| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Resonance cases requiring modified PI |
| Difficulty | Challenging +1.3 This is a Further Maths FP3 question on second-order differential equations with resonance, requiring finding the CF (auxiliary equation), using a modified PI form (given), and applying initial conditions. While it involves multiple steps and Further Maths content, the question explicitly provides the PI form, removing the main conceptual challenge of recognizing resonance. The techniques are systematic and well-practiced in FP3, making this moderately above average but not exceptionally difficult. |
| Spec | 4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| \((2m^2 + 3m - 2 = 0) \Rightarrow m = \frac{1}{2}, -2\) | M1 | For attempt to solve correct auxiliary equation |
| \(\text{CF} = Ae^{1x/2} + Be^{-2x}\) | A1 | For correct CF |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = pe^{-2x} - 2pxe^{-2x}\) | M1 | For differentiating PI twice, using product rule |
| \(\frac{d^2y}{dx^2} = -4pe^{-2x} + 4pxe^{-2x}\) | A1 | For correct \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\) |
| \(\Rightarrow (-8p + 3p + 8px - 6px - 2px)e^{-2x} = 5e^{-2x}\) | M1 | For substituting into DE |
| \(\Rightarrow p = -1\) | A1 | For correct \(p\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{GS} (y =) Ae^{1x/2} + Be^{-2x} - xe^{-2x}\) | B1 FT | For GS soi. FT from CF (2 constants) and \(p\) |
| \((0, 0) \Rightarrow A + B = 0\) | B1 FT | For correct equation. FT from GS of form \(Ae^{\alpha x} + Be^{\beta x} - Cxe^{-2x}\) |
| \(\frac{dy}{dx} = \frac{1}{2}Ae^{1x/2} - 2Be^{-2x} - e^{-2x} + 2xe^{-2x}\) | M1 | For differentiating GS and substituting values, using GS of form \(Ae^{\alpha x} + Be^{\beta x} - Cxe^{-2x}\) |
| \((0, \frac{dy}{dx} = 4) \Rightarrow \frac{1}{2}A - 2B - 1 = 4\) | M1 | For solving for \(A\) and \(B\) (can be gained from incorrect GS) |
| \(\Rightarrow A = 2, B = -2\) | A1 | For correct solution, including \(y =\) |
| [5] |
**(i)**
| $(2m^2 + 3m - 2 = 0) \Rightarrow m = \frac{1}{2}, -2$ | M1 | For attempt to solve correct auxiliary equation |
| $\text{CF} = Ae^{1x/2} + Be^{-2x}$ | A1 | For correct CF |
| | [2] | |
**(ii)**
| $\frac{dy}{dx} = pe^{-2x} - 2pxe^{-2x}$ | M1 | For differentiating PI twice, using product rule |
| $\frac{d^2y}{dx^2} = -4pe^{-2x} + 4pxe^{-2x}$ | A1 | For correct $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ |
| $\Rightarrow (-8p + 3p + 8px - 6px - 2px)e^{-2x} = 5e^{-2x}$ | M1 | For substituting into DE |
| $\Rightarrow p = -1$ | A1 | For correct $p$ |
| | [4] | |
**(iii)**
| $\text{GS} (y =) Ae^{1x/2} + Be^{-2x} - xe^{-2x}$ | B1 FT | For GS soi. FT from CF (2 constants) and $p$ |
| $(0, 0) \Rightarrow A + B = 0$ | B1 FT | For correct equation. FT from GS of form $Ae^{\alpha x} + Be^{\beta x} - Cxe^{-2x}$ |
| $\frac{dy}{dx} = \frac{1}{2}Ae^{1x/2} - 2Be^{-2x} - e^{-2x} + 2xe^{-2x}$ | M1 | For differentiating GS and substituting values, using GS of form $Ae^{\alpha x} + Be^{\beta x} - Cxe^{-2x}$ |
| $(0, \frac{dy}{dx} = 4) \Rightarrow \frac{1}{2}A - 2B - 1 = 4$ | M1 | For solving for $A$ and $B$ (can be gained from incorrect GS) |
| $\Rightarrow A = 2, B = -2$ | A1 | For correct solution, including $y =$ |
| | [5] | |5 The variables $x$ and $y$ satisfy the differential equation
$$2 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 3 \frac { \mathrm {~d} y } { \mathrm {~d} x } - 2 y = 5 \mathrm { e } ^ { - 2 x }$$
(i) Find the complementary function of the differential equation.\\
(ii) Given that there is a particular integral of the form $y = p x \mathrm { e } ^ { - 2 x }$, find the constant $p$.\\
(iii) Find the solution of the equation for which $y = 0$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 4$ when $x = 0$.
\hfill \mbox{\textit{OCR FP3 2012 Q5 [11]}}