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 homogeneous differential equation where the substitution is explicitly given. The method is routine for Further Maths students: substitute y=vx, separate variables, integrate (likely involving inverse trig functions), then apply initial conditions. While it requires careful algebraic manipulation and integration technique, it follows a well-established procedure without requiring novel insight. The 10 marks reflect length rather than conceptual difficulty.
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]{69c540e1-1dad-45a1-9809-7629d16260e0-10_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
Total	10

## 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 | |
| **Total** | **10** | |

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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]{69c540e1-1dad-45a1-9809-7629d16260e0-10_51_1648_527_246}\\

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

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