| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Standard linear first order - constant coefficients |
| Difficulty | Standard +0.3 This is a standard first-order linear differential equation requiring the integrating factor method, which is routine for Further Maths students. Parts (i)-(iii) involve straightforward application of the technique with simple integration, while part (iv) requires basic understanding of asymptotic behavior. The question is slightly above average difficulty due to being Further Maths content and having multiple parts, but it's a textbook-standard exercise with no novel problem-solving required. |
| Spec | 4.10c Integrating factor: first order equations4.10d Second order homogeneous: auxiliary equation method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| CF: \(m=-3\Rightarrow Ae^{-3x}\) | B1 1 | For correct CF |
| \((y=)\ px+q\) | B1 | For stating linear form for PI (may be implied) |
| \(\Rightarrow p+3(px+q)=2x+1\) | M1 | For substituting PI into DE (needs \(y\) and \(\frac{dy}{dx}\)) |
| \(\Rightarrow p=\frac{2}{3},\ q=\frac{1}{9}\) | A1 A1 | For correct values |
| \(\Rightarrow\) GS: \(y=Ae^{-3x}+\frac{2}{3}x+\frac{1}{9}\) | A1\(\sqrt{}\) | For correct GS, f.t. from their CF + PI |
| SR: Integrating factor method may be used, but CF must be stated somewhere to earn the mark in (i) | ||
| IF: \(e^{3x}\Rightarrow\dfrac{d}{dx}(ye^{3x})=(2x+1)e^{3x}\) | B1 | For stating integrating factor |
| \(\Rightarrow ye^{3x}=\frac{1}{3}e^{3x}(2x+1)-\int\frac{2}{3}e^{3x}dx\) | M1 | For attempt at integrating by parts the right way round |
| \(\Rightarrow ye^{3x}=\frac{2}{3}xe^{3x}+\frac{1}{3}e^{3x}-\frac{2}{9}e^{3x}+A\) | A2* | For correct integration, including constant; award A1 for any 2 algebraic terms correct |
| \(\Rightarrow\) GS: \(y=Ae^{-3x}+\frac{2}{3}x+\frac{1}{9}\) | A1\(\sqrt{}\) 5 | For correct GS, f.t. from their * with constant |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| EITHER: \(\dfrac{dy}{dx}=-3Ae^{-3x}+\frac{2}{3}\) | M1 | For differentiating their GS |
| \(\Rightarrow -3A+\frac{2}{3}=0\) | M1 | For putting \(\frac{dy}{dx}=0\) when \(x=0\) |
| \(y=\frac{2}{9}e^{-3x}+\frac{2}{3}x+\frac{1}{9}\) | A1 | For correct solution |
| OR: \(\dfrac{dy}{dx}=0\), \(x=0\Rightarrow 3y=1\) | M1 | For using original DE with \(\frac{dy}{dx}=0\) and \(x=0\) to find \(y\) |
| \(\Rightarrow\frac{1}{3}=A+\frac{1}{9}\) | M1 | For using their GS with \(y\) and \(x=0\) to find \(A\) |
| \(y=\frac{2}{9}e^{-3x}+\frac{2}{3}x+\frac{1}{9}\) | A1 3 | For correct solution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y=\frac{2}{3}x+\frac{1}{9}\) | B1\(\sqrt{}\) 1 | For correct function, f.t. from linear part of (iii) |
## Question 6:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| CF: $m=-3\Rightarrow Ae^{-3x}$ | B1 **1** | For correct CF |
| $(y=)\ px+q$ | B1 | For stating linear form for PI (may be implied) |
| $\Rightarrow p+3(px+q)=2x+1$ | M1 | For substituting PI into DE (needs $y$ and $\frac{dy}{dx}$) |
| $\Rightarrow p=\frac{2}{3},\ q=\frac{1}{9}$ | A1 A1 | For correct values |
| $\Rightarrow$ GS: $y=Ae^{-3x}+\frac{2}{3}x+\frac{1}{9}$ | A1$\sqrt{}$ | For correct GS, f.t. from their CF + PI |
| SR: Integrating factor method may be used, but CF must be stated somewhere to earn the mark in (i) | | |
| IF: $e^{3x}\Rightarrow\dfrac{d}{dx}(ye^{3x})=(2x+1)e^{3x}$ | B1 | For stating integrating factor |
| $\Rightarrow ye^{3x}=\frac{1}{3}e^{3x}(2x+1)-\int\frac{2}{3}e^{3x}dx$ | M1 | For attempt at integrating by parts the right way round |
| $\Rightarrow ye^{3x}=\frac{2}{3}xe^{3x}+\frac{1}{3}e^{3x}-\frac{2}{9}e^{3x}+A$ | A2* | For correct integration, including constant; award A1 for any 2 algebraic terms correct |
| $\Rightarrow$ GS: $y=Ae^{-3x}+\frac{2}{3}x+\frac{1}{9}$ | A1$\sqrt{}$ **5** | For correct GS, f.t. from their * with constant |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| EITHER: $\dfrac{dy}{dx}=-3Ae^{-3x}+\frac{2}{3}$ | M1 | For differentiating their GS |
| $\Rightarrow -3A+\frac{2}{3}=0$ | M1 | For putting $\frac{dy}{dx}=0$ when $x=0$ |
| $y=\frac{2}{9}e^{-3x}+\frac{2}{3}x+\frac{1}{9}$ | A1 | For correct solution |
| OR: $\dfrac{dy}{dx}=0$, $x=0\Rightarrow 3y=1$ | M1 | For using original DE with $\frac{dy}{dx}=0$ and $x=0$ to find $y$ |
| $\Rightarrow\frac{1}{3}=A+\frac{1}{9}$ | M1 | For using their GS with $y$ and $x=0$ to find $A$ |
| $y=\frac{2}{9}e^{-3x}+\frac{2}{3}x+\frac{1}{9}$ | A1 **3** | For correct solution |
### Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=\frac{2}{3}x+\frac{1}{9}$ | B1$\sqrt{}$ **1** | For correct function, f.t. from linear part of (iii) |
---6 The variables $x$ and $y$ satisfy the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } + 3 y = 2 x + 1$$
Find\\
(i) the complementary function,\\
(ii) the general solution.
In a particular case, it is given that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$ when $x = 0$.\\
(iii) Find the solution of the differential equation in this case.\\
(iv) Write down the function to which $y$ approximates when $x$ is large and positive.
\hfill \mbox{\textit{OCR FP3 2007 Q6 [10]}}