Question 6 - A-Level Maths

CAIE Further Paper 2 2022 June — Question 6 10 marks

Exam Board	CAIE
Module	Further Paper 2 (Further Paper 2)
Year	2022
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	First order differential equations (integrating factor)
Type	Homogeneous equation (y = vx substitution)
Difficulty	Challenging +1.2 This is a standard Further Maths homogeneous differential equation with the substitution explicitly given. While it requires multiple steps (substitution, separation of variables, integration using trigonometric substitution, and applying initial conditions), these are all routine techniques for Further Maths students. The algebraic manipulation is moderately involved but follows a well-practiced method, making it above average difficulty but not exceptionally challenging.
Spec	1.08k Separable differential equations: dy/dx = f(x)g(y)4.10a General/particular solutions: of differential equations

6 Use the substitution $y = v x$ to find the solution of the differential equation $$x \frac { d y } { d x } = y + \sqrt { 9 x ^ { 2 } + y ^ { 2 } }$$ for which $y = 0$ when $x = 1$. Give your answer in the form $\mathrm { y } = \mathrm { f } ( \mathrm { x } )$, where $\mathrm { f } ( x )$ is a polynomial in $x$. [10] \includegraphics[max width=\textwidth, alt={}, center]{114ece0d-558d-4c02-8a77-034b3681cff9-10_51_1648_527_246}

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Question 6:

Answer	Marks	Guidance
Answer	Marks	Guidance
$\frac{dy}{dx} = v + x\frac{dv}{dx}$	B1
$x\left(v + x\frac{dv}{dx}\right) = vx + \sqrt{9x^2 + x^2v^2}$	M1	Substitutes and derives first order separable equation
$x\frac{dv}{dx} = \sqrt{v^2 + 9}$	A1
$\int \frac{1}{\sqrt{v^2+9}}\,dv = \int\frac{1}{x}\,dx$	M1	Separates variables and integrates both sides
$\sinh^{-1}\left(\frac{v}{3}\right) = \ln x + C$	A1
$C = 0$	B1	Substitutes initial conditions
$\ln\left(\frac{v}{3} + \sqrt{\frac{v^2}{9}+1}\right) = \ln x$	M1	Converts to logarithmic form
$y + \sqrt{y^2 + 9x^2} = 3x^2$	A1
$9x^4 - 6x^2y + y^2 - y^2 - 9x^2 = 0$	M1	Squares to eliminate radical
$y = \frac{3}{2}(x^2 - 1)$	A1

## Question 6:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = v + x\frac{dv}{dx}$ | B1 | |
| $x\left(v + x\frac{dv}{dx}\right) = vx + \sqrt{9x^2 + x^2v^2}$ | M1 | Substitutes and derives first order separable equation |
| $x\frac{dv}{dx} = \sqrt{v^2 + 9}$ | A1 | |
| $\int \frac{1}{\sqrt{v^2+9}}\,dv = \int\frac{1}{x}\,dx$ | M1 | Separates variables and integrates both sides |
| $\sinh^{-1}\left(\frac{v}{3}\right) = \ln x + C$ | A1 | |
| $C = 0$ | B1 | Substitutes initial conditions |
| $\ln\left(\frac{v}{3} + \sqrt{\frac{v^2}{9}+1}\right) = \ln x$ | M1 | Converts to logarithmic form |
| $y + \sqrt{y^2 + 9x^2} = 3x^2$ | A1 | |
| $9x^4 - 6x^2y + y^2 - y^2 - 9x^2 = 0$ | M1 | Squares to eliminate radical |
| $y = \frac{3}{2}(x^2 - 1)$ | A1 | |

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6 Use the substitution $y = v x$ to find the solution of the differential equation

$$x \frac { d y } { d x } = y + \sqrt { 9 x ^ { 2 } + y ^ { 2 } }$$

for which $y = 0$ when $x = 1$. Give your answer in the form $\mathrm { y } = \mathrm { f } ( \mathrm { x } )$, where $\mathrm { f } ( x )$ is a polynomial in $x$. [10]\\
\includegraphics[max width=\textwidth, alt={}, center]{114ece0d-558d-4c02-8a77-034b3681cff9-10_51_1648_527_246}\\

\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q6 [10]}}

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