Question 7 - A-Level Maths

CAIE FP1 2013 June — Question 7 10 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2013
Session	June
Marks	10
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 Further Maths question on resonance cases in second-order differential equations, requiring students to recognize why the standard PI fails (since e^{-x} is part of the complementary function) and use the modified form λxe^{-x}. While conceptually more advanced than standard A-level, the execution is mechanical: substitute the given form, differentiate twice, match coefficients to find λ, then apply initial conditions. The resonance concept elevates it above average difficulty, but the structured approach and clear guidance make it accessible for FP1 students.
Spec	4.10a General/particular solutions: of differential equations 4.10e Second order non-homogeneous: complementary + particular integral

7 Find the value of the constant $\lambda$ such that $\lambda x \mathrm { e } ^ { - x }$ is a particular integral of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 6 \mathrm { e } ^ { - x }$$ Find the solution of the differential equation for which $y = 2$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 3$ when $x = 0$.

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Question 7:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$y' = \lambda e^{-x} - \lambda x e^{-x}$, $y'' = -2\lambda e^{-x} + \lambda x e^{-x}$	B1B1
$y'' + 5y' + 4y = 3\lambda e^{-x} = 6e^{-x} \Rightarrow \lambda = 2$	M1A1	4 marks
$(m+1)(m+4) = 0 \Rightarrow m = -1$ or $-4$	M1
C.F.: $y = Ae^{-x} + Be^{-4x}$	A1
G.S.: $y = Ae^{-x} + Be^{-4x} + 2xe^{-x}$	A1
$\Rightarrow y' = -Ae^{-x} - 4Be^{-4x} + 2e^{-x} - 2xe^{-x}$	B1$\sqrt{}$
$y(0) \Rightarrow A + B = 2$, $y'(0) \Rightarrow 2 - A - 4B = 3$	M1
$\Rightarrow A = 3$ and $B = -1$
$\Rightarrow y = 3e^{-x} - e^{-4x} + 2xe^{-x}$	A1	6 marks, [10]

# Question 7:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $y' = \lambda e^{-x} - \lambda x e^{-x}$, $y'' = -2\lambda e^{-x} + \lambda x e^{-x}$ | B1B1 | |
| $y'' + 5y' + 4y = 3\lambda e^{-x} = 6e^{-x} \Rightarrow \lambda = 2$ | M1A1 | 4 marks |
| $(m+1)(m+4) = 0 \Rightarrow m = -1$ or $-4$ | M1 | |
| C.F.: $y = Ae^{-x} + Be^{-4x}$ | A1 | |
| G.S.: $y = Ae^{-x} + Be^{-4x} + 2xe^{-x}$ | A1 | |
| $\Rightarrow y' = -Ae^{-x} - 4Be^{-4x} + 2e^{-x} - 2xe^{-x}$ | B1$\sqrt{}$ | |
| $y(0) \Rightarrow A + B = 2$, $y'(0) \Rightarrow 2 - A - 4B = 3$ | M1 | |
| $\Rightarrow A = 3$ and $B = -1$ | |
| $\Rightarrow y = 3e^{-x} - e^{-4x} + 2xe^{-x}$ | A1 | 6 marks, **[10]** |

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7 Find the value of the constant $\lambda$ such that $\lambda x \mathrm { e } ^ { - x }$ is a particular integral of the differential equation

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 6 \mathrm { e } ^ { - x }$$

Find the solution of the differential equation for which $y = 2$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 3$ when $x = 0$.

\hfill \mbox{\textit{CAIE FP1 2013 Q7 [10]}}

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