Challenging +1.2 This is a standard Bernoulli equation problem from Further Maths FP2. Part (a) requires straightforward substitution and differentiation using the chain rule, while part (b) involves solving a linear first-order ODE using integrating factor—both are textbook techniques. However, it's harder than average A-level due to being Further Maths content and requiring careful algebraic manipulation across multiple steps.
7. (a) Show that the substitution \(v = y ^ { - 3 }\) transforms the differential equation
$$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = 2 x ^ { 4 } y ^ { 4 }$$
into the differential equation
$$\begin{aligned}
& \frac { \mathrm { d } v } { \mathrm {~d} x } - \frac { 3 v } { x } = - 6 x ^ { 3 } \\
& \text { ration (II), find a general solution of differential equation (I) }
\end{aligned}$$
in the form \(y ^ { 3 } = \mathrm { f } ( x )\).
dM1: Substitutes correctly for \(\frac{dv}{dx}\) and \(v\) in equation (II) to obtain a D.E. in terms of \(x\) and \(y\) only; A1: Correct completion to obtain equation (I) with no errors seen
M1: \(e^{\int \pm\frac{3}{x}dx}\) and attempt integration; if not correct, \(\ln x\) must be seen; A1: \(\frac{1}{x^3}\)
\(\frac{v}{x^3} = \int -6\,dx = -6x(+c)\)
dM1A1
M1: \(v \times\) their \(I = \int -6x^3 \times\) their \(I\,dx\); A1: Correct equation with or without \(+c\)
\(\frac{1}{y^3 x^3} = -6x + c \Rightarrow y^3 = \ldots\)
ddM1
Include the constant, substitute for \(y\), and attempt to rearrange to \(y^3 = \ldots\) or \(y = \ldots\) with constant treated correctly; dependent on both M marks of (b)
\(y^3 = \frac{1}{cx^3-6x^4}\)
A1
Or equivalent
Total: 11 marks
## Question 7(a):
**Way 1:**
| Working | Mark | Guidance |
|---------|------|----------|
| $v = y^{-3} \Rightarrow \frac{dv}{dy} = -3y^{-4}$ | B1 | Correct derivative |
| $\frac{dy}{dx} = \frac{dy}{dv}\cdot\frac{dv}{dx} = -\frac{y^4}{3}\frac{dv}{dx}$ or $-3y^{-4}\frac{dy}{dx}x - 3y^{-3} = -6x^4$ | M1A1 | M1: Correct use of chain rule; A1: Correct equation |
| $-\frac{y^4}{3}\frac{dv}{dx}x + y = 2x^4y^4 \Rightarrow \frac{dv}{dx} - \frac{3v}{x} = -6x^3$ | dM1A1 | dM1: Substitutes to obtain equation in $v$ and $x$; A1: Correct completion with no errors seen |
**Way 2:**
| Working | Mark | Guidance |
|---------|------|----------|
| $y = v^{-\frac{1}{3}} \Rightarrow \frac{dy}{dv} = -\frac{1}{3}v^{-\frac{4}{3}}$ | B1 | Correct derivative |
| $\frac{dy}{dx} = \frac{dy}{dv}\cdot\frac{dv}{dx} = -\frac{1}{3}v^{-\frac{4}{3}}\frac{dv}{dx}$ | M1A1 | M1: Correct use of chain rule; A1: Correct equation |
| $-\frac{v^{-\frac{4}{3}}}{3}\frac{dv}{dx}x + v^{-\frac{1}{3}} = 2x^4v^{-\frac{4}{3}}$ | dM1 | dM1: Substitutes to obtain equation in $v$ and $x$ |
| $\frac{dv}{dx} - \frac{3v}{x} = -6x^3$ | A1 | Correct completion with no errors seen |
**Way 3 (Working in reverse):**
| Working | Mark | Guidance |
|---------|------|----------|
| $v = y^{-3} \Rightarrow \frac{dv}{dy} = -3y^{-4}$ | B1 | Correct derivative |
| $\frac{dv}{dx} = \frac{dv}{dy}\cdot\frac{dy}{dx} = -3y^{-4}\frac{dy}{dx}$ | M1A1 | M1: Correct use of chain rule; A1: Correct expression for $\frac{dv}{dx}$ |
| $-3y^{-4}\frac{dy}{dx} - \frac{3y^{-3}}{x} = -6x^3$ | dM1A1 | dM1: Substitutes correctly for $\frac{dv}{dx}$ and $v$ in equation (II) to obtain a D.E. in terms of $x$ and $y$ only; A1: Correct completion to obtain equation (I) with no errors seen |
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## Question 7(b):
| Working | Mark | Guidance |
|---------|------|----------|
| $I = e^{\int -\frac{3}{x}dx} = e^{-3\ln x} = \frac{1}{x^3}$ | M1A1 | M1: $e^{\int \pm\frac{3}{x}dx}$ and attempt integration; if not correct, $\ln x$ must be seen; A1: $\frac{1}{x^3}$ |
| $\frac{v}{x^3} = \int -6\,dx = -6x(+c)$ | dM1A1 | M1: $v \times$ their $I = \int -6x^3 \times$ their $I\,dx$; A1: Correct equation with or without $+c$ |
| $\frac{1}{y^3 x^3} = -6x + c \Rightarrow y^3 = \ldots$ | ddM1 | Include the constant, substitute for $y$, and attempt to rearrange to $y^3 = \ldots$ or $y = \ldots$ with constant treated correctly; dependent on both M marks of (b) |
| $y^3 = \frac{1}{cx^3-6x^4}$ | A1 | Or equivalent |
**Total: 11 marks**
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7. (a) Show that the substitution $v = y ^ { - 3 }$ transforms the differential equation
$$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = 2 x ^ { 4 } y ^ { 4 }$$
into the differential equation
$$\begin{aligned}
& \frac { \mathrm { d } v } { \mathrm {~d} x } - \frac { 3 v } { x } = - 6 x ^ { 3 } \\
& \text { ration (II), find a general solution of differential equation (I) }
\end{aligned}$$
in the form $y ^ { 3 } = \mathrm { f } ( x )$.\\
\hfill \mbox{\textit{Edexcel FP2 2014 Q7 [11]}}