| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Standard linear first order - variable coefficients |
| Difficulty | Standard +0.8 This is a Further Maths first-order linear ODE requiring the integrating factor method with a parameter k. While the technique is standard, the algebraic manipulation with the general parameter k and the fractional power x^(1/2) makes this more demanding than typical A-level questions. The integration step requires careful handling of the right-hand side after applying the integrating factor, likely involving substitution or recognition of standard forms. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} + \frac{ky}{(1+x)} = \frac{x^{\frac{1}{2}}(1+x)^{2-k}}{(1+x)}\) | M1 | Divides by \((1+x)\) including the \(ky\) term |
| \(I = e^{\int \frac{k}{1+x}dx} = (1+x)^k\) | dM1 | Attempt integrating factor. \(I = e^{\int \frac{k}{1+x}dx}\) is sufficient for this mark but must include the \(k\). Condone omission of "\(dx\)". |
| \((1+x)^k\) | A1 | Correct integrating factor |
| \(y(1+x)^k = \int x^{\frac{1}{2}}(1+x)\,dx\) | M1 | Reaches \(y \times (\text{their } I) = \int x^{\frac{1}{2}}(1+x)^{1-k} \times (\text{their } I)\,dx\) |
| \(\int x^{\frac{1}{2}}(1+x)\,dx = \frac{2}{3}x^{\frac{3}{2}} + \frac{2}{5}x^{\frac{5}{2}} (+c)\) or by parts \(= \frac{2}{3}x^{\frac{3}{2}}(1+x) - \frac{4}{15}x^{\frac{5}{2}} (+c)\) | A1 | Correct integration |
| \(y = \dfrac{\frac{2}{3}x^{\frac{3}{2}} + \frac{2}{5}x^{\frac{5}{2}} + c}{(1+x)^k}\) or equivalent forms e.g. \(y = \frac{2}{3}x^{\frac{3}{2}}(1+x)^{1-k} - \frac{4}{15}x^{\frac{5}{2}}(1+x)^{-k} + c(1+x)^{-k}\) | A1 | Correct answer with constant correctly placed. Allow any equivalent correct answer. |
# Question 3:
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} + \frac{ky}{(1+x)} = \frac{x^{\frac{1}{2}}(1+x)^{2-k}}{(1+x)}$ | M1 | Divides by $(1+x)$ including the $ky$ term |
| $I = e^{\int \frac{k}{1+x}dx} = (1+x)^k$ | dM1 | Attempt integrating factor. $I = e^{\int \frac{k}{1+x}dx}$ is sufficient for this mark but must include the $k$. Condone omission of "$dx$". |
| $(1+x)^k$ | A1 | Correct integrating factor |
| $y(1+x)^k = \int x^{\frac{1}{2}}(1+x)\,dx$ | M1 | Reaches $y \times (\text{their } I) = \int x^{\frac{1}{2}}(1+x)^{1-k} \times (\text{their } I)\,dx$ |
| $\int x^{\frac{1}{2}}(1+x)\,dx = \frac{2}{3}x^{\frac{3}{2}} + \frac{2}{5}x^{\frac{5}{2}} (+c)$ or by parts $= \frac{2}{3}x^{\frac{3}{2}}(1+x) - \frac{4}{15}x^{\frac{5}{2}} (+c)$ | A1 | Correct integration |
| $y = \dfrac{\frac{2}{3}x^{\frac{3}{2}} + \frac{2}{5}x^{\frac{5}{2}} + c}{(1+x)^k}$ or equivalent forms e.g. $y = \frac{2}{3}x^{\frac{3}{2}}(1+x)^{1-k} - \frac{4}{15}x^{\frac{5}{2}}(1+x)^{-k} + c(1+x)^{-k}$ | A1 | Correct answer with constant correctly placed. Allow any equivalent correct answer. |\begin{enumerate}
\item
\item Find, in terms of $k$, where $k$ is a positive integer, the general solution of the differential equation
\end{enumerate}
$$( 1 + x ) \frac { \mathrm { d } y } { \mathrm {~d} x } + k y = x ^ { \frac { 1 } { 2 } } ( 1 + x ) ^ { 2 - k } , \quad x > 0$$
giving your answer in the form $y = \mathrm { f } ( x )$.\\
(6)\\
\hfill \mbox{\textit{Edexcel F2 2016 Q3 [6]}}