3. Second-Order Homogeneous

Video coming soon3 worked examples

Theory

A second-order homogeneous equation $a\,y'' + b\,y' + c\,y = 0$ is solved via its auxiliary equation $am^2 + bm + c = 0$ . The nature of the roots gives the general solution:

Real distinct roots $m_1, m_2$ : $y = Ae^{m_1 x} + Be^{m_2 x}$
Real repeated root $m$ : $y = (A + Bx)e^{mx}$
Complex roots $p \pm qi$ : $y = e^{px}(A\cos qx + B\sin qx)$

The Golden Rule: solve the auxiliary equation first — the type of roots (distinct real, repeated, or complex) determines which of the three solution forms to use.

⚠️ Common Examiner Traps

Repeated root: you need the $(A + Bx)$ form, not just $Ae^{mx}$ .
Complex roots: use the $e^{px}(A\cos qx + B\sin qx)$ form.
Two constants: a second-order equation always has two arbitrary constants.

Worked examples

Example 1

Find the general solution of $y'' - 5y' + 6y = 0$ .

Step 1: Form and solve the auxiliary equation:

m^2 - 5m + 6 = 0 \implies (m-2)(m-3) = 0 \implies m = 2,\ 3

Step 2: Real distinct roots, so:

y = Ae^{2x} + Be^{3x}

Example 2

Find the general solution of $y'' - 4y' + 4y = 0$ .

Step 1: The auxiliary equation has a repeated root:

m^2 - 4m + 4 = (m-2)^2 = 0 \implies m = 2 \ (\text{repeated})

Step 2: Use the repeated-root form:

y = (A + Bx)e^{2x}

Example 3

Find the general solution of $y'' + 4y = 0$ .

Step 1: The auxiliary equation gives complex roots:

m^2 + 4 = 0 \implies m = \pm 2i \quad (p = 0,\ q = 2)

Step 2: Use the complex-root form (here $e^{0x} = 1$ ):

y = A\cos 2x + B\sin 2x