4. Second-Order Non-Homogeneous

Video coming soon2 worked examples

Theory

For $a\,y'' + b\,y' + c\,y = f(x)$ , the general solution is the complementary function (CF) — the solution of the homogeneous equation — plus a particular integral (PI), a trial function shaped like $f(x)$ :

y = \text{CF} + \text{PI}

Trial forms: a constant or polynomial for a polynomial $f(x)$ , $Ce^{kx}$ for an exponential, and $C\cos + D\sin$ for a trig term. If the trial clashes with the CF, multiply it by $x$ .

The Golden Rule: find the CF from the auxiliary equation, find a PI by substituting a suitable trial function, add them — and apply any conditions to the full solution, not just the CF.

⚠️ Common Examiner Traps

Missing a part: the general solution is CF plus PI — both are required.
Wrong trial: match the PI trial to the form of $f(x)$ .
Conditions applied too early: fix the constants using the full CF + PI, not the CF alone.

Worked examples

Example 1

Find the general solution of $y'' - 5y' + 6y = 12$ .

Step 1: The complementary function comes from $m^2 - 5m + 6 = 0$ , giving $m = 2, 3$ :

\text{CF} = Ae^{2x} + Be^{3x}

Step 2: The right side is a constant, so try $y = C$ (then $y' = y'' = 0$ ):

6C = 12 \implies C = 2 \quad\Rightarrow\quad \text{PI} = 2

Step 3: The general solution is CF + PI:

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

Example 2

Find the general solution of $y'' - 3y' + 2y = e^{3x}$ .

Step 1: The auxiliary equation $m^2 - 3m + 2 = 0$ gives $m = 1, 2$ :

\text{CF} = Ae^{x} + Be^{2x}

Step 2: Try $y = Ce^{3x}$ , so $y' = 3Ce^{3x}$ and $y'' = 9Ce^{3x}$ . Substitute:

(9C - 9C + 2C)e^{3x} = e^{3x} \implies 2C = 1 \implies C = \tfrac{1}{2}

Step 3: The general solution is:

y = Ae^{x} + Be^{2x} + \tfrac{1}{2}e^{3x}