Non-homogeneous — method of undetermined coefficients (annihilator method)

What is an annihilator?

WHY do annihilators even exist? Because the functions appearing in "nice" $g(x)$ — polynomials, exponentials, sines, cosines, and their products — are exactly the solutions of constant-coefficient homogeneous ODEs. Each such solution is born from a root of a characteristic polynomial, so reading the function backwards gives the operator.

HOW to read the table backwards (derivation of one row). Take $g=e^{\alpha x}$. The ODE with characteristic root $r=\alpha$ is $(r-\alpha)=0 \Rightarrow (D-\alpha)y=0$, whose solution is $Ce^{\alpha x}$. So $(D-\alpha)$ annihilates $e^{\alpha x}$: $(D-\alpha)e^{\alpha x} = \alpha e^{\alpha x} - \alpha e^{\alpha x} = 0.\ \checkmark$ For $x^n e^{\alpha x}$ we need the root $\alpha$ repeated $n+1$ times (repeated roots give $x^k e^{\alpha x}$ terms), hence $(D-\alpha)^{n+1}$.

The method, step by step

Worked Example 1 — exponential, no overlap

Solve $y'' - 3y' + 2y = 4e^{3x}$, i.e. $(D^2-3D+2)y = 4e^{3x}$.

Step 1 (Why): factor $D^2-3D+2=(D-1)(D-2)$, roots $1,2$. So $y_h = c_1 e^{x} + c_2 e^{2x}.$

Step 2 (Why): $g=4e^{3x}$ comes from root $3$, so annihilator $A=(D-3)$. Apply it: $(D-3)(D-1)(D-2)\,y = 0.$

Step 3–4 (Why this step?): roots now $1,2,3$. General solution $c_1e^x+c_2e^{2x}+c_3e^{3x}$. The first two are $y_h$, so the new term gives the form: $y_p = A e^{3x}.$

Step 5 (Why): substitute. $y_p''-3y_p'+2y_p = (9-9+2)Ae^{3x}=2Ae^{3x}=4e^{3x}\Rightarrow A=2.$

Step 6: $\boxed{y = c_1e^x + c_2e^{2x} + 2e^{3x}.}$

Worked Example 2 — overlap (resonance) forces an $x$

Solve $y'' - 3y' + 2y = 5e^{2x}$.

Step 1: same $y_h = c_1e^x + c_2e^{2x}$.

Step 2 (Why the overlap matters): $g=5e^{2x}$ has root $2$, but $2$ is already a root of $L$. Annihilator $A=(D-2)$: $(D-2)\,(D-1)(D-2)\,y = (D-1)(D-2)^2 y = 0.$

Step 3–4 (Why this step?): roots $1,\,2,\,2$ (double). General solution $c_1e^x + c_2e^{2x} + c_3\,x e^{2x}$. The first two are $y_h$; the genuinely new term is $xe^{2x}$. Hence $y_p = A\,x e^{2x}.$ (See — the $x$ appeared automatically because the root became double.)

Step 5 (Why): with $y_p=Axe^{2x}$, $y_p' = A e^{2x}(1+2x)$, $y_p'' = A e^{2x}(4+4x)$. Then $y_p''-3y_p'+2y_p = Ae^{2x}\big[(4+4x)-3(1+2x)+2x\big]=Ae^{2x}(1)=5e^{2x}\Rightarrow A=5.$

Step 6: $\boxed{y=c_1e^x+c_2e^{2x}+5xe^{2x}.}$

Worked Example 3 — trig with the $D^2+\beta^2$ annihilator

Solve $y'' + y = \cos 2x$.

Step 1: $D^2+1=0\Rightarrow r=\pm i$, so $y_h=c_1\cos x + c_2\sin x$.

Step 2 (Why): $\cos 2x$ comes from roots $\pm 2i$, annihilator $A=D^2+4$. No overlap (since $i\ne 2i$). $(D^2+4)(D^2+1)y=0.$

Step 3–4: roots $\pm i,\pm 2i$. New terms: $\cos2x,\sin2x$. So $y_p = A\cos 2x + B\sin 2x.$

Step 5 (Why): $y_p'' = -4A\cos2x - 4B\sin2x$, so $y_p''+y_p = -3A\cos2x -3B\sin2x = \cos2x \Rightarrow A=-\tfrac13,\ B=0.$

Step 6: $\boxed{y=c_1\cos x + c_2\sin x -\tfrac13\cos 2x.}$

Common mistakes (Steel-man + fix)

Active recall

Flashcards

Connections

• Linear differential operators and the D-operator
• Homogeneous linear ODEs with constant coefficients
• Characteristic equation and repeated roots
• Variation of parameters (works when $g$ is not annihilator-friendly, e.g. $\sec x$)
• Superposition principle for linear ODEs
• Resonance in forced oscillations (physical meaning of the $x$-factor)

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Hinglish (regional understanding)