MTH2032 Chap.3 General Theory of Second-Order Linear ODEs
General Theory of Second-Order Linear ODEs
Week 3 treats second-order linear equations as linear algebra. Superposition makes the solution set of a homogeneous equation a space, the Basis theorem shows that two independent solutions span it, and the Wronskian decides independence.
For constant coefficients, the characteristic equation produces the basis, and this unit writes the real-root case with cosh and sinh, a choice that pays off whenever a boundary value at zero must be imposed. Worked examples use initial data at zero, where the hyperbolic basis removes a constant in one step.
What this chapter covers
- 01
Standard form and Theorem 7.1
- 02
General solution y_h + y_p
- 03
Superposition principle
- 04
Basis theorem
- 05
Wronskian theorem
- 06
Characteristic roots and Table 1
Worked example · free
Repeated characteristic root with initial data
- 1The characteristic equation λ² − 6λ + 9 = 0 has the double root λ = 3.
- 1The basis is e^(3x) and xe^(3x), so y = (c1 + c2x)e^(3x).
- 1y(0) = c1 = 2.
- 1y′(0) = c2 + 3c1 = 1 gives c2 = −5, so y = (2 − 5x)e^(3x).
Key terms
- Linear Independence
- The property that c1y1 + c2y2 = 0 on an interval only when both constants are zero.
- Basis of Solutions
- A pair of independent solutions whose combinations give every solution of a homogeneous second-order equation.
- Superposition Principle
- The fact that any linear combination of solutions of a homogeneous linear equation is again a solution.
- Hyperbolic Functions
- The functions cosh x = (e^x + e^−x)/2 and sinh x = (e^x − e^−x)/2, used for the real-root basis.
- Standard Form
- The arrangement y″ + py′ + qy = r, with coefficient functions of x, in which the coefficient of the second derivative is one.
- Fundamental Theorem III
- The result that a second-order linear initial value problem with continuous coefficients has exactly one solution on the whole interval.
General Theory of Second-Order Linear ODEs FAQ
What does the Wronskian theorem let me conclude?
For two solutions of a homogeneous equation with continuous coefficients on an interval, the Wronskian is either never zero, in which case the solutions are independent, or identically zero. So a pair whose Wronskian vanishes at one point but not at another cannot both solve such an equation, which is the argument behind several released verification problems.
Why are there exactly two constants in the general solution?
Theorem 7.1 says an initial value problem with given y and y′ at a point has exactly one solution. The notes use it to show that two specific solutions span every solution, and the Basis theorem then states that any independent pair does the same. Two pieces of initial data need two constants.
When should I use cosh and sinh instead of two exponentials?
Use the hyperbolic form whenever a condition is given at zero, especially y(0) = 0 or y′(0) = 0. Because sinh vanishes at zero while cosh equals one there, one constant disappears at once. The unit's table and its problem-set solutions write the real-root case this way, so matching it also makes your work easier to mark.
How are midterm questions on this week usually structured?
Expect a short theory part, such as stating why a pair of functions is or is not a basis, followed by a calculation with a constant-coefficient equation and initial data. Midterm Test 1 covers lecture-note Chapters 1 to 10, so these second-order ideas sit alongside first-order methods and the free oscillation models.
Exam move
Build fluency in three directions. First, the theory: be able to state the superposition principle, the Basis theorem and the Wronskian theorem, and explain in two sentences how Theorem 7.1 shows that two solutions span the space.
Second, the computation: for constant-coefficient equations, practise writing the characteristic equation, its discriminant, the roots in the form −b/2 ± ρ and the matching basis from the unit's table, then fitting initial data. Do at least two problems in each of the three cases and redo one real-root problem twice, once with exponentials and once with cosh and sinh, to see where the hyperbolic form saves work.
Third, the Wronskian argument: compute W for pairs such as x and x³, or x² and x³ ln x, and explain why they cannot both solve an equation with continuous coefficients on an interval containing a zero of W. These are short, high-yield items in the verification tasks and midterms.
Keep a list of the theorem numbers used in this week and one line on what each one guarantees, since short theory questions often ask you to name the result you rely on.
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