In A-Level Further Mathematics, solving second-order differential equations is one of the highest-scoring areas, yet it contains several subtle details that trip students up. We will look at equations of the form:
where $a, b,$ and $c$ are constants, and $f(x)$ is a non-zero function. The solution consists of finding two components: the **Complementary Function (CF)** and the **Particular Integral (PI)**.
- General Solution Structure: $y = y_{CF} + y_{PI}$.
- CF Step: Solve homogeneous form $am^2 + bm + c = 0$ using quadratic roots.
- PI Step: Match trial function form to $f(x)$ on RHS.
- The Overlap Rule: If the trial function is already a term in the CF, multiply the trial function by $x$ (or $x^2$) to make it unique.
The Standard Method
Find the Complementary Function (CF)
Set $f(x) = 0$ and solve the auxiliary quadratic equation $am^2 + bm + c = 0$. Depending on the roots, choose the correct CF form:
- Two real roots $m_1, m_2$: $y_{CF} = Ae^{m_1 x} + Be^{m_2 x}$
- One repeated root $m$: $y_{CF} = (A + Bx)e^{mx}$
- Complex roots $p \pm qi$: $y_{CF} = e^{px}(A\cos(qx) + B\sin(qx))$
Propose a Particular Integral (PI)
Choose a trial function $y_{PI}$ that matches the form of $f(x)$. Check standard guides: linear $f(x) = px + q \implies \text{trial } Cx + D$, exponential $f(x) = ke^{rx} \implies \text{trial } Ce^{rx}$.
Solve for constants in the PI
Differentiate your proposed $y_{PI}$ to find $y'_{PI}$ and $y''_{PI}$. Substitute these back into the original equation and equate coefficients to find the values of your constants.
Write down the General Solution
Combine the two parts: $y(x) = y_{CF} + y_{PI}$. If boundary conditions are given (e.g. $y(0) = k$, $y'(0) = j$), solve for constants $A$ and $B$ only after combining.
Understanding the Overlap Case
The "overlap case" is the most common place students lose marks. If your standard trial PI function is identical to a term in your CF, substituting it into the LHS will result in zero, rendering it impossible to solve for your constants.
| RHS Function $f(x)$ | CF Terms Contain | Standard Trial PI | Correct Overlap Trial PI |
|---|---|---|---|
| $e^{kx}$ | No $e^{kx}$ terms | $Ce^{kx}$ | (No change) $Ce^{kx}$ |
| $e^{kx}$ | $Ae^{kx}$ (distinct root) | $Ce^{kx}$ | $Cxe^{kx}$ |
| $e^{kx}$ | $(A+Bx)e^{kx}$ (repeated root) | $Ce^{kx}$ | $Cx^2 e^{kx}$ |
Worked Example: Handling the Overlap Case
Solve the following differential equation:
Step 1: Solve for the CF
Set LHS to 0. The auxiliary equation is:
Since we have two distinct real roots, the CF is:
Step 2: Propose the PI (Check for Overlap)
The RHS function is $f(x) = e^{2x}$. Normally, we would try $y_{PI} = Ce^{2x}$.
However, notice that $Ae^{2x}$ is already part of the CF! If we try $Ce^{2x}$, the LHS simplifies to 0. To fix this overlap, we multiply by $x$:
Step 3: Differentiate and Substitute
Using the product rule, differentiate $y_{PI}$ twice:
Substitute $y_{PI}, y'_{PI},$ and $y''_{PI}$ back into the original equation:
Divide through by $e^{2x}$ and expand:
Thus, the Particular Integral is $y_{PI} = -xe^{2x}$.
Step 4: Write General Solution
Combine the CF and PI:
Common Pitfall: Applying boundary conditions too early
Students often try to find the constants $A$ and $B$ using the boundary values on the CF alone. This is incorrect! You must write the full general solution ($y = y_{CF} + y_{PI}$) first, differentiate that combined equation, and only then apply the conditions to find $A$ and $B$.
Frequently Asked Questions
Even if the RHS is only $\sin(kx)$, your trial PI must contain both sine and cosine terms: $y_{PI} = C\sin(kx) + D\cos(kx)$. This is because differentiating sine produces cosine, and both are required to balance the LHS coefficients.
If your auxiliary equation has a repeated root (e.g. $m=2, 2$), the CF is $(A+Bx)e^{2x}$. If the RHS is $e^{2x}$, multiplying by $x$ still leaves an overlap with the $Bxe^{2x}$ term. Therefore, you must multiply by $x^2$, giving a trial PI of $y_{PI} = Cx^2 e^{2x}$.
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