AQA Further Paper 2 2020 June — Question 13 10 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2020
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeResonance cases requiring modified PI
DifficultyChallenging +1.2 This is a standard second-order linear differential equation with a resonance case. Part (a) requires substituting Charlotte's attempt to show it yields 0=10e^{-2x} (straightforward algebra). Part (b) requires recognizing that -2 is a root of the auxiliary equation, so the particular integral needs the form Ξ»xe^{-2x} insteadβ€”a well-known technique taught in Further Maths. While it requires understanding of resonance/repeated roots, this is a textbook application rather than novel problem-solving. The 10 marks reflect length rather than exceptional difficulty.
Spec4.10e Second order non-homogeneous: complementary + particular integral
Charlotte is trying to solve this mathematical problem: Find the general solution of the differential equation $$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 10e^{-2x}$$ Charlotte's solution starts as follows: Particular integral: \(y = \lambda e^{-2x}\) so $$\frac{dy}{dx} = -2\lambda e^{-2x}$$ and $$\frac{d^2y}{dx^2} = 4\lambda e^{-2x}$$
  1. Show that Charlotte's method will fail to find a particular integral for the differential equation. [2 marks]
  2. Explain how Charlotte should have started her solution differently and find the general solution of the differential equation. [8 marks]
Question 13:
AnswerMarks
13(a)Evaluates Charlotte’s method
by substituting her particular
integral and its derivatives into
AnswerMarks Guidance
the differential equation.2.3 M1
2
d 𝑦𝑦 d𝑦𝑦 βˆ’2π‘₯π‘₯ βˆ’2π‘₯π‘₯ βˆ’2π‘₯π‘₯
2 + βˆ’2𝑦𝑦 = 4πœ†πœ†e βˆ’2πœ†πœ†e βˆ’2πœ†πœ†e
dπ‘₯π‘₯ dπ‘₯π‘₯
This would make 1=0 equal to zero,
w hich is impossible, sβˆ’o2 π‘₯π‘₯the method fails.
0e
Explains why Charlotte’s
method fails to find a
particular integral for the
AnswerMarks Guidance
differential equation.2.3 E1
AnswerMarks
13(b)Evaluates Charlotte’s method
by explaining that she should
first have found the
complementary function or the
AnswerMarks Guidance
auxiliary equation.2.3 E1
complementary function first.
2 or
π‘šπ‘š +π‘šπ‘šβˆ’2 = 0
Cπ‘šπ‘šF: = 1 π‘šπ‘š = βˆ’2
π‘₯π‘₯ βˆ’2π‘₯π‘₯
The RHS has a𝑦𝑦 s=im𝐴𝐴ilear +fo𝐡𝐡rme to the CF
and so we need to introduce a factor of
into the particular integral.
π‘₯π‘₯ PI:
βˆ’2π‘₯π‘₯
𝑦𝑦 = πœ†πœ†π‘₯π‘₯e
βˆ’2π‘₯π‘₯
𝑦𝑦′ = πœ†πœ†e (βˆ’2π‘₯π‘₯+1)
βˆ’2π‘₯π‘₯
𝑦𝑦′′ = πœ†πœ†e (4π‘₯π‘₯βˆ’4)
βˆ’2π‘₯π‘₯ βˆ’2π‘₯π‘₯
πœ†πœ†e (4π‘₯π‘₯βˆ’4βˆ’2π‘₯π‘₯+1βˆ’2π‘₯π‘₯) = 10e
10
πœ†πœ† = βˆ’
General solution: 3
π‘₯π‘₯ βˆ’2π‘₯π‘₯ 10 βˆ’2π‘₯π‘₯
𝑦𝑦 = 𝐴𝐴e +𝐡𝐡e βˆ’ π‘₯π‘₯e
3
Obtains the auxiliary equation
AnswerMarks Guidance
and its solutions1.1b B1
Writes down the 𝑒𝑒 = βˆ’2,1
complementary function.
FT their solutions of their
AnswerMarks Guidance
auxiliary equation.1.1b B1F
Selects a method to solve the
differential equation by stating
AnswerMarks Guidance
the correct PI3.1a B1
βˆ’2π‘₯π‘₯
Differentiates t𝑦𝑦he=irπœ†πœ† Pπ‘₯π‘₯I𝑒𝑒 twice
(must be different from
AnswerMarks Guidance
Charlotte’s PI)1.1a M1
