| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Integrating factor with non-standard form |
| Difficulty | Challenging +1.2 This is a Further Maths integrating factor question requiring recognition that 1/(1-x²) = 1/((1-x)(1+x)) integrates to (1/2)ln((1+x)/(1-x)), then applying the standard method. While it involves algebraic manipulation and a formula book lookup, it follows a well-established procedure with clear signposting. The non-standard integrating factor form elevates it slightly above average A-level difficulty, but it remains a routine Further Maths exercise. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int e^{\frac{1}{1-x^2}} dx = e^{\frac{1}{1-x^2}} \cdot \frac{1}{1-x^2} = \left(\frac{1+x}{1-x}\right)^{\frac{1}{2}}\) | M1, A1, 2 | For IF stated or implied. Allow \(\pm\int\) and omission of \(dx\). For integration and simplification to AG (intermediate step must be seen) |
| \(\frac{d}{dx}\left(\sqrt{\frac{1+x}{1-x}}\right) = (1+x)^{\frac{1}{2}}\) | M1* | For multiplying both sides by IF |
| \(y\sqrt{\frac{1+x}{1-x}} = \frac{2}{3}(1+x)^{\frac{3}{2}} + c\) | M1, A1 | For integrating RHS to \(k(1+x)^n\). For correct equation (including \(+ c\)) |
| \((0, 2) \Rightarrow \frac{2}{3} + c = c = \frac{4}{3}\) | M1 (*dep), M1 (*dep) | For substituting \((0, 2)\) into their GS (including \(+c\)). For dividing solution through by IF, including dividing \(c\) or their numerical value for \(c\) |
| \(y = \frac{2}{3}(1+x)(1-x)^{\frac{1}{2}} + \frac{4}{3}\left(\frac{1-x}{1+x}\right)^{\frac{1}{2}}\) | A1, 6 | For correct solution aef (even unsimplified) in form \(y = f(x)\) |
**Part (i)**
| $\int e^{\frac{1}{1-x^2}} dx = e^{\frac{1}{1-x^2}} \cdot \frac{1}{1-x^2} = \left(\frac{1+x}{1-x}\right)^{\frac{1}{2}}$ | M1, A1, 2 | For IF stated or implied. Allow $\pm\int$ and omission of $dx$. For integration and simplification to AG (intermediate step must be seen) |
|---|---|---|
| $\frac{d}{dx}\left(\sqrt{\frac{1+x}{1-x}}\right) = (1+x)^{\frac{1}{2}}$ | M1* | For multiplying both sides by IF |
| $y\sqrt{\frac{1+x}{1-x}} = \frac{2}{3}(1+x)^{\frac{3}{2}} + c$ | M1, A1 | For integrating RHS to $k(1+x)^n$. For correct equation (including $+ c$) |
| $(0, 2) \Rightarrow \frac{2}{3} + c = c = \frac{4}{3}$ | M1 (*dep), M1 (*dep) | For substituting $(0, 2)$ into their GS (including $+c$). For dividing solution through by IF, including dividing $c$ or their numerical value for $c$ |
| $y = \frac{2}{3}(1+x)(1-x)^{\frac{1}{2}} + \frac{4}{3}\left(\frac{1-x}{1+x}\right)^{\frac{1}{2}}$ | A1, 6 | For correct solution aef (even unsimplified) in form $y = f(x)$ |
**Part (ii)**
[Content from part (ii) follows similar format from the mark scheme]
---4 The differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 1 } { 1 - x ^ { 2 } } y = ( 1 - x ) ^ { \frac { 1 } { 2 } } , \quad \text { where } | x | < 1$$
can be solved by the integrating factor method.\\
(i) Use an appropriate result given in the List of Formulae (MF1) to show that the integrating factor can be written as $\left( \frac { 1 + x } { 1 - x } \right) ^ { \frac { 1 } { 2 } }$.\\
(ii) Hence find the solution of the differential equation for which $y = 2$ when $x = 0$, giving your answer in the form $y = \mathrm { f } ( x )$.
\hfill \mbox{\textit{OCR FP3 2009 Q4 [8]}}