| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Session | Specimen |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Verify particular integral form |
| Difficulty | Challenging +1.2 This is a standard Further Maths second-order differential equation question with repeated roots (auxiliary equation (m+2)²=0). Part (i) requires routine differentiation and substitution to find k. Part (ii) involves finding the complementary function and applying initial conditions—standard FP3 technique. Part (iii) adds a nice twist requiring use of the DE itself to find the second derivative and prove a bound, which elevates it slightly above routine but remains within expected FP3 problem-solving scope. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(y = kx^2 e^{-2x} \Rightarrow y' = 2kxe^{-2x} - 2kx^2 e^{-2x}\) and \(y'' = 2ke^{-2x} - 8kxe^{-2x} + 4kx^2 e^{-2x}\) | M1 | For differentiation at least once |
| A1 | For both \(y'\) and \(y''\) correct | |
| \((2k - 8kx + 4kx^2 + 8k - 8kx^2 + 4kx^2)e^{-2x} = 2e^{-2x}\) | M1 | For substituting completely in D.E. |
| Hence \(k = 1\) | A1 | For correct value of \(k\) |
| 4 | ||
| (ii) Auxiliary equation is \(m^2 + 4m + 4 = 0 \Rightarrow m = -2\) | B1 | For correct repeated root |
| B1 | For correct form of C.F. | |
| C.F. is \((A + Bx)e^{-2x}\) | B1√ | For sum of C.F. and P.I. |
| G.S. is \(y = (A + Bx)e^{-2x} + x^2 e^{-2x}\) | M1 | For using given values of \(x\) and \(y\) in G.S. |
| \(x = 0, y = 1 \Rightarrow 1 = A\) | M1 | For differentiating the G.S. |
| \(y' = Be^{-2x} - 2(A + Bx)e^{-2x} + 2xe^{-2x} - 2x^2 e^{-2x}\) | M1 | For using given values of \(x\) and \(y'\) in G.S. |
| \(x = 0, y' = 0 \Rightarrow 0 = B - 2A \Rightarrow B = 2\) | A1 | For correct answer |
| Hence solution is \(y = (1 + x)^2 e^{-2x}\) | 7 | |
| (iii) \(\frac{d^2y}{dx^2} = -2 - 4 = -2\) when \(x = 0\) | B1 | For correct value \(-2\) |
| Hence \((0, 1)\) is a maximum point | B1 | For statement of maximum at \(x = 0\) |
| \(\frac{dy}{dx} = 2(1 + x)e^{-2x} - 2(1 + x)^2 e^{-2x} = -2x(1 + x)e^{-2x}\), | M1 | For investigation of turning points, or equiv |
| so there are no turning points for \(x > 0\) | A1 | For complete proof of given result |
| Hence \(0 < y \leq 1\), since \(y \to 0\) as \(x \to \infty\) | 4 | |
| 15 |
**(i)** $y = kx^2 e^{-2x} \Rightarrow y' = 2kxe^{-2x} - 2kx^2 e^{-2x}$ and $y'' = 2ke^{-2x} - 8kxe^{-2x} + 4kx^2 e^{-2x}$ | M1 | For differentiation at least once
| A1 | For both $y'$ and $y''$ correct
$(2k - 8kx + 4kx^2 + 8k - 8kx^2 + 4kx^2)e^{-2x} = 2e^{-2x}$ | M1 | For substituting completely in D.E.
Hence $k = 1$ | A1 | For correct value of $k$
| **4** |
**(ii)** Auxiliary equation is $m^2 + 4m + 4 = 0 \Rightarrow m = -2$ | B1 | For correct repeated root
| B1 | For correct form of C.F.
C.F. is $(A + Bx)e^{-2x}$ | B1√ | For sum of C.F. and P.I.
G.S. is $y = (A + Bx)e^{-2x} + x^2 e^{-2x}$ | M1 | For using given values of $x$ and $y$ in G.S.
$x = 0, y = 1 \Rightarrow 1 = A$ | M1 | For differentiating the G.S.
$y' = Be^{-2x} - 2(A + Bx)e^{-2x} + 2xe^{-2x} - 2x^2 e^{-2x}$ | M1 | For using given values of $x$ and $y'$ in G.S.
$x = 0, y' = 0 \Rightarrow 0 = B - 2A \Rightarrow B = 2$ | A1 | For correct answer
Hence solution is $y = (1 + x)^2 e^{-2x}$ | **7** |
**(iii)** $\frac{d^2y}{dx^2} = -2 - 4 = -2$ when $x = 0$ | B1 | For correct value $-2$
Hence $(0, 1)$ is a maximum point | B1 | For statement of maximum at $x = 0$
$\frac{dy}{dx} = 2(1 + x)e^{-2x} - 2(1 + x)^2 e^{-2x} = -2x(1 + x)e^{-2x}$, | M1 | For investigation of turning points, or equiv
so there are no turning points for $x > 0$ | A1 | For complete proof of given result
Hence $0 < y \leq 1$, since $y \to 0$ as $x \to \infty$ | **4** |
| **15** |8 (i) Find the value of the constant $k$ such that $y = k x ^ { 2 } \mathrm { e } ^ { - 2 x }$ is a particular integral of the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 2 \mathrm { e } ^ { - 2 x }$$
(ii) Find the solution of this differential equation for which $y = 1$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$ when $x = 0$.\\
(iii) Use the differential equation to determine the value of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ when $x = 0$. Hence prove that $0 < y \leqslant 1$ for $x \geqslant 0$.
\hfill \mbox{\textit{OCR FP3 Q8 [15]}}