AQA FP3 2014 June — Question 4 10 marks

Exam BoardAQA
ModuleFP3 (Further Pure Mathematics 3)
Year2014
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeParticular solution with initial conditions
DifficultyChallenging +1.2 This is a standard second-order linear differential equation with constant coefficients requiring complementary function (auxiliary equation with roots 3 and -1), particular integral (trial solution Ae^{-x}), and applying two initial conditions. While it involves multiple steps and Further Maths content, it follows a completely routine algorithmic procedure taught in FP3 with no novel insight required. The asymptotic condition x→∞ adds slight complexity but is a standard technique for determining which constant must be zero.
Spec4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral
4 Solve the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } - 3 y = 2 \mathrm { e } ^ { - x }$$ given that \(y \rightarrow 0\) as \(x \rightarrow \infty\) and that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - 3\) when \(x = 0\).
[0pt] [10 marks] \includegraphics[max width=\textwidth, alt={}, center]{0eb3e96e-528c-4a99-b164-31cc865f0d68-08_46_145_829_159}
Question 4:
Complementary Function:
AnswerMarks Guidance
Auxiliary equation: \(m^2 - 2m - 3 = 0\)M1 Attempt auxiliary equation
\((m-3)(m+1) = 0\), so \(m = 3\) or \(m = -1\)A1
CF: \(y = Ae^{3x} + Be^{-x}\)A1 Both terms required
Particular Integral:
AnswerMarks Guidance
Try \(y = ke^{-x}\)... but \(e^{-x}\) is in CF, so try \(y = kxe^{-x}\)M1 Recognise need for \(kxe^{-x}\)
\(\frac{dy}{dx} = ke^{-x} - kxe^{-x}\)
\(\frac{d^2y}{dx^2} = -ke^{-x} - ke^{-x} + kxe^{-x} = -2ke^{-x} + kxe^{-x}\)
AnswerMarks Guidance
Substituting: \((-2ke^{-x} + kxe^{-x}) - 2(ke^{-x} - kxe^{-x}) - 3kxe^{-x} = 2e^{-x}\)M1 Correct substitution
\(-4ke^{-x} = 2e^{-x}\), so \(k = -\frac{1}{2}\)A1
GS: \(y = Ae^{3x} + Be^{-x} - \frac{1}{2}xe^{-x}\)A1ft
Applying Conditions:
AnswerMarks
\(y \to 0\) as \(x \to \infty \Rightarrow A = 0\)B1
\(y = Be^{-x} - \frac{1}{2}xe^{-x}\)
\(\frac{dy}{dx} = -Be^{-x} - \frac{1}{2}e^{-x} + \frac{1}{2}xe^{-x}\)
AnswerMarks
At \(x = 0\): \(\frac{dy}{dx} = -B - \frac{1}{2} = -3\), so \(B = \frac{5}{2}\)M1 A1
\(y = \frac{5}{2}e^{-x} - \frac{1}{2}xe^{-x}\) 
# Question 4:

**Complementary Function:**

Auxiliary equation: $m^2 - 2m - 3 = 0$ | M1 | Attempt auxiliary equation

$(m-3)(m+1) = 0$, so $m = 3$ or $m = -1$ | A1 |

CF: $y = Ae^{3x} + Be^{-x}$ | A1 | Both terms required

**Particular Integral:**

Try $y = ke^{-x}$... but $e^{-x}$ is in CF, so try $y = kxe^{-x}$ | M1 | Recognise need for $kxe^{-x}$

$\frac{dy}{dx} = ke^{-x} - kxe^{-x}$

$\frac{d^2y}{dx^2} = -ke^{-x} - ke^{-x} + kxe^{-x} = -2ke^{-x} + kxe^{-x}$

Substituting: $(-2ke^{-x} + kxe^{-x}) - 2(ke^{-x} - kxe^{-x}) - 3kxe^{-x} = 2e^{-x}$ | M1 | Correct substitution

$-4ke^{-x} = 2e^{-x}$, so $k = -\frac{1}{2}$ | A1 |

GS: $y = Ae^{3x} + Be^{-x} - \frac{1}{2}xe^{-x}$ | A1ft |

**Applying Conditions:**

$y \to 0$ as $x \to \infty \Rightarrow A = 0$ | B1 |

$y = Be^{-x} - \frac{1}{2}xe^{-x}$

$\frac{dy}{dx} = -Be^{-x} - \frac{1}{2}e^{-x} + \frac{1}{2}xe^{-x}$

At $x = 0$: $\frac{dy}{dx} = -B - \frac{1}{2} = -3$, so $B = \frac{5}{2}$ | M1 A1 |

$y = \frac{5}{2}e^{-x} - \frac{1}{2}xe^{-x}$ |  |

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4 Solve the differential equation

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

given that $y \rightarrow 0$ as $x \rightarrow \infty$ and that $\frac { \mathrm { d } y } { \mathrm {~d} x } = - 3$ when $x = 0$.\\[0pt]
[10 marks]\\

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\hfill \mbox{\textit{AQA FP3 2014 Q4 [10]}}
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Q1 5 Q2 8 Q3 4 Q4 10 Q5 4 Q6 8 Q7 4 Q8 1