Question 9 - A-Level Maths

CAIE FP1 2014 November — Question 9 11 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2014
Session	November
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Solve via substitution then back-substitute
Difficulty	Challenging +1.3 This is a structured Further Maths question with clear guidance (substitution given explicitly). Students must differentiate v=xy to find relationships, verify the simplified DE (routine algebra), then solve a standard constant-coefficient second-order linear DE with particular integral. While multi-step, each component is textbook-standard for FP1, making it moderately above average difficulty but not requiring novel insight.
Spec	4.10c Integrating factor: first order equations 4.10d Second order homogeneous: auxiliary equation method 4.10e Second order non-homogeneous: complementary + particular integral

9 Given that $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 2 x + 2 ) \frac { \mathrm { d } y } { \mathrm {~d} x } + ( 2 - 3 x ) y = 10 \mathrm { e } ^ { 2 x }$$ and that $v = x y$, show that $$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} v } { \mathrm {~d} x } - 3 v = 10 \mathrm { e } ^ { 2 x }$$ Find the general solution for $y$ in terms of $x$.

Show mark scheme Show mark scheme source

Question 9:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$v = xy \Rightarrow \frac{dv}{dx} = y + x\frac{dy}{dx} \Rightarrow \frac{d^2v}{dx^2} = 2\frac{dy}{dx} + x\frac{d^2y}{dx^2}$	B1B1
Substitution leading to $\frac{d^2v}{dx^2} + 2\frac{dv}{dx} - 3v = 10e^{2x}$	M1A1 (4)
$m^2 + 2m - 3 = 0 \Rightarrow (m+3)(m-1) = 0 \Rightarrow m = -3, 1$	M1
CF: $Ae^{-3x} + Be^{x}$	A1
$v = ke^{2x} \Rightarrow v' = 2ke^{2x} \Rightarrow v'' = 4ke^{2x}$	M1
$\Rightarrow 4ke^{2x} + 4ke^{2x} - 3ke^{2x} = 10e^{2x} \Rightarrow 5k = 10 \Rightarrow k = 2$, PI: $2e^{2x}$	M1A1
GS: $v = Ae^{-3x} + Be^{x} + 2e^{2x} \Rightarrow y = \frac{1}{x}(Ae^{-3x} + Be^{x} + 2e^{2x})$	M1A1 (7) [11]

# Question 9:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $v = xy \Rightarrow \frac{dv}{dx} = y + x\frac{dy}{dx} \Rightarrow \frac{d^2v}{dx^2} = 2\frac{dy}{dx} + x\frac{d^2y}{dx^2}$ | B1B1 | |
| Substitution leading to $\frac{d^2v}{dx^2} + 2\frac{dv}{dx} - 3v = 10e^{2x}$ | M1A1 (4) | |
| $m^2 + 2m - 3 = 0 \Rightarrow (m+3)(m-1) = 0 \Rightarrow m = -3, 1$ | M1 | |
| CF: $Ae^{-3x} + Be^{x}$ | A1 | |
| $v = ke^{2x} \Rightarrow v' = 2ke^{2x} \Rightarrow v'' = 4ke^{2x}$ | M1 | |
| $\Rightarrow 4ke^{2x} + 4ke^{2x} - 3ke^{2x} = 10e^{2x} \Rightarrow 5k = 10 \Rightarrow k = 2$, PI: $2e^{2x}$ | M1A1 | |
| GS: $v = Ae^{-3x} + Be^{x} + 2e^{2x} \Rightarrow y = \frac{1}{x}(Ae^{-3x} + Be^{x} + 2e^{2x})$ | M1A1 (7) **[11]** | |

---

Show LaTeX source

9 Given that

$$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 2 x + 2 ) \frac { \mathrm { d } y } { \mathrm {~d} x } + ( 2 - 3 x ) y = 10 \mathrm { e } ^ { 2 x }$$

and that $v = x y$, show that

$$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} v } { \mathrm {~d} x } - 3 v = 10 \mathrm { e } ^ { 2 x }$$

Find the general solution for $y$ in terms of $x$.

\hfill \mbox{\textit{CAIE FP1 2014 Q9 [11]}}

This paper (12 questions)

View full paper

Q1 5 Q2 6 Q3 7 Q4 7 Q5 8 Q6 9 Q7 10 Q8 11 Q9 11 Q10 12 Q11 EITHER Q11 OR