Complementary Function and Particular Integral in Linear Differential equations

The structure. A second-order linear ODE with constant coefficients has the form $a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + c\,y = f(x).$ Its general solution always splits into two pieces: $y = \underbrace{y_c}_{\text{complementary function}} + \underbrace{y_p}_{\text{particular integral}}.$

What each piece does.

• The complementary function (CF) $y_c$ is the general solution of the homogeneous equation $a\,y'' + b\,y' + c\,y = 0$. It carries the two arbitrary constants $A$ and $B$. (Solving the auxiliary equation for $y_c$ — real, repeated or complex roots — is covered in the sibling subtopic; here we simply quote it.)
• The particular integral (PI) $y_p$ is any single function that satisfies the full equation with $f(x)$ on the right. It has no arbitrary constants.

Why adding them works. Substitute $y = y_c + y_p$ into the left-hand side. Because the operator $a\frac{d^2}{dx^2}+b\frac{d}{dx}+c$ is linear, it distributes: $a(y_c+y_p)'' + b(y_c+y_p)' + c(y_c+y_p) = \underbrace{\big(a y_c'' + b y_c' + c y_c\big)}_{=\,0} + \underbrace{\big(a y_p'' + b y_p' + c y_p\big)}_{=\,f(x)} = f(x).$ So $y_c + y_p$ satisfies the full equation, and because $y_c$ holds two free constants, this captures every solution.

Cambridge tip. A PI is any solution that works — you do not need the "simplest" one, but the standard trial forms give you the simplest automatically, so use them.

The PI is found by trying a form that looks like $f(x)$, with unknown coefficients, then substituting to pin those coefficients down. The standard menu (not in MF19 — memorise it):

$f(x)$ on the right	Trial PI $y_p$
constant $k$	$C$
polynomial degree $n$	full polynomial degree $n$
$e^{kx}$	$C e^{kx}$
$\cos px$ or $\sin px$	$C\cos px + D\sin px$
sum of the above	sum of the matching trials

Two rules that trap students.

• Trig: always include BOTH $\cos$ and $\sin$. Even if $f(x)$ is only $\cos px$, the $y'$ term mixes $\cos$ and $\sin$, so you need $C\cos px + D\sin px$ to balance both.
• Polynomial: use the FULL polynomial. For $f(x)=2x$ try $y_p = ax + b$, not just $ax$ — the lower-degree terms are needed to absorb the contributions from $y'$ and $y$.

The method, every time.

1. Write the trial $y_p$ with unknown coefficients.
2. Differentiate to get $y_p'$ and $y_p''$.
3. Substitute all three into $a y'' + b y' + c y = f(x)$.
4. Compare coefficients (or compare $\cos$/$\sin$ separately) and solve for the unknowns.

Cambridge tip. Substituting back is not optional verification — it is how you find the coefficients. Show the substitution and the comparison explicitly; mark schemes award method marks for it.

The problem. Sometimes the natural trial PI is already part of the CF. Then substituting it gives $0 = f(x)$ — a contradiction — and the trial fails. This is a clash (or resonance).

The fix. Multiply the failing trial by $x$. If the clashing CF term comes from a repeated root (so the CF already contains both $e^{kx}$ and $x e^{kx}$), multiply by $x^2$ instead.

Worked logic. Suppose the CF is $y_c = Ae^{2x} + Be^{-x}$ and the equation has $f(x) = e^{2x}$. The natural trial $Ce^{2x}$ is already in the CF, so it solves the homogeneous equation and substitutes to $0$. Instead try $y_p = Cx\,e^{2x}.$ This is not in the CF, so it can balance the $e^{2x}$ on the right. (For a repeated root, e.g. $y_c = (A+Bx)e^{2x}$ with $f(x)=e^{2x}$, you would need $y_p = Cx^2 e^{2x}$.)

How to spot a clash quickly. Before differentiating, line your trial up against the CF terms. If any term of $y_p$ matches a CF term up to a constant multiple, it clashes. The same applies to trig: if $f(x)=\sin px$ and the CF already contains $\cos px$ and $\sin px$ (i.e. the auxiliary roots are $\pm ip$), use $y_p = x(C\cos px + D\sin px)$.

Cambridge tip. Always write the CF down first, then choose the trial — that ordering makes the clash obvious and is exactly how examiners expect the working laid out.

Put the pieces together. Once you have the CF (with $A, B$) and the PI (coefficients now numerical), the general solution is simply $y = y_c + y_p.$ State this line explicitly before going further — it is a marked step.

Apply the conditions to the WHOLE solution. A second-order ODE needs two conditions, typically $y(x_0)$ and $y'(x_0)$. The fatal error is applying them to the CF alone; the constants live in the CF, but the value of $y$ at a point depends on CF and PI together.

1. Differentiate the full general solution to get $\dfrac{dy}{dx}$ (you will need both $A$ and $B$ in it).
2. Substitute the first condition (e.g. $y=y_0$ when $x=0$) to get one equation in $A, B$.
3. Substitute the second condition (e.g. $y'=v_0$ when $x=0$) to get a second equation.
4. Solve the simultaneous pair for $A$ and $B$.
5. Write the particular solution with the numerical constants substituted in.

Check. A quick sanity check: differentiate your final answer and substitute back into the original ODE — it should reproduce $f(x)$ exactly, and satisfy both conditions.

Cambridge tip. Differentiate before substituting the second condition, and keep $A$ and $B$ symbolic until you have both equations — solving prematurely loses the link between the two conditions.