| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Particular solution with initial conditions |
| Difficulty | Challenging +1.2 This is a standard Further Maths second-order differential equation with repeated roots (auxiliary equation gives (m-4)²=0) and polynomial particular integral. While it requires multiple techniques (complementary function, particular integral by inspection, applying initial conditions), these are all routine procedures taught systematically in F2. The final part involves straightforward substitution. More challenging than typical A-level questions due to being Further Maths content, but follows a well-practiced algorithm without requiring novel insight. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (AE:) \(m^2-8m+16=0 \Rightarrow (m-4)^2=0 \Rightarrow m=4\) | M1 | Forms auxiliary equation; solves 3TQ; one consistent solution |
| (CF: \(y=\)) \((A+Bx)e^{4x}\) | A1 | Correct complementary function |
| (PI: \(y=\)) \(\lambda x^2+\mu x+\nu\) | B1 | Correct form for PI |
| \(2\lambda-8(2\lambda x+\mu)+16(\lambda x^2+\mu x+\nu)=48x^2-34\); \(\lambda=3\), \(\mu=3\), \(\nu=-1\) | M1 | Differentiates twice, substitutes, equates terms, solves |
| \(y=(A+Bx)e^{4x}+3x^2+3x-1\) | A1ft | Correct GS following their CF only; must have "\(y=\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((0,4) \Rightarrow 4=A-1 \Rightarrow (A=5)\) | M1 | Uses \((0,4)\) in answer to (a) |
| \(\frac{dy}{dx}=4(A+Bx)e^{4x}+Be^{4x}+6x+3\) | M1 | Differentiates GS; product rule used; \(Bxe^{kx}\) term present |
| \(21=4A+B+3 \Rightarrow B=-2\), \(A=5\); \(y=(5-2x)e^{4x}+3x^2+3x-1\) | M1A1 | Substitutes \(x=0\), \(\frac{dy}{dx}=21\); correct particular solution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((x=-2 \Rightarrow y=) \ 9e^{-8}+5\) | M1 A1 | Substitutes \(x=-2\) into particular solution; correct expression |
# Question 4:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| (AE:) $m^2-8m+16=0 \Rightarrow (m-4)^2=0 \Rightarrow m=4$ | M1 | Forms auxiliary equation; solves 3TQ; one consistent solution |
| (CF: $y=$) $(A+Bx)e^{4x}$ | A1 | Correct complementary function |
| (PI: $y=$) $\lambda x^2+\mu x+\nu$ | B1 | Correct form for PI |
| $2\lambda-8(2\lambda x+\mu)+16(\lambda x^2+\mu x+\nu)=48x^2-34$; $\lambda=3$, $\mu=3$, $\nu=-1$ | M1 | Differentiates twice, substitutes, equates terms, solves |
| $y=(A+Bx)e^{4x}+3x^2+3x-1$ | A1ft | Correct GS following their CF only; must have "$y=$" |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(0,4) \Rightarrow 4=A-1 \Rightarrow (A=5)$ | M1 | Uses $(0,4)$ in answer to (a) |
| $\frac{dy}{dx}=4(A+Bx)e^{4x}+Be^{4x}+6x+3$ | M1 | Differentiates GS; product rule used; $Bxe^{kx}$ term present |
| $21=4A+B+3 \Rightarrow B=-2$, $A=5$; $y=(5-2x)e^{4x}+3x^2+3x-1$ | M1A1 | Substitutes $x=0$, $\frac{dy}{dx}=21$; correct particular solution |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(x=-2 \Rightarrow y=) \ 9e^{-8}+5$ | M1 A1 | Substitutes $x=-2$ into particular solution; correct expression |\begin{enumerate}
\item (a) Determine the general solution of the differential equation
\end{enumerate}
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 8 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 16 y = 48 x ^ { 2 } - 34$$
Given that $y = 4$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 21$ at $x = 0$\\
(b) determine the particular solution of the differential equation.\\
(c) Hence find the value of $y$ at $x = - 2$, giving your answer in the form $p \mathrm { e } ^ { q } + r$ where $p , q$ and $r$ are integers to be determined.
\hfill \mbox{\textit{Edexcel F2 2023 Q4 [11]}}