| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Verify particular integral form |
| Difficulty | Standard +0.8 This is a Further Maths second-order DE question requiring verification of a particular integral form (involving the 'resonance' case where the RHS matches a complementary function root), then finding a complete solution with initial conditions. While methodical, it demands careful differentiation of a product, substitution, algebraic manipulation, and integration of complementary function plus particular integral—significantly above standard A-level but routine for Further Maths students who have practiced this topic. |
| Spec | 4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = kxe^{-x}\) leading to \(y' = ke^{-x}(1-x)\) leading to \(y'' = ke^{-x}(x-2)\) | B1 B1 | Differentiates particular integral |
| \(ke^{-x}(x-2) - 2ke^{-x}(1-x) - 3kxe^{-x} = 4e^{-x}\) leading to \(-4k = 4\) | M1 | Substitutes |
| \(k = -1\) | A1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(m^2 - 2m - 3 = 0\) leading to \(m = -1, 3\) | M1 | Auxiliary equation |
| \(y = Ae^{-x} + Be^{3x}\) | A1 | Complementary function (can be awarded if seen in general solution) |
| \(y = Ae^{-x} + Be^{3x} - xe^{-x} = (A-x)e^{-x} + Be^{3x}\) | A1 FT | General solution. FT on their \(k\). Must have "\(y=\)" |
| \(y' = -(A-x)e^{-x} - e^{-x} + 3Be^{3x}\) | B1 | Derivative of general solution |
| \(A + B = \frac{1}{2}\), \(-A - 1 + 3B = \frac{1}{2}\) leading to \(A = 0\), \(B = \frac{1}{2}\) | M1 | Substitutes initial conditions into general solution |
| \(y = \frac{1}{2}e^{3x} - xe^{-x}\) | A1 | Must have "\(y=\)" |
| Total: 6 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = kxe^{-x}$ leading to $y' = ke^{-x}(1-x)$ leading to $y'' = ke^{-x}(x-2)$ | B1 B1 | Differentiates particular integral |
| $ke^{-x}(x-2) - 2ke^{-x}(1-x) - 3kxe^{-x} = 4e^{-x}$ leading to $-4k = 4$ | M1 | Substitutes |
| $k = -1$ | A1 | |
| **Total: 4** | | |
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## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $m^2 - 2m - 3 = 0$ leading to $m = -1, 3$ | M1 | Auxiliary equation |
| $y = Ae^{-x} + Be^{3x}$ | A1 | Complementary function (can be awarded if seen in general solution) |
| $y = Ae^{-x} + Be^{3x} - xe^{-x} = (A-x)e^{-x} + Be^{3x}$ | A1 FT | General solution. FT on their $k$. Must have "$y=$" |
| $y' = -(A-x)e^{-x} - e^{-x} + 3Be^{3x}$ | B1 | Derivative of general solution |
| $A + B = \frac{1}{2}$, $-A - 1 + 3B = \frac{1}{2}$ leading to $A = 0$, $B = \frac{1}{2}$ | M1 | Substitutes initial conditions into general solution |
| $y = \frac{1}{2}e^{3x} - xe^{-x}$ | A1 | Must have "$y=$" |
| **Total: 6** | | |
---5 The variables $x$ and $y$ are related by the differential equation
$$\frac { d ^ { 2 } y } { d x ^ { 2 } } - 2 \frac { d y } { d x } - 3 y = 4 e ^ { - x }$$
\begin{enumerate}[label=(\alph*)]
\item Find the value of the constant $k$ such that $\mathrm { y } = \mathrm { kxe } ^ { - \mathrm { x } }$ is a particular integral of the differential equation.
\item Find the solution of the differential equation for which $\mathrm { y } = \frac { \mathrm { dy } } { \mathrm { dx } } = \frac { 1 } { 2 }$ when $x = 0$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q5 [10]}}