| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Find higher derivatives from equation |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question requiring product rule differentiation and Taylor series construction. Part (a) is routine differentiation with given structure, and part (b) involves standard successive differentiation to find coefficients. While it's Further Maths content, the techniques are mechanical and well-practiced, making it slightly easier than average overall. |
| Spec | 1.07q Product and quotient rules: differentiation4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^3y}{dx^3} = e^x\left(2y\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^2 + 2y\frac{dy}{dx}\right) + e^x\left(2y\frac{dy}{dx} + y^2 + 1\right)\) | M1 A1 | |
| \(\frac{d^3y}{dx^3} = e^x\left(2y\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^2 + 4y\frac{dy}{dx} + y^2 + 1\right)\) | A1 | \((k = 4)\) |
| (3) | 1st M1 for evidence of Product Rule. 1st A1 for completely correct expression or equivalent. 2nd A1 for correct expression or \(k = 4\) stated. |
| Answer | Marks |
|---|---|
| \(\left[\frac{d^2y}{dx^2}\right]_0 = e^0(4 + 1 + 1) = 6\) | B1 |
| \(\left[\frac{d^3y}{dx^3}\right]_0 = e^0(12 + 8 + 8 + 1 + 1) = 30\) | B1 |
| \(y = 1 + 2x + \frac{6x^2}{2} + \frac{30x^3}{6} = 1 + 2x + 3x^2 + 5x^3\) | M1 A1ft |
| (4) 7 | 2nd M1 require four terms and denominators of 2 and 6 (might be implied). A1 follow through from their values in the final answer. |
**Part (a):**
| $\frac{d^3y}{dx^3} = e^x\left(2y\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^2 + 2y\frac{dy}{dx}\right) + e^x\left(2y\frac{dy}{dx} + y^2 + 1\right)$ | M1 A1 | |
| $\frac{d^3y}{dx^3} = e^x\left(2y\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^2 + 4y\frac{dy}{dx} + y^2 + 1\right)$ | A1 | $(k = 4)$ |
| | (3) | 1st M1 for evidence of Product Rule. 1st A1 for completely correct expression or equivalent. 2nd A1 for correct expression or $k = 4$ stated. |
**Part (b):**
| $\left[\frac{d^2y}{dx^2}\right]_0 = e^0(4 + 1 + 1) = 6$ | B1 | |
| $\left[\frac{d^3y}{dx^3}\right]_0 = e^0(12 + 8 + 8 + 1 + 1) = 30$ | B1 | |
| $y = 1 + 2x + \frac{6x^2}{2} + \frac{30x^3}{6} = 1 + 2x + 3x^2 + 5x^3$ | M1 A1ft | |
| | (4) 7 | 2nd M1 require four terms and denominators of 2 and 6 (might be implied). A1 follow through from their values in the final answer. |2.
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \mathrm { e } ^ { x } \left( 2 y \frac { \mathrm {~d} y } { \mathrm {~d} x } + y ^ { 2 } + 1 \right)$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = \mathrm { e } ^ { x } \left[ 2 y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + k y \frac { \mathrm {~d} y } { \mathrm {~d} x } + y ^ { 2 } + 1 \right]$$
where $k$ is a constant to be found.
Given that, at $x = 0 , y = 1$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 2$,
\item find a series solution for $y$ in ascending powers of $x$, up to and including the term in $x ^ { 3 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2011 Q2 [7]}}