| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Euler-Cauchy equations via exponential substitution |
| Difficulty | Challenging +1.2 This is a standard Euler-Cauchy equation problem from FP2 requiring systematic application of the exponential substitution x=e^u, solving a constant-coefficient equation, then back-substituting. Part (a) is guided verification using chain rule, parts (b) and (c) follow standard procedures. While requiring multiple techniques and careful algebra, this is a textbook example with no novel insight needed, making it moderately above average difficulty for Further Maths students. |
| Spec | 4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x = e^u \Rightarrow \frac{dx}{du} = e^u\) or \(\frac{du}{dx} = e^{-u}\) or \(\frac{dx}{du} = x\) or \(\frac{du}{dx} = \frac{1}{x}\) | B1 | For \(\frac{dx}{du} = e^u\) seen explicitly or used |
| \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = e^{-u}\frac{dy}{du}\) | M1 | Obtaining \(\frac{dy}{dx}\) using chain rule here or seen later |
| \(\frac{d^2y}{dx^2} = -e^{-u}\frac{du}{dx}\frac{dy}{du} + e^{-u}\frac{d^2y}{du^2}\frac{du}{dx} = e^{-2u}\left(-\frac{dy}{du} + \frac{d^2y}{du^2}\right)\) | M1A1 | Obtaining \(\frac{d^2y}{dx^2}\) using product rule (penalise lack of chain rule by the A mark); correct expression in any equivalent form |
| \(e^{2u} \times e^{-2u}\left(-\frac{dy}{du} + \frac{d^2y}{du^2}\right) - 7e^u \times e^{-u}\frac{dy}{du} + 16y = 2\ln(e^u)\) | dM1 | Substituting to eliminate \(x\); only \(u\) and \(y\) present; depends on 2nd M mark |
| \(\frac{d^2y}{du^2} - 8\frac{dy}{du} + 16y = 2u\) ★ | A1cso | Obtaining given result from completely correct work |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x = e^u\), \(\frac{dx}{du} = e^u = x\) | B1 | As above |
| \(\frac{dy}{du} = \frac{dy}{dx} \times \frac{dx}{du} = x\frac{dy}{dx}\) | M1 | Obtaining \(\frac{dy}{du}\) using chain rule |
| \(\frac{d^2y}{du^2} = 1\cdot\frac{dx}{du}\cdot\frac{dy}{dx} + x\frac{d^2y}{dx^2}\cdot\frac{dx}{du} = x\frac{dy}{dx} + x^2\frac{d^2y}{dx^2}\) | M1A1 | Using product rule; correct expression |
| \(x^2\frac{d^2y}{dx^2} = \frac{d^2y}{du^2} - \frac{dy}{du}\) | — | — |
| \(\left(\frac{d^2y}{du^2} - \frac{dy}{du}\right) - 7x \times \frac{1}{x}\frac{dy}{du} + 16y = 2\ln(e^u)\) | — | — |
| \(\frac{d^2y}{du^2} - 8\frac{dy}{du} + 16y = 2u\) ★ | dM1A1cso | — |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(u = \ln x\), \(\frac{du}{dx} = \frac{1}{x}\) | B1 | — |
| \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = \frac{1}{x}\frac{dy}{du}\) | M1 | Chain rule |
| \(\frac{d^2y}{dx^2} = -\frac{1}{x^2}\frac{dy}{du} + \frac{1}{x}\frac{d^2y}{du^2}\cdot\frac{du}{dx} = -\frac{1}{x^2}\frac{dy}{du} + \frac{1}{x^2}\frac{d^2y}{du^2}\) | M1A1 | Product rule; correct expression |
| \(x^2\left(-\frac{1}{x^2}\frac{dy}{du} + \frac{1}{x^2}\frac{d^2y}{du^2}\right) - 7x \times \frac{1}{x}\frac{dy}{du} + 16y = 2u\) | — | — |
| \(\frac{d^2y}{du^2} - 8\frac{dy}{du} + 16y = 2u\) ★ | dM1A1cso | — |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(m^2 - 8m + 16 = 0\) | — | — |
| \((m-4)^2 = 0\), \(m = 4\) | M1A1 | Writing correct auxiliary equation and solving; \(m=4\) correct |
| CF \(= (A + Bu)e^{4u}\) | A1 | Correct CF; can use any single variable |
| PI: try \(y = au + b\); \(\frac{dy}{du} = a\), \(\frac{d^2y}{du^2} = 0\) | M1 | Using appropriate PI and finding \(\frac{dy}{du}\) and \(\frac{d^2y}{du^2}\); note \(y = \lambda u\) scores M0 |
| \(0 - 8a + 16(au+b) = 2u\) | — | — |
| \(a = \frac{1}{8}\), \(b = \frac{1}{16}\) | dM1A1 | Substituting to find unknowns; correct values (decimals: 0.125 and 0.0625); depends on second M1 |
| \(y = (A+Bu)e^{4u} + \frac{1}{8}u + \frac{1}{16}\) | B1ft | Complete solution following through their CF and PI; must have \(y\) as function of \(u\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(y = (A + B\ln x)x^4 + \frac{1}{8}\ln x + \frac{1}{16}\) | B1 | Reverse substitution to obtain correct \(y\) in terms of \(x\); \(x^4\) or \(e^{4\ln x}\) allowed; must start \(y = \ldots\); no follow-through |
