9 min readConcept sheet1-mark MCQ on the number of arbitrary constants; 2-mark or 3-mark 'verify that y = ... is a solution'

General and particular solutions

A solution is a function that satisfies the DE. The general solution has as many arbitrary constants as the order; a particular solution has none.

SolutionA function y = φ(x) that makes the DE true when y and its derivatives are substituted

General solution (primitive)Contains arbitrary constants; for a DE of order n it has n independent arbitrary constants

Particular solutionNo arbitrary constants; found by giving the constants definite values, usually from conditions like y(0) = 1

Exampley = A cos x + B sin x is the general solution of y″ + y = 0; y = cos x is a particular one (A = 1, B = 0)

What a solution is

A function y = φ(x) is a solution of a DE if it makes the DE true. When you substitute y and its derivatives, the left side equals the right side for every x in the domain. Solving a DE means finding such functions. A solution with arbitrary constants (C, A, B, ...) is the general solution, also called the primitive. Each value of the constants gives one curve, so the general solution is a family of curves.

General vs particular solution

General solutionContains arbitrary constants. Number of independent constants = order of the DE. Describes a whole family of curves.

Particular solutionContains no arbitrary constants. One member of the family, fixed by extra conditions such as y = 2 when x = 0.

A DE of order n needs n conditions to fix all n constants.

How to verify that a function is a solution

1
Differentiate the given function as many times as the order of the DE.
2
Substitute y, y′, y″, ... into the left side of the DE.
3
Simplify. If it equals the right side for every x, the function is a solution.
4
Count the arbitrary constants: as many as the order means general; none means particular.

Example 1: verify a general solution

Problem

Show that y = Ae³ˣ + Be⁻⁴ˣ is a solution of y″ + y′ − 12y = 0. Is it the general solution?

Working

y = Ae³ˣ + Be⁻⁴ˣ
y′ = 3Ae³ˣ − 4Be⁻⁴ˣ
y″ = 9Ae³ˣ + 16Be⁻⁴ˣ
y″ + y′ − 12y = (9A + 3A − 12A)e³ˣ + (16B − 4B − 12B)e⁻⁴ˣ
= 0 × e³ˣ + 0 × e⁻⁴ˣ = 0, which is the right side.
The DE has order 2 and y has two arbitrary constants A and B.

Answer

y = Ae³ˣ + Be⁻⁴ˣ satisfies the DE, and it is the general solution.

Example 2: find a particular solution

Problem

y = A cos 2x + B sin 2x is the general solution of y″ + 4y = 0. Find the particular solution with y(0) = 3 and y′(0) = 4.

Working

Put x = 0 in y: y(0) = A cos 0 + B sin 0 = A. So A = 3.
Differentiate: y′ = −2A sin 2x + 2B cos 2x.
Put x = 0: y′(0) = 2B = 4, so B = 2.
Check: y″ = −12 cos 2x − 8 sin 2x = −4y, so y″ + 4y = 0.

Answer

y = 3 cos 2x + 2 sin 2x

In real life

Every possible flight of a cricket ball fits one general solution of the equations of motion. The speed and angle at which the bowler releases it pick out the one particular path it actually takes.

Exam angle

1 marks: “How many arbitrary constants are there in the general solution of a second order DE? How many in a particular solution?” 2 in the general solution (equal to the order); 0 in a particular solution.

2 marks: “Verify that y = 3e⁻²ˣ + 1 is a solution of dy/dx + 2y = 2.” y′ = −6e⁻²ˣ. LHS = −6e⁻²ˣ + 2(3e⁻²ˣ + 1) = −6e⁻²ˣ + 6e⁻²ˣ + 2 = 2 = RHS. It is a particular solution (no arbitrary constant).

3 marks: “Show that y = Ax + B/x (x ≠ 0) is the general solution of x²y″ + xy′ − y = 0.” y′ = A − B/x², y″ = 2B/x³. x²y″ + xy′ − y = 2B/x + Ax − B/x − Ax − B/x = 0. Two constants and order 2, so it is the general solution.

Remember it

Constants = order for the general solution; zero constants for a particular one.