make* it exact.

Rules for Finding an IF:

Rule 1: If [ (∂M/∂y) - (∂N/∂x) ] / N = f(x) (a function of x alone)

IF = e∫ f(x) dx

Rule 2: If [ (∂N/∂x) - (∂M/∂y) ] / M = g(y) (a function of y alone)

IF = e∫ g(y) dy

Rule 3: If the equation is homogeneous and (Mx + Ny) ≠ 0

IF = 1 / (Mx + Ny)

Rule 4: If the equation is of the form y·f(xy)dx + x·g(xy)dy = 0 and (Mx - Ny) ≠ 0

IF = 1 / (Mx - Ny)
Note: The IF for a linear equation (dy/dx + P(x)y = Q(x)) is a special case of Rule 1.

5. Change of Variables

This is a general technique where a substitution is used to transform a complicated DE into a simpler, standard form (like separable or linear).

General Strategy: 1. Look for a repeated group in the equation, e.g., (x + y). 2. Substitute this group: z = x + y. 3. Find the derivative: dz/dx = 1 + dy/dx => dy/dx = (dz/dx) - 1. 4. Substitute 'z' and 'dy/dx' into the original equation. 5. The new equation in 'z' and 'x' should be simpler (e.g., variables separable). 6. Solve for 'z', then substitute back z = x + y.

This method is highly flexible and depends on spotting the right substitution.


6. Higher Order Linear Differential Equations

This is an introduction to the equations that will be solved in Unit 5. A linear DE of order 'n' has the general form:

an(x) · (dny/dxn) + an-1(x) · (dn-1y/dxn-1) + ... + a1(x) · (dy/dx) + a0(x) · y = R(x)

Unit 5 will focus on solving these equations, particularly when the coefficients are constants.