| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Solve via substitution then back-substitute |
| Difficulty | Challenging +1.2 This is a structured FP3 differential equations question with clear guidance. Part (a) is routine partial fractions (A-level standard). Part (b) uses a given substitution to reduce to first-order, requires integration with logarithms, then back-substitution and a second integration. While it involves multiple steps and FP3 content, the substitution is provided, the method is standard for this topic, and the algebraic manipulations are straightforward. It's harder than typical C3/C4 questions due to being Further Maths content, but it's a textbook example of this technique rather than requiring novel insight. |
| Spec | 1.02y Partial fractions: decompose rational functions4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{1+x} - \frac{1}{2+x}\) | B1 | Condone \(A=1\), \(B=-1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{du}{dx} + \frac{1}{(1+x)(2+x)}u = \ldots\) | M1 | Replacing \(\frac{d^2y}{dx^2}\) by \(\frac{du}{dx}\) and \(\frac{dy}{dx}\) by \(u\) |
| I.F. \(e^{\int \frac{1}{(1+x)(2+x)}dx}\) | M1 | \(e^{\int \frac{1}{(1+x)(2+x)}dx}\) OE PI |
| \(= e^{\ln(1+x)-\ln(2+x)}\) | A1F | \(e^{A\ln(1+x)+B\ln(2+x)}\) OE ft non-zero \(A\) and \(B\) values |
| \(= \frac{1+x}{2+x}\) | A1 | Correct IF in form \(\frac{\lambda(1+x)}{2+x}\) |
| \(\frac{d}{dx}\left[\frac{(1+x)u}{2+x}\right] = 1\) | m1 | Dep on both previous M1s; LHS as \(\frac{d}{dx}[u \times \text{candidate's IF}]\) PI |
| \(\frac{1+x}{2+x}u = x \; (+c)\) | A1 | |
| \(x=0, u=4 \Rightarrow c=2\) so \(\frac{1+x}{2+x}u = x+2\) | A1 | \(\frac{1+x}{2+x}u = x+2\) OE |
| \(\frac{dy}{dx} = \frac{(2+x)^2}{1+x}\) | m1 | Dep on previous MMm, replacing \(u\) to form \(\frac{dy}{dx}=g(x)\) seen or used (\(c\) could still be present) |
| \(\frac{dy}{dx} = x+3+\frac{1}{1+x}\) | m1 | \(\frac{(2+x)^2}{1+x}\) in form \(x+p+\frac{q}{1+x}\) or other valid method to integrate \(\frac{(2+x)^2}{(1+x)}\) |
| \(y = \frac{1}{2}x^2 + 3x + \ln(1+x) \; (+d)\) | A1 | ACF |
| \(x=0, y=1 \Rightarrow d=1\) | ||
| \(y = \frac{1}{2}x^2 + 3x + \ln(1+x) + 1\) | A1 | ACF eg \(y = \ln(1+x)+2(1+x)+\frac{(1+x)^2}{2}-\frac{3}{2}\) |
# Question 5:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{1+x} - \frac{1}{2+x}$ | B1 | Condone $A=1$, $B=-1$ |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{du}{dx} + \frac{1}{(1+x)(2+x)}u = \ldots$ | M1 | Replacing $\frac{d^2y}{dx^2}$ by $\frac{du}{dx}$ and $\frac{dy}{dx}$ by $u$ |
| I.F. $e^{\int \frac{1}{(1+x)(2+x)}dx}$ | M1 | $e^{\int \frac{1}{(1+x)(2+x)}dx}$ OE PI |
| $= e^{\ln(1+x)-\ln(2+x)}$ | A1F | $e^{A\ln(1+x)+B\ln(2+x)}$ OE ft non-zero $A$ and $B$ values |
| $= \frac{1+x}{2+x}$ | A1 | Correct IF in form $\frac{\lambda(1+x)}{2+x}$ |
| $\frac{d}{dx}\left[\frac{(1+x)u}{2+x}\right] = 1$ | m1 | Dep on both previous M1s; LHS as $\frac{d}{dx}[u \times \text{candidate's IF}]$ PI |
| $\frac{1+x}{2+x}u = x \; (+c)$ | A1 | |
| $x=0, u=4 \Rightarrow c=2$ so $\frac{1+x}{2+x}u = x+2$ | A1 | $\frac{1+x}{2+x}u = x+2$ OE |
| $\frac{dy}{dx} = \frac{(2+x)^2}{1+x}$ | m1 | Dep on previous MMm, replacing $u$ to form $\frac{dy}{dx}=g(x)$ seen or used ($c$ could still be present) |
| $\frac{dy}{dx} = x+3+\frac{1}{1+x}$ | m1 | $\frac{(2+x)^2}{1+x}$ in form $x+p+\frac{q}{1+x}$ or other valid method to integrate $\frac{(2+x)^2}{(1+x)}$ |
| $y = \frac{1}{2}x^2 + 3x + \ln(1+x) \; (+d)$ | A1 | ACF |
| $x=0, y=1 \Rightarrow d=1$ | | |
| $y = \frac{1}{2}x^2 + 3x + \ln(1+x) + 1$ | A1 | ACF eg $y = \ln(1+x)+2(1+x)+\frac{(1+x)^2}{2}-\frac{3}{2}$ |
---5
\begin{enumerate}[label=(\alph*)]
\item Express $\frac { 1 } { ( 1 + x ) ( 2 + x ) }$ in the form $\frac { A } { 1 + x } + \frac { B } { 2 + x }$, where $A$ and $B$ are integers.
\item Use the substitution $u = \frac { \mathrm { d } y } { \mathrm {~d} x }$ to solve the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \frac { 1 } { ( 1 + x ) ( 2 + x ) } \frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 + x } { 1 + x }$$
given that $y = 1$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 4$ when $x = 0$. Give your answer in the form $y = \mathrm { f } ( x )$.\\[0pt]
[11 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2016 Q5 [12]}}