Question 7 - A-Level Maths

Edexcel FP2 2002 June — Question 7 14 marks

Exam Board	Edexcel
Module	FP2 (Further Pure Mathematics 2)
Year	2002
Session	June
Marks	14
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 FP2 second-order linear differential equation with constant coefficients. Part (a) requires finding the complementary function (solving the auxiliary equation) and a particular integral (polynomial trial solution), part (b) applies initial conditions to find constants, and part (c) is simple substitution. While it's Further Maths content, it follows a completely routine algorithmic procedure with no novel insight required, making it slightly easier than an average A-level question overall.
Spec	4.10e Second order non-homogeneous: complementary + particular integral

7. (a) Find the general solution of the differential equation $$2 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 7 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 y = 3 t ^ { 2 } + 11 t$$ (b) Find the particular solution of this differential equation for which $y = 1$ and $\frac { \mathrm { d } y } { \mathrm {~d} t } = 1$ when $t = 0$.
(c) For this particular solution, calculate the value of $y$ when $t = 1$.

Show mark scheme Show mark scheme source

Question 7:

Part (a)

Answer	Marks	Guidance
Answer	Marks	Guidance
$2m^2+7m+3=0$, $(2m+1)(m+3)=0$, $m=-\frac{1}{2},-3$
C.F. is $y=Ae^{-\frac{1}{2}t}+Be^{-3t}$	M1, A1
P.I. $y=at^2+bt+c$	B1
$y'=2at+b$, $y''=2a$
$2(2a)+7(2at+b)+3(at^2+bt+c)\equiv 3t^2+11t$	M1
$3a=3$, $a=1$; $14+3b=11$, $b=-1$	A1
$4-7+3c=0$, $c=1$	M1, A1
General solution: $y=Ae^{-\frac{1}{2}t}+Be^{-3t}+(t^2-t+1)$	A1 ft	(8)

Part (b)

Answer	Marks	Guidance
Answer	Marks	Guidance
$y'=-\frac{1}{2}Ae^{-\frac{1}{2}t}-3Be^{-3t}+(2t-1)$	M1
$t=0$, $y'=1$: $1=-1-\frac{1}{2}A-3B$
$t=0$, $y=1$: $1=1+A+B$	M1, A1	one of these
Solve: $A+B=0$, $A+6B=-4$
$A=\frac{4}{5}$, $B=-\frac{4}{5}$	M1
$y=(t^2-t+1)+\frac{4}{5}(e^{-\frac{1}{2}t}-e^{-3t})$	A1	(5)

Part (c)

Answer	Marks	Guidance
Answer	Marks	Guidance
$t=1$: $y=\frac{4}{5}(e^{-\frac{1}{2}}-e^{-3})+1\quad (=1.445...)$	B1	(1)
(14 marks)

# Question 7:

## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2m^2+7m+3=0$, $(2m+1)(m+3)=0$, $m=-\frac{1}{2},-3$ | | |
| C.F. is $y=Ae^{-\frac{1}{2}t}+Be^{-3t}$ | M1, A1 | |
| P.I. $y=at^2+bt+c$ | B1 | |
| $y'=2at+b$, $y''=2a$ | | |
| $2(2a)+7(2at+b)+3(at^2+bt+c)\equiv 3t^2+11t$ | M1 | |
| $3a=3$, $a=1$; $14+3b=11$, $b=-1$ | A1 | |
| $4-7+3c=0$, $c=1$ | M1, A1 | |
| General solution: $y=Ae^{-\frac{1}{2}t}+Be^{-3t}+(t^2-t+1)$ | A1 ft | **(8)** |

## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y'=-\frac{1}{2}Ae^{-\frac{1}{2}t}-3Be^{-3t}+(2t-1)$ | M1 | |
| $t=0$, $y'=1$: $1=-1-\frac{1}{2}A-3B$ | | |
| $t=0$, $y=1$: $1=1+A+B$ | M1, A1 | one of these |
| Solve: $A+B=0$, $A+6B=-4$ | | |
| $A=\frac{4}{5}$, $B=-\frac{4}{5}$ | M1 | |
| $y=(t^2-t+1)+\frac{4}{5}(e^{-\frac{1}{2}t}-e^{-3t})$ | A1 | **(5)** |

## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $t=1$: $y=\frac{4}{5}(e^{-\frac{1}{2}}-e^{-3})+1\quad (=1.445...)$ | B1 | **(1)** |
| | | **(14 marks)** |

---

Show LaTeX source

7. (a) Find the general solution of the differential equation

$$2 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 7 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 y = 3 t ^ { 2 } + 11 t$$

(b) Find the particular solution of this differential equation for which $y = 1$ and $\frac { \mathrm { d } y } { \mathrm {~d} t } = 1$ when $t = 0$.\\
(c) For this particular solution, calculate the value of $y$ when $t = 1$.\\

\hfill \mbox{\textit{Edexcel FP2 2002 Q7 [14]}}

This paper (10 questions)

View full paper

Q1 5 Q2 10 Q3 13 Q4 18 Q5 7 Q6 11 Q7 14 Q8 15 Q9 7 Q10 12

Answer	Marks	Guidance
Answer	Marks	Guidance
\(2m^2+7m+3=0\), \((2m+1)(m+3)=0\), \(m=-\frac{1}{2},-3\)
C.F. is \(y=Ae^{-\frac{1}{2}t}+Be^{-3t}\)	M1, A1
P.I. \(y=at^2+bt+c\)	B1
\(y'=2at+b\), \(y''=2a\)
\(2(2a)+7(2at+b)+3(at^2+bt+c)\equiv 3t^2+11t\)	M1
\(3a=3\), \(a=1\); \(14+3b=11\), \(b=-1\)	A1
\(4-7+3c=0\), \(c=1\)	M1, A1
General solution: \(y=Ae^{-\frac{1}{2}t}+Be^{-3t}+(t^2-t+1)\)	A1 ft	(8)

Answer	Marks	Guidance
Answer	Marks	Guidance
\(y'=-\frac{1}{2}Ae^{-\frac{1}{2}t}-3Be^{-3t}+(2t-1)\)	M1
\(t=0\), \(y'=1\): \(1=-1-\frac{1}{2}A-3B\)
\(t=0\), \(y=1\): \(1=1+A+B\)	M1, A1	one of these
Solve: \(A+B=0\), \(A+6B=-4\)
\(A=\frac{4}{5}\), \(B=-\frac{4}{5}\)	M1
\(y=(t^2-t+1)+\frac{4}{5}(e^{-\frac{1}{2}t}-e^{-3t})\)	A1	(5)