| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| 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 structured Bernoulli equation problem from Further Maths with the substitution provided. Part (a) requires careful differentiation using the chain rule, part (b) is a standard integrating factor application, and part (c) is simple back-substitution. While it requires multiple techniques and careful algebra, the roadmap is completely given and each step follows standard procedures without requiring insight or novel problem-solving. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(z = y^{-2} \Rightarrow y = z^{-\frac{1}{2}}\) | ||
| \(\frac{dy}{dx} = -\frac{1}{2}z^{-\frac{3}{2}}\frac{dz}{dx}\) | M1A1 | 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 \(dy/dx\) |
| \(\frac{dz}{dx} - 4xz = -2xe^{-x^2}\) * | A1cso | Correct completion to printed answer with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dz}{dy} = -2y^{-3}\) | M1A1 | 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 |
| \(\frac{dz}{dx} = -2y^{-3}\frac{dy}{dx}\) | M1A1 | 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 |
| \(I = e^{\int -4x\,dx} = e^{-2x^2}\) | M1A1 | 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:
$$\frac{dy}{dx} + 2xy = xe^{-x^2}y^3$$
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = y^{-2} \Rightarrow y = z^{-\frac{1}{2}}$ | | |
| $\frac{dy}{dx} = -\frac{1}{2}z^{-\frac{3}{2}}\frac{dz}{dx}$ | M1A1 | 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 $dy/dx$ |
| $\frac{dz}{dx} - 4xz = -2xe^{-x^2}$ * | A1cso | Correct completion to printed answer with no errors seen |
**Part (a) Alternative 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dz}{dy} = -2y^{-3}$ | M1A1 | 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 |
**Part (a) Alternative 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dz}{dx} = -2y^{-3}\frac{dy}{dx}$ | M1A1 | 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 |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I = e^{\int -4x\,dx} = e^{-2x^2}$ | M1A1 | 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 |
**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 | $y^2 = \frac{1}{(b)}\left(= \frac{3e^{x^2}}{1+ke^{3x^2}}\right)$ |
---\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 $y ^ { 2 } = \mathrm { f } ( x )$.\\
\hfill \mbox{\textit{Edexcel F2 2015 Q3 [10]}}