Linear ODE with constant coefficients

At a Glance

• Frequency: 4 sub-parts across 4 of 13 years (2017, 2018, 2021, 2023)
• Priority tier: T3
• Marks (count): 10 (1), 15 (2), 7 (1)
• Average solve time: ~10 min
• Difficulty mix: medium 2, easy 1, hard 1
• Section: B | Dominant type: computation

Why This Chapter Matters

Constant-coefficient linear ODEs appear in Section B at 7–15 marks. They always follow the same structure: find the complementary function (CF) from the auxiliary equation, find a particular integral (PI) for the right-hand side, and combine. The CF is determined entirely by the roots of the auxiliary polynomial; the PI technique depends on the form of the forcing. The key decisions are (a) which root type gives which CF form, and (b) whether the forcing term resonates with the CF (requiring a modified PI). Master the resonance rule and these questions become routine.

Minimum Theory

Auxiliary equation and CF. For $a_n y^{(n)}+\cdots+a_1 y'+a_0 y = f(x)$, substitute $y=e^{mx}$ to get the auxiliary polynomial $p(m)=0$. The CF depends on the roots:

Particular integral — resonance rule. If the forcing $f(x)$ contains a term that matches a term in the CF:

• $f = e^{ax}g(x)$ and $a$ is a root of multiplicity $s$: multiply the standard PI by $x^s$.
• Otherwise (no resonance): use the standard PI directly.

Applying initial conditions. Compute $y_p(0)$ and $y_p'(0)$ from the particular integral, then set up equations for the CF constants from the total $y(0)$ and $y'(0)$.

Substitution to constant coefficients. Some variable-coefficient ODEs can be converted to constant-coefficient by a substitution (e.g. $t=\sin x$ for certain oscillatory coefficients). When the coefficient of $y'$ involves $\tan x$ and the coefficient of $y$ involves $\cos^2 x$, try $t=\sin x$.

Question Archetypes

constant-coeff-ivp (2 question(s); 2017, 2018)

Recognition Cues

• Constant-coefficient homogeneous or non-homogeneous ODE with explicit initial conditions $y(0)$, $y'(0)$.
• Usually second-order; 7–10 marks.

Solution Template

1. Auxiliary equation: substitute $y=e^{mx}$; solve for roots.
2. Write CF from the root type.
3. Find PI (if non-homogeneous): standard or operator method.
4. Apply ICs: differentiate $y_c+y_p$; evaluate at $x=0$; solve for the constants.

Worked Example 1

2017 Paper 1, 2017-P1-Q6b-ii (7 marks)

Solve $20y''+4y'+y=0$, $y(0)=3.2$, $y'(0)=0$.

Step 1. Auxiliary: $20m^2+4m+1=0$; discriminant $16-80=-64<0$. $m = \frac{-4\pm8i}{40} = -\frac{1}{10}\pm\frac{i}{5}.$

Step 2. CF: $y=e^{-x/10}(A\cos\tfrac{x}{5}+B\sin\tfrac{x}{5})$.

Step 3. Homogeneous; no PI.

Step 4. $y(0)=A=3.2$. Differentiate: $y'(0)=-A/10+B/5=0 \Rightarrow B=A/2=1.6$. $\boxed{y = e^{-x/10}\!\left(3.2\cos\tfrac{x}{5}+1.6\sin\tfrac{x}{5}\right).}$

Worked Example 2

2018 Paper 1, 2018-P1-Q7c (13 marks)

Solve $y''-5y'+4y=e^{2t}$, $y(0)=19/12$, $y'(0)=8/3$.

Step 1. Auxiliary: $m^2-5m+4=(m-1)(m-4)=0$; roots $m=1,4$.

Step 2. CF: $y_c=Ae^t+Be^{4t}$.

Step 3. PI: try $y_p=Ce^{2t}$. $4C-10C+4C=-2C=1 \Rightarrow C=-1/2$. So $y_p=-e^{2t}/2$.

Step 4. $y=Ae^t+Be^{4t}-e^{2t}/2$.

• $y(0)=A+B-1/2=19/12 \Rightarrow A+B=19/12+1/2=25/12$.
• $y'(0)=A+4B-1=8/3 \Rightarrow A+4B=11/3=44/12$.

Subtracting: $3B=44/12-25/12=19/12 \Rightarrow B=19/36$; $A=25/12-19/36=75/36-19/36=56/36=14/9$. $\boxed{y = \frac{14}{9}e^t+\frac{19}{36}e^{4t}-\frac{1}{2}e^{2t}.}$

Common Traps

• Include $y_p'(0)$ when applying ICs. The derivative of $y=y_c+y_p$ at $x=0$ includes $y_p'(0)$; in the 2018 example, $y_p'(0)=-1$. Forgetting this shifts the ICs and gives wrong constants.
• Complex roots: include the product rule. When differentiating $y_c=e^{-x/10}(\ldots)$, use the product rule — the derivative picks up the $-1/10$ factor from $e^{-x/10}$, giving a cross term that is essential for the IC equation.

higher-order-constant-coeff (1 question(s); 2023)

Recognition Cues

• Third-order or higher equation with multiple forcing terms ($e^x + \cos x$, etc.).
• Each forcing term must be checked independently for resonance.

