Question 7 - A-Level Maths

Edexcel FP2 2013 June — Question 7 7 marks

Exam Board	Edexcel
Module	FP2 (Further Pure Mathematics 2)
Year	2013
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Resonance cases requiring modified PI
Difficulty	Standard +0.8 This is a resonance case in second-order DEs where the standard PI form fails because e^(2x) is part of the complementary function. Students must recognize this and use the modified form λxe^(2x), then differentiate twice and substitute. While systematic, it requires conceptual understanding of resonance and careful algebraic manipulation, placing it moderately above average difficulty.
Spec	4.10d Second order homogeneous: auxiliary equation method 4.10e Second order non-homogeneous: complementary + particular integral

7. (a) Find the value of the constant $\lambda$ for which $y = \lambda x \mathrm { e } ^ { 2 x }$ is a particular integral of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 y = 6 \mathrm { e } ^ { 2 x }$$ (b) Hence, or otherwise, find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 y = 6 \mathrm { e } ^ { 2 x }$$

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Question 7:

Part (a):

Answer	Marks	Guidance
Differentiate twice obtaining $\frac{dy}{dx} = 2\lambda xe^{2x} + \lambda e^{2x}$ and $\frac{d^2y}{dx^2} = 4\lambda xe^{2x} + 2\lambda e^{2x} + 2\lambda e^{2x}$	M1 A1
Substitute to give $\lambda = \frac{3}{2}$	M1 A1	(4 marks)

Part (b):

Answer	Marks	Guidance
Complementary function is $y = Ae^{2x} + Be^{-2x}$	M1 A1
So general solution is $y = Ae^{2x} + Be^{-2x} + \frac{3}{2}xe^{2x}$	A1	(3 marks, 7 total)

## Question 7:

### Part (a):
| Differentiate twice obtaining $\frac{dy}{dx} = 2\lambda xe^{2x} + \lambda e^{2x}$ and $\frac{d^2y}{dx^2} = 4\lambda xe^{2x} + 2\lambda e^{2x} + 2\lambda e^{2x}$ | M1 A1 | |
| Substitute to give $\lambda = \frac{3}{2}$ | M1 A1 | (4 marks) |

### Part (b):
| Complementary function is $y = Ae^{2x} + Be^{-2x}$ | M1 A1 | |
| So general solution is $y = Ae^{2x} + Be^{-2x} + \frac{3}{2}xe^{2x}$ | A1 | (3 marks, 7 total) |

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7. (a) Find the value of the constant $\lambda$ for which $y = \lambda x \mathrm { e } ^ { 2 x }$ is a particular integral of the differential equation

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

(b) Hence, or otherwise, find the general solution of the differential equation

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

\hfill \mbox{\textit{Edexcel FP2 2013 Q7 [7]}}

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Q1 7 Q2 9 Q3 9 Q4 9 Q5 7 Q6 8 Q7 7 Q8 8 Q9 11

Answer	Marks	Guidance
Differentiate twice obtaining \(\frac{dy}{dx} = 2\lambda xe^{2x} + \lambda e^{2x}\) and \(\frac{d^2y}{dx^2} = 4\lambda xe^{2x} + 2\lambda e^{2x} + 2\lambda e^{2x}\)	M1 A1
Substitute to give \(\lambda = \frac{3}{2}\)	M1 A1	(4 marks)

Answer	Marks	Guidance
Complementary function is \(y = Ae^{2x} + Be^{-2x}\)	M1 A1
So general solution is \(y = Ae^{2x} + Be^{-2x} + \frac{3}{2}xe^{2x}\)	A1	(3 marks, 7 total)