| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Asymptotic behavior for large values |
| Difficulty | Standard +0.3 This is a standard second-order linear differential equation with constant coefficients and a polynomial forcing term. Part (i) requires finding complementary function (solving auxiliary equation) and particular integral (trying y=ax+b), which is routine FP3 technique. Parts (ii) and (iii) are straightforward applications. While Further Maths content is inherently harder, this is a textbook example with no novel insight required, placing it slightly above average difficulty overall. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Aux. equation \(3m^2 + 5m - 2 = ( \text{0})\) | M1 | For correct auxiliary equation seen and solution attempted |
| \(\Rightarrow m = \frac{1}{3}, -2\) | A1 | For correct roots |
| CF \((y = ) Ae^{\frac{1}{3}x} + Be^{-2x}\) | A1∇ | For correct CF t.t. from m with 2 arbitrary constants |
| PI \((y = ) px + q \Rightarrow 5p - 2(px + q) = -2x + 13\) | M1 | For stating and substituting PI of correct form |
| \(\Rightarrow p = 1, \quad q = -4\) | A1 A1 | For correct value of \(p\), and of \(q\) |
| GS \((y = ) Ae^{\frac{1}{3}x} + Be^{-2x} + x - 4\) | B1∇ 7 | For GS t.t. from CF+PI with 2 arbitrary constants (0 marks if PI incorrect or none in PI) |
| Answer | Marks | Guidance |
|---|---|---|
| \((0, -\frac{7}{2}) \Rightarrow A + B = \frac{1}{2}\) | M1 | For substituting \((0, -\frac{7}{2})\) in their GS and obtaining an equation in \(A\) and \(B\) |
| \(y' = \frac{1}{3}Ae^{\frac{1}{3}x} - 2Be^{-2x} + 1, \quad (0, 0) \Rightarrow A - 6B = -3\) | M1 | For finding \(y'\), substituting \((0,0)\) and obtaining an equation in \(A\) and \(B\) |
| \(\Rightarrow A = 0, \quad B = \frac{1}{2}\) | M1 | For solving their 2 equations in \(A\) and \(B\) |
| \(\Rightarrow (y = ) \frac{1}{2}e^{-2x} + x - 4\) | A1 | For correct \(A\) and \(B\) CAO |
| B1∇ 5 | For correct solution t.t. with their \(A\) and \(B\) in their GS |
| Answer | Marks | Guidance |
|---|---|---|
| \(x\) large \(\Rightarrow (y = ) x - 4\) | B1∇ 1 | For correct equation or function (allow \(\approx\) and \(\rightarrow\)) www t.t. from (ii) if valid |
## (i)
Aux. equation $3m^2 + 5m - 2 = ( \text{0})$ | M1 | For correct auxiliary equation seen and solution attempted
$\Rightarrow m = \frac{1}{3}, -2$ | A1 | For correct roots
CF $(y = ) Ae^{\frac{1}{3}x} + Be^{-2x}$ | A1∇ | For correct CF t.t. from m with 2 arbitrary constants
PI $(y = ) px + q \Rightarrow 5p - 2(px + q) = -2x + 13$ | M1 | For stating and substituting PI of correct form
$\Rightarrow p = 1, \quad q = -4$ | A1 A1 | For correct value of $p$, and of $q$
GS $(y = ) Ae^{\frac{1}{3}x} + Be^{-2x} + x - 4$ | B1∇ 7 | For GS t.t. from CF+PI with 2 arbitrary constants (0 marks if PI incorrect or none in PI)
## (ii)
$(0, -\frac{7}{2}) \Rightarrow A + B = \frac{1}{2}$ | M1 | For substituting $(0, -\frac{7}{2})$ in their GS and obtaining an equation in $A$ and $B$
$y' = \frac{1}{3}Ae^{\frac{1}{3}x} - 2Be^{-2x} + 1, \quad (0, 0) \Rightarrow A - 6B = -3$ | M1 | For finding $y'$, substituting $(0,0)$ and obtaining an equation in $A$ and $B$
$\Rightarrow A = 0, \quad B = \frac{1}{2}$ | M1 | For solving their 2 equations in $A$ and $B$
$\Rightarrow (y = ) \frac{1}{2}e^{-2x} + x - 4$ | A1 | For correct $A$ and $B$ **CAO**
| B1∇ 5 | For correct solution t.t. with their $A$ and $B$ in their GS
## (iii)
$x$ large $\Rightarrow (y = ) x - 4$ | B1∇ 1 | For correct equation or function (allow $\approx$ and $\rightarrow$) www t.t. from (ii) if valid
---\begin{enumerate}[label=(\roman*)]
\item Find the general solution of the differential equation
$$3\frac{d^2y}{dx^2} + 5\frac{dy}{dx} - 2y = -2x + 13.$$ [7]
\item Find the particular solution for which $y = -\frac{7}{2}$ and $\frac{dy}{dx} = 0$ when $x = 0$. [5]
\item Write down the function to which $y$ approximates when $x$ is large and positive. [1]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2011 Q5 [13]}}