AQA Paper 2 2019 June — Question 5 7 marks

Exam BoardAQA
ModulePaper 2 (Paper 2)
Year2019
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferential equations
TypeSeparable variables - standard (polynomial/exponential x-side)
DifficultyStandard +0.3 This is a separable differential equation requiring standard technique: separate variables, integrate both sides (using substitution for ln(x)/x²), apply initial condition, and rearrange to the required form. Slightly above average due to the integration requiring substitution and careful algebraic manipulation, but follows a well-practiced A-level procedure with no novel insight required.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)

Solve the differential equation $$\frac{dt}{dx} = \frac{\ln x}{x^2 t} \quad \text{for } x > 0$$ given \(x = 1\) when \(t = 2\) Write your answer in the form \(t^2 = f(x)\) [7 marks]

Question 5:
AnswerMarks
5Separates the variables – one
side correct
Condone missing integral signs
AnswerMarks Guidance
PI by correct integration3.1a M1
∫ ∫
ln xdx = tdt
x2
t2
tdt = + c
2
u = ln x
1
u′ =
x
v′ = x−2
v = −x−1
− 1 ln x − ∫ 1 ( −x−1 ) dx
x x
1 1
− ln x + dx
x x2
1 1
− ln x −
x x
1 1 t2
− ln x − = + c
x x 2
t = 2,x =1⇒ −1= 2+ c
c = −3
1+ln x 
t2 = 6−2
 x 
AnswerMarks Guidance
Integrates their tdtcorrectly1.1b A1F
1 1
Obtains u′= and v =− OE
AnswerMarks Guidance
x x1.1b B1
∫ 1
Integrates lnxdx
x2
Substitutes their u, u’, v and v’
into the correct formula for
integration by parts
AnswerMarks Guidance
Condone sign errors in formula1.1a M1
Obtains
1 1
− lnx−
AnswerMarks Guidance
x x1.1b A1
Substitutes t = 2 and x = 1 into
their integrated equation to find
AnswerMarks Guidance
their +c1.1a M1
Obtains correct solution must
t2
have = ….
AnswerMarks Guidance
ACF2.5 A1
Total7
QMarking instructions AO
Question 5:
5 | Separates the variables – one
side correct
Condone missing integral signs
PI by correct integration | 3.1a | M1 | 1
∫ ∫
ln xdx = tdt
x2
t2
∫
tdt = + c
2
u = ln x
1
u′ =
x
v′ = x−2
v = −x−1
− 1 ln x − ∫ 1 ( −x−1 ) dx
x x
1 1
∫
− ln x + dx
x x2
1 1
− ln x −
x x
1 1 t2
− ln x − = + c
x x 2
t = 2,x =1⇒ −1= 2+ c
c = −3
1+ln x 
t2 = 6−2

 x 
∫
Integrates their tdtcorrectly | 1.1b | A1F
1 1
Obtains u′= and v =− OE
x x | 1.1b | B1
∫ 1
Integrates lnxdx
x2
Substitutes their u, u’, v and v’
into the correct formula for
integration by parts
Condone sign errors in formula | 1.1a | M1
Obtains
1 1
− lnx−
x x | 1.1b | A1
Substitutes t = 2 and x = 1 into
their integrated equation to find
their +c | 1.1a | M1
Obtains correct solution must
t2
have = ….
ACF | 2.5 | A1
Total | 7
Q | Marking instructions | AO | Mark | Typical solution
Solve the differential equation
$$\frac{dt}{dx} = \frac{\ln x}{x^2 t} \quad \text{for } x > 0$$

given $x = 1$ when $t = 2$

Write your answer in the form $t^2 = f(x)$
[7 marks]

\hfill \mbox{\textit{AQA Paper 2 2019 Q5 [7]}}