Question 9 - A-Level Maths

CAIE FP1 2016 June — Question 9 11 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2016
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Resonance cases requiring modified PI
Difficulty	Challenging +1.2 This is a standard Further Maths resonance case where the PI form is given, requiring differentiation, substitution to find k, then applying initial conditions. While it involves second-order DEs and the modified PI due to repeated roots (resonance), the structure is entirely routine for FP1 students with no novel problem-solving required—just careful algebraic manipulation across multiple steps.
Spec	4.10d Second order homogeneous: auxiliary equation method 4.10e Second order non-homogeneous: complementary + particular integral

9 Find the value of the constant $k$ such that $y = k x ^ { 2 } \mathrm { e } ^ { 2 x }$ is a particular integral of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 4 \mathrm { e } ^ { 2 x }$$ Hence find the general solution of ( $*$ ). Find the particular solution of ( $*$ ) such that $y = 3$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = - 2$ when $x = 0$.

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Question 9:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$y' = 2kxe^{2x} + 2kx^2e^{2x}$	B1
$y'' = 2ke^{2x} + 8kxe^{2x} + 4kx^2e^{2x}$	B1$\checkmark$
$2ke^{2x}+8kxe^{2x}+4kx^2e^{2x}-8kxe^{2x}-8kx^2e^{2x}+4kx^2e^{2x}=4e^{2x}$ $\Rightarrow 2k=4 \Rightarrow k=2$	M1, A1 [4]
$m^2-4m+4=0 \Rightarrow (m-2)^2=0 \Rightarrow m=2$	M1
CF: $y = Ae^{2x}+Bxe^{2x}$	A1
$y = Ae^{2x}+Bxe^{2x}+2x^2e^{2x}$	A1$\checkmark$ [3]
$y=3$ when $x=0 \Rightarrow A=3$	B1
$y' = 2Ae^{2x}+B(e^{2x}+2xe^{2x})+(4xe^{2x}+4x^2e^{2x})$	M1
$y'=-2$ when $x=0$ and $A=3 \Rightarrow B=-8$	A1
$\Rightarrow y = 3e^{2x}-8xe^{2x}+2x^2e^{2x}$	A1$\checkmark$ [4]

# Question 9:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $y' = 2kxe^{2x} + 2kx^2e^{2x}$ | B1 | |
| $y'' = 2ke^{2x} + 8kxe^{2x} + 4kx^2e^{2x}$ | B1$\checkmark$ | |
| $2ke^{2x}+8kxe^{2x}+4kx^2e^{2x}-8kxe^{2x}-8kx^2e^{2x}+4kx^2e^{2x}=4e^{2x}$ $\Rightarrow 2k=4 \Rightarrow k=2$ | M1, A1 [4] | |
| $m^2-4m+4=0 \Rightarrow (m-2)^2=0 \Rightarrow m=2$ | M1 | |
| CF: $y = Ae^{2x}+Bxe^{2x}$ | A1 | |
| $y = Ae^{2x}+Bxe^{2x}+2x^2e^{2x}$ | A1$\checkmark$ [3] | |
| $y=3$ when $x=0 \Rightarrow A=3$ | B1 | |
| $y' = 2Ae^{2x}+B(e^{2x}+2xe^{2x})+(4xe^{2x}+4x^2e^{2x})$ | M1 | |
| $y'=-2$ when $x=0$ and $A=3 \Rightarrow B=-8$ | A1 | |
| $\Rightarrow y = 3e^{2x}-8xe^{2x}+2x^2e^{2x}$ | A1$\checkmark$ [4] | |

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Show LaTeX source

9 Find the value of the constant $k$ such that $y = k x ^ { 2 } \mathrm { e } ^ { 2 x }$ is a particular integral of the differential equation

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 4 \mathrm { e } ^ { 2 x }$$

Hence find the general solution of ( $*$ ).

Find the particular solution of ( $*$ ) such that $y = 3$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = - 2$ when $x = 0$.

\hfill \mbox{\textit{CAIE FP1 2016 Q9 [11]}}

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Q1 4 Q2 6 Q3 6 Q4 8 Q5 9 Q6 9 Q7 10 Q8 11 Q9 11 Q10 12 Q11 EITHER Q11 OR Q15

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
\(y' = 2kxe^{2x} + 2kx^2e^{2x}\)	B1
\(y'' = 2ke^{2x} + 8kxe^{2x} + 4kx^2e^{2x}\)	B1\(\checkmark\)
\(2ke^{2x}+8kxe^{2x}+4kx^2e^{2x}-8kxe^{2x}-8kx^2e^{2x}+4kx^2e^{2x}=4e^{2x}\) \(\Rightarrow 2k=4 \Rightarrow k=2\)	M1, A1 [4]
\(m^2-4m+4=0 \Rightarrow (m-2)^2=0 \Rightarrow m=2\)	M1
CF: \(y = Ae^{2x}+Bxe^{2x}\)	A1
\(y = Ae^{2x}+Bxe^{2x}+2x^2e^{2x}\)	A1\(\checkmark\) [3]
\(y=3\) when \(x=0 \Rightarrow A=3\)	B1
\(y' = 2Ae^{2x}+B(e^{2x}+2xe^{2x})+(4xe^{2x}+4x^2e^{2x})\)	M1
\(y'=-2\) when \(x=0\) and \(A=3 \Rightarrow B=-8\)	A1
\(\Rightarrow y = 3e^{2x}-8xe^{2x}+2x^2e^{2x}\)	A1\(\checkmark\) [4]