Question 8 - A-Level Maths

Edexcel P4 2021 June — Question 8 9 marks

Exam Board	Edexcel
Module	P4 (Pure Mathematics 4)
Year	2021
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Differential equations
Type	Separable variables - standard (requires integration by parts)
Difficulty	Standard +0.3 This is a straightforward separable variables question requiring standard techniques: separate variables, integrate both sides (using integration by parts for the right side), apply initial condition, and rearrange. Part (b) requires finding a limit as x→∞, which is routine. While it involves multiple steps and integration by parts, it follows a completely standard template with no novel insight required, making it slightly easier than average.
Spec	1.08i Integration by parts 1.08k Separable differential equations: dy/dx = f(x)g(y)

8. (a) Given that $y = 1$ at $x = 0$, solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 x y ^ { \frac { 1 } { 3 } } } { \mathrm { e } ^ { 2 x } } \quad y \geqslant 0$$ giving your answer in the form $y ^ { 2 } = \mathrm { g } ( x )$.
(b) Hence find the equation of the horizontal asymptote to the curve with equation $y ^ { 2 } = \mathrm { g } ( x )$. \includegraphics[max width=\textwidth, alt={}, center]{960fe82f-c180-422c-b409-a5cdc5fae924-27_2644_1840_118_111} \includegraphics[max width=\textwidth, alt={}, center]{960fe82f-c180-422c-b409-a5cdc5fae924-29_2646_1838_121_116}

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Question 8:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\int y^{-1/3}dy = \int 6xe^{-2x}dx$	B1	Separates variables; $dx$ and $dy$ must be present in correct positions
$\frac{3}{2}y^{2/3} = -3xe^{-2x} + \int 3e^{-2x}dx$	M1 M1	M1: integrates LHS $y^{-1/3}\to y^{2/3}$; M1: integration by parts RHS correct direction
$\frac{3}{2}y^{2/3} = -3xe^{-2x} - \frac{3}{2}e^{-2x} + c$	dM1 A1	dM1: fully integrates RHS to $...xe^{-2x}\pm...e^{-2x}$; A1: correct integration
Substitutes $(0,1)\Rightarrow c=3$	M1	Must have $+c$; dependent on at least one integration M being awarded
$y^2 = \left(-2xe^{-2x}-e^{-2x}+2\right)^3$	A1	CSO

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
As $x\to\infty$, $e^{-2x}\to 0$ and $y^2=(2)^3 \Rightarrow y=2^{3/2}$	M1 A1	M1: for $e^{-2x}\to 0$, giving $y^2=("2")^3$; A1: $y=2^{3/2}$ o.e. such as $y=\sqrt{8}$; condone $y=\pm 2^{3/2}$

# Question 8:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int y^{-1/3}dy = \int 6xe^{-2x}dx$ | B1 | Separates variables; $dx$ and $dy$ must be present in correct positions |
| $\frac{3}{2}y^{2/3} = -3xe^{-2x} + \int 3e^{-2x}dx$ | M1 M1 | M1: integrates LHS $y^{-1/3}\to y^{2/3}$; M1: integration by parts RHS correct direction |
| $\frac{3}{2}y^{2/3} = -3xe^{-2x} - \frac{3}{2}e^{-2x} + c$ | dM1 A1 | dM1: fully integrates RHS to $...xe^{-2x}\pm...e^{-2x}$; A1: correct integration |
| Substitutes $(0,1)\Rightarrow c=3$ | M1 | Must have $+c$; dependent on at least one integration M being awarded |
| $y^2 = \left(-2xe^{-2x}-e^{-2x}+2\right)^3$ | A1 | CSO |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| As $x\to\infty$, $e^{-2x}\to 0$ and $y^2=(2)^3 \Rightarrow y=2^{3/2}$ | M1 A1 | M1: for $e^{-2x}\to 0$, giving $y^2=("2")^3$; A1: $y=2^{3/2}$ o.e. such as $y=\sqrt{8}$; condone $y=\pm 2^{3/2}$ |

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8. (a) Given that $y = 1$ at $x = 0$, solve the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 x y ^ { \frac { 1 } { 3 } } } { \mathrm { e } ^ { 2 x } } \quad y \geqslant 0$$

giving your answer in the form $y ^ { 2 } = \mathrm { g } ( x )$.\\
(b) Hence find the equation of the horizontal asymptote to the curve with equation $y ^ { 2 } = \mathrm { g } ( x )$.\\

\includegraphics[max width=\textwidth, alt={}, center]{960fe82f-c180-422c-b409-a5cdc5fae924-27_2644_1840_118_111}

\includegraphics[max width=\textwidth, alt={}, center]{960fe82f-c180-422c-b409-a5cdc5fae924-29_2646_1838_121_116}\\

\hfill \mbox{\textit{Edexcel P4 2021 Q8 [9]}}

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