Question 7 - A-Level Maths

CAIE FP1 2018 June — Question 7 10 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2018
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Particular solution with initial conditions
Difficulty	Standard +0.3 This is a standard second-order linear differential equation with constant coefficients requiring the complementary function (repeated root case with auxiliary equation 49m² + 14m + 1 = 0) and particular integral (polynomial form). While it involves multiple steps, the method is entirely routine for Further Maths students: solve auxiliary equation, find PI by inspection, apply initial conditions. The arithmetic is straightforward and no novel insight is required.
Spec	4.10e Second order non-homogeneous: complementary + particular integral

7 Find the particular solution of the differential equation $$49 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 14 \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = 49 x + 735$$ given that when $x = 0 , y = 0$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$.

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Question 7:

Answer	Marks	Guidance
The auxiliary equation is $\left(n+\frac{1}{7}\right)^2=0 \Rightarrow n=-\frac{1}{7}$	M1
CF: $y=(A+Bx)e^{-\frac{1}{7}x}$	A1
PI: $y=px+q$, $y'=p$ and $y''=0$	M1	Correct form of PI and derivatives
$14p+px+q=49x+735 \Rightarrow p=49,\ q=49$	M1 A1	Substitutes in equation correctly
Thus $y=(A+Bx)e^{-\frac{1}{7}x}+49(x+1)$	A1 FT	FT only on correct form of CF AEF
$y=0$ when $x=0$ gives $A+49=0 \Rightarrow A=-49$	B1
$y'=-\frac{1}{7}(A+Bx)e^{-\frac{1}{7}x}+Be^{-\frac{1}{7}x}+49$	M1	Differentiating their $y$
$y'=0$ when $x=0$ gives $-\frac{1}{7}A+B+49=0 \Rightarrow B=-56$	A1	AEF
Thus $y=-(49+56x)e^{-\frac{1}{7}x}+49(x+1)$	A1
Total: 10

## Question 7:

| The auxiliary equation is $\left(n+\frac{1}{7}\right)^2=0 \Rightarrow n=-\frac{1}{7}$ | M1 | |
|---|---|---|
| CF: $y=(A+Bx)e^{-\frac{1}{7}x}$ | A1 | |
| PI: $y=px+q$, $y'=p$ and $y''=0$ | M1 | Correct form of PI and derivatives |
| $14p+px+q=49x+735 \Rightarrow p=49,\ q=49$ | M1 A1 | Substitutes in equation correctly |
| Thus $y=(A+Bx)e^{-\frac{1}{7}x}+49(x+1)$ | A1 FT | FT only on correct form of CF AEF |
| $y=0$ when $x=0$ gives $A+49=0 \Rightarrow A=-49$ | B1 | |
| $y'=-\frac{1}{7}(A+Bx)e^{-\frac{1}{7}x}+Be^{-\frac{1}{7}x}+49$ | M1 | Differentiating their $y$ |
| $y'=0$ when $x=0$ gives $-\frac{1}{7}A+B+49=0 \Rightarrow B=-56$ | A1 | AEF |
| Thus $y=-(49+56x)e^{-\frac{1}{7}x}+49(x+1)$ | A1 | |
| **Total: 10** | | |

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Show LaTeX source

7 Find the particular solution of the differential equation

$$49 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 14 \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = 49 x + 735$$

given that when $x = 0 , y = 0$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$.\\

\hfill \mbox{\textit{CAIE FP1 2018 Q7 [10]}}

This paper (12 questions)

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Q1 5 Q2 6 Q3 8 Q4 8 Q5 8 Q6 9 Q7 10 Q8 10 Q9 10 Q10 12 Q11 EITHER Q11 OR

Answer	Marks	Guidance
The auxiliary equation is \(\left(n+\frac{1}{7}\right)^2=0 \Rightarrow n=-\frac{1}{7}\)	M1
CF: \(y=(A+Bx)e^{-\frac{1}{7}x}\)	A1
PI: \(y=px+q\), \(y'=p\) and \(y''=0\)	M1	Correct form of PI and derivatives
\(14p+px+q=49x+735 \Rightarrow p=49,\ q=49\)	M1 A1	Substitutes in equation correctly
Thus \(y=(A+Bx)e^{-\frac{1}{7}x}+49(x+1)\)	A1 FT	FT only on correct form of CF AEF
\(y=0\) when \(x=0\) gives \(A+49=0 \Rightarrow A=-49\)	B1
\(y'=-\frac{1}{7}(A+Bx)e^{-\frac{1}{7}x}+Be^{-\frac{1}{7}x}+49\)	M1	Differentiating their \(y\)
\(y'=0\) when \(x=0\) gives \(-\frac{1}{7}A+B+49=0 \Rightarrow B=-56\)	A1	AEF
Thus \(y=-(49+56x)e^{-\frac{1}{7}x}+49(x+1)\)	A1
Total: 10