Obtains correct 1st and 2nd
AnswerMarks Guidance
derivative of the correct PI1.1b A1
Substitutes their PI and its
derivatives into the differential
AnswerMarks Guidance
equation.1.1a M1
Correctly reasons that the
general solution of the
differential equation is
AnswerMarks Guidance
π‘₯π‘₯ βˆ’2π‘₯π‘₯ 10 βˆ’2π‘₯π‘₯2.1 R1
𝑦𝑦 = 𝐴𝐴𝑒𝑒 +𝐡𝐡𝑒𝑒 βˆ’ π‘₯π‘₯𝑒𝑒Total10
QMarking Instructions AO 
Question 13:
--- 13(a) ---
13(a) | Evaluates Charlotte’s method
by substituting her particular
integral and its derivatives into
the differential equation. | 2.3 | M1 | Continuing Charlotte’s method gives
2
d 𝑦𝑦 d𝑦𝑦 βˆ’2π‘₯π‘₯ βˆ’2π‘₯π‘₯ βˆ’2π‘₯π‘₯
2 + βˆ’2𝑦𝑦 = 4πœ†πœ†e βˆ’2πœ†πœ†e βˆ’2πœ†πœ†e
dπ‘₯π‘₯ dπ‘₯π‘₯
This would make 1=0 equal to zero,
w hich is impossible, sβˆ’o2 π‘₯π‘₯the method fails.
0e
Explains why Charlotte’s
method fails to find a
particular integral for the
differential equation. | 2.3 | E1
--- 13(b) ---
13(b) | Evaluates Charlotte’s method
by explaining that she should
first have found the
complementary function or the
auxiliary equation. | 2.3 | E1 | Charlotte needs to find the
complementary function first.
2 or
π‘šπ‘š +π‘šπ‘šβˆ’2 = 0
Cπ‘šπ‘šF: = 1 π‘šπ‘š = βˆ’2
π‘₯π‘₯ βˆ’2π‘₯π‘₯
The RHS has a𝑦𝑦 s=im𝐴𝐴ilear +fo𝐡𝐡rme to the CF
and so we need to introduce a factor of
into the particular integral.
π‘₯π‘₯ PI:
βˆ’2π‘₯π‘₯
𝑦𝑦 = πœ†πœ†π‘₯π‘₯e
βˆ’2π‘₯π‘₯
𝑦𝑦′ = πœ†πœ†e (βˆ’2π‘₯π‘₯+1)
βˆ’2π‘₯π‘₯
𝑦𝑦′′ = πœ†πœ†e (4π‘₯π‘₯βˆ’4)
βˆ’2π‘₯π‘₯ βˆ’2π‘₯π‘₯
πœ†πœ†e (4π‘₯π‘₯βˆ’4βˆ’2π‘₯π‘₯+1βˆ’2π‘₯π‘₯) = 10e
10
πœ†πœ† = βˆ’
General solution: 3
π‘₯π‘₯ βˆ’2π‘₯π‘₯ 10 βˆ’2π‘₯π‘₯
𝑦𝑦 = 𝐴𝐴e +𝐡𝐡e βˆ’ π‘₯π‘₯e
3
Obtains the auxiliary equation
and its solutions | 1.1b | B1
Writes down the 𝑒𝑒 = βˆ’2,1
complementary function.
FT their solutions of their
auxiliary equation. | 1.1b | B1F
Selects a method to solve the
differential equation by stating
the correct PI | 3.1a | B1
βˆ’2π‘₯π‘₯
Differentiates t𝑦𝑦he=irπœ†πœ† Pπ‘₯π‘₯I𝑒𝑒 twice
(must be different from
Charlotte’s PI) | 1.1a | M1
Obtains correct 1st and 2nd
derivative of the correct PI | 1.1b | A1
Substitutes their PI and its
derivatives into the differential
equation. | 1.1a | M1
Correctly reasons that the
general solution of the
differential equation is
π‘₯π‘₯ βˆ’2π‘₯π‘₯ 10 βˆ’2π‘₯π‘₯ | 2.1 | R1
𝑦𝑦 = 𝐴𝐴𝑒𝑒 +𝐡𝐡𝑒𝑒 βˆ’ π‘₯π‘₯𝑒𝑒Total | 10
Q | Marking Instructions | AO | Marks | Typical Solution
Charlotte is trying to solve this mathematical problem:

Find the general solution of the differential equation
$$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 10e^{-2x}$$

Charlotte's solution starts as follows:

Particular integral: $y = \lambda e^{-2x}$

so
$$\frac{dy}{dx} = -2\lambda e^{-2x}$$

and
$$\frac{d^2y}{dx^2} = 4\lambda e^{-2x}$$

\begin{enumerate}[label=(\alph*)]
\item Show that Charlotte's method will fail to find a particular integral for the differential equation.
[2 marks]

\item Explain how Charlotte should have started her solution differently and find the general solution of the differential equation.
[8 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 2 2020 Q13 [10]}}
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