Solution Template

1. Find all roots of the auxiliary polynomial (check integer roots by trial).
2. For each forcing term, determine whether its frequency matches a root.
3. For resonant terms, multiply the PI by $x^s$ (where $s$ = multiplicity of the matching root).
4. Combine all PIs; write the general solution.

Worked Example

2023 Paper 1, 2023-P1-Q6a (15 marks)

Solve $y'''-3y''+4y'-2y=e^x+\cos x$.

Step 1. Auxiliary: $m^3-3m^2+4m-2$. Try $m=1$: $1-3+4-2=0$ ✓. Factor: $(m-1)(m^2-2m+2)=0$. Roots: $m=1$ and $m=1\pm i$.

Step 2. CF: $y_c=c_1e^x + e^x(c_2\cos x+c_3\sin x)$.

Step 3. Check resonance:

• For $e^x$: $m=1$ is a simple root → multiply PI by $x$. Since $(D-1)(D^2-2D+2)(xe^x) = e^x$ (after applying the operator), we get $y_{p,1}=xe^x$.
• For $\cos x$: $m=\pm i$ are NOT roots (roots are $1\pm i$, not $\pm i$) → no resonance. Standard ansatz $B\cos x+C\sin x$.

Substituting $B\cos x+C\sin x$: coefficient matching gives $B=1/10$, $C=3/10$.

Step 4. $\boxed{y = c_1e^x+e^x(c_2\cos x+c_3\sin x)+xe^x+\tfrac{\cos x+3\sin x}{10}.}$

Common Traps

• $1\pm i$ are not $\pm i$. The roots of $m^2-2m+2$ are $1\pm i$, not $\pm i$. Confusing these leads to incorrectly declaring resonance for $\cos x$ (which has frequency $m=\pm i$).
• Resonance multiplicity. $m=1$ is a simple root, so multiply $e^x$ PI by $x^1$ (not $x^2$). Multiplying by $x^2$ when the root has multiplicity 1 gives the wrong coefficient.

reducible-to-constant-coeff (1 question(s); 2021)

Recognition Cues

• Variable-coefficient equation where the coefficient of $y'$ involves $\tan x$ and the coefficient of $y$ involves $\cos^2 x$.
• The substitution $t=\sin x$ reduces it to a constant-coefficient equation.

Solution Template

1. Try $t = \sin x$ (suggested by the coefficient structure): $dy/dx = (dy/dt)\cos x$.
2. Compute $d^2y/dx^2 = -\sin x\,(dy/dt) + \cos^2x\,(d^2y/dt^2)$.
3. Substitute; the cross terms involving $\sin x$ cancel.
4. Divide by $\cos^2x$ to obtain a constant-coefficient ODE in $t$.
5. Solve and convert back to $x$.

Worked Example

2021 Paper 1, 2021-P1-Q6b (15 marks)

Solve $y''+(\tan x-3\cos x)y'+2y\cos^2 x=\cos^4 x$.

Step 1–3. Substitute $t=\sin x$; $d^2y/dx^2 = -\sin x\,y_t + \cos^2x\,y_{tt}$. The $-\sin x\,y_t$ from $y''$ and the $+\sin x\,y_t$ from $\tan x\cdot y'$ cancel, giving: $\cos^2x\,y_{tt} - 3\cos^2x\,y_t + 2\cos^2x\,y = \cos^4x.$

Step 4. Divide by $\cos^2x$: $y_{tt}-3y_t+2y = 1-t^2$ (using $\cos^2x = 1-t^2$).

Step 5. Auxiliary: $(D-1)(D-2)$; CF $= C_1e^t+C_2e^{2t}$. PI: $at^2+bt+c$ — matching coefficients gives $a=-1/2$, $b=-3/2$, $c=-5/4$.

Convert back ($t=\sin x$): $\boxed{y = C_1e^{\sin x}+C_2e^{2\sin x}-\tfrac{\sin^2x}{2}-\tfrac{3\sin x}{2}-\tfrac{5}{4}.}$

Common Traps

• The cancellation is the key. The $-\sin x\,y_t$ from $d^2y/dx^2$ must cancel with $+\sin x\,y_t$ from $\tan x\cdot y'$. If they don’t cancel, the substitution is wrong.
• RHS in terms of $t$. After dividing by $\cos^2x$, write $\cos^2x = 1-\sin^2x = 1-t^2$ on the right side.

Marks-Aware Writing

For 7-mark IVP: show the auxiliary equation, state all roots, write the CF, differentiate once, apply ICs in two steps. No steps can be skipped.

For 15-mark problems: show the full operator working for each PI separately, and verify each PI by substitution (or state the operator result). The ICs must be applied to $y_c+y_p$, not just $y_c$.

For the reducible case: the cancellation step (showing the $\sin x$ terms cancel) is the “demonstration” the question asks for — show it explicitly.

Practice Set

• 2018-P1-Q5c (10 m) — — Hint: auxiliary $(m-2)^3$; $e^{2x}$ resonates with multiplicity 3; multiply PI by $x^3$.
• 2023-P1-Q6a already worked above.