## Question 8:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x = e^u \Rightarrow \frac{dx}{du} = e^u$ or $\frac{du}{dx} = e^{-u}$ or $\frac{dx}{du} = x$ or $\frac{du}{dx} = \frac{1}{x}$ | B1 | For $\frac{dx}{du} = e^u$ seen explicitly or used |
| $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = e^{-u}\frac{dy}{du}$ | M1 | Obtaining $\frac{dy}{dx}$ using chain rule here or seen later |
| $\frac{d^2y}{dx^2} = -e^{-u}\frac{du}{dx}\frac{dy}{du} + e^{-u}\frac{d^2y}{du^2}\frac{du}{dx} = e^{-2u}\left(-\frac{dy}{du} + \frac{d^2y}{du^2}\right)$ | M1A1 | Obtaining $\frac{d^2y}{dx^2}$ using product rule (penalise lack of chain rule by the A mark); correct expression in any equivalent form |
| $e^{2u} \times e^{-2u}\left(-\frac{dy}{du} + \frac{d^2y}{du^2}\right) - 7e^u \times e^{-u}\frac{dy}{du} + 16y = 2\ln(e^u)$ | dM1 | Substituting to eliminate $x$; only $u$ and $y$ present; depends on 2nd M mark |
| $\frac{d^2y}{du^2} - 8\frac{dy}{du} + 16y = 2u$ ★ | A1cso | Obtaining given result from completely correct work |
**Alternative 1:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x = e^u$, $\frac{dx}{du} = e^u = x$ | B1 | As above |
| $\frac{dy}{du} = \frac{dy}{dx} \times \frac{dx}{du} = x\frac{dy}{dx}$ | M1 | Obtaining $\frac{dy}{du}$ using chain rule |
| $\frac{d^2y}{du^2} = 1\cdot\frac{dx}{du}\cdot\frac{dy}{dx} + x\frac{d^2y}{dx^2}\cdot\frac{dx}{du} = x\frac{dy}{dx} + x^2\frac{d^2y}{dx^2}$ | M1A1 | Using product rule; correct expression |
| $x^2\frac{d^2y}{dx^2} = \frac{d^2y}{du^2} - \frac{dy}{du}$ | — | — |
| $\left(\frac{d^2y}{du^2} - \frac{dy}{du}\right) - 7x \times \frac{1}{x}\frac{dy}{du} + 16y = 2\ln(e^u)$ | — | — |
| $\frac{d^2y}{du^2} - 8\frac{dy}{du} + 16y = 2u$ ★ | dM1A1cso | — |
**Alternative 2:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $u = \ln x$, $\frac{du}{dx} = \frac{1}{x}$ | B1 | — |
| $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = \frac{1}{x}\frac{dy}{du}$ | M1 | Chain rule |
| $\frac{d^2y}{dx^2} = -\frac{1}{x^2}\frac{dy}{du} + \frac{1}{x}\frac{d^2y}{du^2}\cdot\frac{du}{dx} = -\frac{1}{x^2}\frac{dy}{du} + \frac{1}{x^2}\frac{d^2y}{du^2}$ | M1A1 | Product rule; correct expression |
| $x^2\left(-\frac{1}{x^2}\frac{dy}{du} + \frac{1}{x^2}\frac{d^2y}{du^2}\right) - 7x \times \frac{1}{x}\frac{dy}{du} + 16y = 2u$ | — | — |
| $\frac{d^2y}{du^2} - 8\frac{dy}{du} + 16y = 2u$ ★ | dM1A1cso | — |
---
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $m^2 - 8m + 16 = 0$ | — | — |
| $(m-4)^2 = 0$, $m = 4$ | M1A1 | Writing correct auxiliary equation and solving; $m=4$ correct |
| CF $= (A + Bu)e^{4u}$ | A1 | Correct CF; can use any single variable |
| PI: try $y = au + b$; $\frac{dy}{du} = a$, $\frac{d^2y}{du^2} = 0$ | M1 | Using appropriate PI and finding $\frac{dy}{du}$ and $\frac{d^2y}{du^2}$; note $y = \lambda u$ scores M0 |
| $0 - 8a + 16(au+b) = 2u$ | — | — |
| $a = \frac{1}{8}$, $b = \frac{1}{16}$ | dM1A1 | Substituting to find unknowns; correct values (decimals: 0.125 and 0.0625); depends on second M1 |
| $y = (A+Bu)e^{4u} + \frac{1}{8}u + \frac{1}{16}$ | B1ft | Complete solution following through their CF and PI; must have $y$ as function of $u$ |
---
### Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $y = (A + B\ln x)x^4 + \frac{1}{8}\ln x + \frac{1}{16}$ | B1 | Reverse substitution to obtain correct $y$ in terms of $x$; $x^4$ or $e^{4\ln x}$ allowed; must start $y = \ldots$; no follow-through |\begin{enumerate}
\item (a) Show that the transformation $x = \mathrm { e } ^ { u }$ transforms the differential equation
\end{enumerate}
$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 7 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 16 y = 2 \ln x , \quad x > 0$$
into the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} u ^ { 2 } } - 8 \frac { \mathrm {~d} y } { \mathrm {~d} u } + 16 y = 2 u$$
(b) Find the general solution of the differential equation (II), expressing $y$ as a function of $u$.\\
(c) Hence obtain the general solution of the differential equation (I).\\
\hfill \mbox{\textit{Edexcel FP2 2015 Q8 [14]}}