| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Bernoulli equation |
| Difficulty | Challenging +1.2 This is a standard Bernoulli equation problem from Further Maths with the substitution explicitly given. Part (a) requires routine differentiation and algebraic manipulation to verify the transformation. Parts (b) and (c) involve applying the integrating factor method and back-substitution—all mechanical steps following a well-established algorithm. While it's Further Maths content (inherently harder), the question is highly structured with no novel insight required, making it moderately above average difficulty. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.10c Integrating factor: first order equations |
| VIIIV SIHI NI J14M 10N OC | VIIN SIHI NI III HM ION OO | VERV SIHI NI JIIIM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(z = y^{-2} \Rightarrow y = z^{-\frac{1}{2}}\) | ||
| \(\frac{dy}{dx} = -\frac{1}{2}z^{-\frac{3}{2}}\frac{dz}{dx}\) | M1 A1 | M1: \(\frac{dy}{dx} = kz^{-\frac{3}{2}}\frac{dz}{dx}\); A1: Correct differentiation |
| \(-\frac{1}{2}z^{-\frac{3}{2}}\frac{dz}{dx} + \frac{2x}{z^{\frac{1}{2}}} = xe^{-x^2}z^{-\frac{3}{2}}\) | M1 | Substitutes for \(dy/dx\) |
| \(\frac{dz}{dx} - 4xz = -2xe^{-x^2}\) * | A1 cso | Correct completion to printed answer with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dz}{dy} = -2y^{-3}\) | M1 A1 | M1: \(\frac{dz}{dy} = ky^{-3}\); A1: Correct differentiation |
| \(-\frac{1}{2}y^3\frac{dz}{dx} + 2xy = xe^{-x^2}y^3\) | M1 | Substitutes for \(dy/dx\) |
| \(\frac{dz}{dx} - 4xz = -2xe^{-x^2}\) * | A1 | Correct completion to printed answer with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dz}{dx} = -2y^{-3}\frac{dy}{dx}\) | M1 A1 | M1: \(\frac{dz}{dx} = ky^{-3}\frac{dy}{dx}\) inc chain rule; A1: Correct differentiation |
| \(-\frac{1}{2}y^3\frac{dz}{dx} + 2xy = xe^{-x^2}y^3\) | M1 | Substitutes for \(dy/dx\) |
| \(\frac{dz}{dx} - 4xz = -2xe^{-x^2}\) * | A1 | Correct completion to printed answer with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I = e^{\int -4x\,dx} = e^{-2x^2}\) | M1 A1 | M1: \(I = e^{\int \pm 4x\,dx}\); A1: \(e^{-2x^2}\) |
| \(ze^{-2x^2} = \int -2xe^{-3x^2}\,dx\) | dM1 | \(z \times I = \int -2xe^{-x^2} I\,dx\) |
| \(\frac{1}{3}e^{-3x^2}(+c)\) | M1 | \(\int xe^{qx^2}\,dx = pe^{qx^2}(+c)\) |
| \(z = ce^{2x^2} + \frac{1}{3}e^{-x^2}\) | A1 | Or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{y^2} = ce^{2x^2} + \frac{1}{3}e^{-x^2} \Rightarrow y^2 = \frac{1}{ce^{2x^2} + \frac{1}{3}e^{-x^2}}\) | B1ft | \(y^2 = \frac{1}{(b)}\left(= \frac{3e^{x^2}}{1+ke^{3x^2}}\right)\) |
# Question 3:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = y^{-2} \Rightarrow y = z^{-\frac{1}{2}}$ | | |
| $\frac{dy}{dx} = -\frac{1}{2}z^{-\frac{3}{2}}\frac{dz}{dx}$ | M1 A1 | M1: $\frac{dy}{dx} = kz^{-\frac{3}{2}}\frac{dz}{dx}$; A1: Correct differentiation |
| $-\frac{1}{2}z^{-\frac{3}{2}}\frac{dz}{dx} + \frac{2x}{z^{\frac{1}{2}}} = xe^{-x^2}z^{-\frac{3}{2}}$ | M1 | Substitutes for $dy/dx$ |
| $\frac{dz}{dx} - 4xz = -2xe^{-x^2}$ * | A1 cso | Correct completion to printed answer with no errors seen |
**Alternative 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dz}{dy} = -2y^{-3}$ | M1 A1 | M1: $\frac{dz}{dy} = ky^{-3}$; A1: Correct differentiation |
| $-\frac{1}{2}y^3\frac{dz}{dx} + 2xy = xe^{-x^2}y^3$ | M1 | Substitutes for $dy/dx$ |
| $\frac{dz}{dx} - 4xz = -2xe^{-x^2}$ * | A1 | Correct completion to printed answer with no errors seen |
**Alternative 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dz}{dx} = -2y^{-3}\frac{dy}{dx}$ | M1 A1 | M1: $\frac{dz}{dx} = ky^{-3}\frac{dy}{dx}$ inc chain rule; A1: Correct differentiation |
| $-\frac{1}{2}y^3\frac{dz}{dx} + 2xy = xe^{-x^2}y^3$ | M1 | Substitutes for $dy/dx$ |
| $\frac{dz}{dx} - 4xz = -2xe^{-x^2}$ * | A1 | Correct completion to printed answer with no errors seen |
**Total: (4)**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I = e^{\int -4x\,dx} = e^{-2x^2}$ | M1 A1 | M1: $I = e^{\int \pm 4x\,dx}$; A1: $e^{-2x^2}$ |
| $ze^{-2x^2} = \int -2xe^{-3x^2}\,dx$ | dM1 | $z \times I = \int -2xe^{-x^2} I\,dx$ |
| $\frac{1}{3}e^{-3x^2}(+c)$ | M1 | $\int xe^{qx^2}\,dx = pe^{qx^2}(+c)$ |
| $z = ce^{2x^2} + \frac{1}{3}e^{-x^2}$ | A1 | Or equivalent |
**Total: (5)**
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{y^2} = ce^{2x^2} + \frac{1}{3}e^{-x^2} \Rightarrow y^2 = \frac{1}{ce^{2x^2} + \frac{1}{3}e^{-x^2}}$ | B1ft | $y^2 = \frac{1}{(b)}\left(= \frac{3e^{x^2}}{1+ke^{3x^2}}\right)$ |
**Total: (1)**
---\begin{enumerate}
\item (a) Show that the substitution $z = y ^ { - 2 }$ transforms the differential equation
\end{enumerate}
$$\frac { \mathrm { d } y } { \mathrm {~d} x } + 2 x y = x \mathrm { e } ^ { - x ^ { 2 } } y ^ { 3 }$$
into the differential equation
$$\frac { \mathrm { d } z } { \mathrm {~d} x } - 4 x z = - 2 x \mathrm { e } ^ { - x ^ { 2 } }$$
(b) Solve differential equation (II) to find $z$ as a function of $x$.\\
(c) Hence find the general solution of differential equation (I), giving your answer in the form $y ^ { 2 } = \mathrm { f } ( x )$.\\
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VIIIV SIHI NI J14M 10N OC & VIIN SIHI NI III HM ION OO & VERV SIHI NI JIIIM ION OO \\
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\hfill \mbox{\textit{Edexcel F2 2018 Q3 [10]}}