| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Implicit differentiation for d²y/dx² |
| Difficulty | Challenging +1.2 This is a standard Further Maths FP1 question on successive differentiation of implicit differential equations. While it requires careful bookkeeping through multiple differentiations and application of the product rule, it follows a well-established algorithmic procedure taught in FP1. Part (b) is routine substitution into a Maclaurin series. The question is harder than typical A-level maths due to being Further Maths content, but it's a textbook exercise within FP1 rather than requiring novel insight. |
| Spec | 4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d^2y}{dx^2} = 1 - 2y\frac{dy}{dx} \Rightarrow \frac{d^3y}{dx^3} = -2y\frac{d^2y}{dx^2} - 2\left(\frac{dy}{dx}\right)^2\) | M1, A1 | M1: Attempts second and third derivatives, requiring \(\frac{d^2y}{dx^2} = 1 \pm 2y\frac{dy}{dx}\) followed by \(\frac{d^3y}{dx^3} = \pm 2y\frac{d^2y}{dx^2} \pm \ldots\). A1: Correct second and third derivatives. |
| \(\frac{d^4y}{dx^4} = -2\frac{dy}{dx}\frac{d^2y}{dx^2} - 2y\frac{d^3y}{dx^3} - 4\frac{dy}{dx}\frac{d^2y}{dx^2} = -6\frac{dy}{dx}\frac{d^2y}{dx^2} - 2y\frac{d^3y}{dx^3}\) | dM1 | Continues to differentiate to reach 5th derivative. Dependent on first M mark. No need to check detail as long as 5th derivative is reached. |
| \(\frac{d^5y}{dx^5} = -6\frac{dy}{dx}\frac{d^3y}{dx^3} - 6\left(\frac{d^2y}{dx^2}\right)^2 - 2y\frac{d^4y}{dx^4} - 2\frac{dy}{dx}\frac{d^3y}{dx^3}\) \(= -2y\frac{d^4y}{dx^4} - 8\frac{dy}{dx}\frac{d^3y}{dx^3} - 6\left(\frac{d^2y}{dx^2}\right)^2\) | A1 | Correct expression. NB \(a = -2,\ b = -8,\ c = -6\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=0,\ y=1 \Rightarrow \left(\frac{dy}{dx}\right)_0 = 0 - 1^2 = -1\) | B1 | Deduces correct value for \(y'(0)\) |
| \(\left(\frac{d^2y}{dx^2}\right)_0 = 1-2(1)(-1) = 3\) \(\left(\frac{d^3y}{dx^3}\right)_0 = -2(1)(3)-2(-1)^2 = -8\) \(\left(\frac{d^4y}{dx^4}\right)_0 = -6(-1)(3)-2(1)(-8) = 34\) \(\left(\frac{d^5y}{dx^5}\right)_0 = -2(1)(34)-8(-1)(-8)-6(3)^2 = -186\) | M1, A1 | M1: Finds values of all derivatives at \(x=0\) up to 5th. A1: All values correct as single simplified values. |
| \(y = y(0) + x\left(\frac{dy}{dx}\right)_0 + \frac{x^2}{2!}\left(\frac{d^2y}{dx^2}\right)_0 + \frac{x^3}{3!}\left(\frac{d^3y}{dx^3}\right)_0 + \frac{x^4}{4!}\left(\frac{d^4y}{dx^4}\right)_0 + \frac{x^5}{5!}\left(\frac{d^5y}{dx^5}\right)_0 + \ldots\) | M1 | Applies correct Maclaurin series with factorials up to term in \(x^5\) |
| \(y = 1 - x + \frac{3}{2}x^2 - \frac{4}{3}x^3 + \frac{17}{12}x^4 - \frac{31}{20}x^5 + \ldots\) | A1ft | Correct expansion, follow through on their derivative values. Factorials must be evaluated. Once correct expression seen apply isw. |
# Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d^2y}{dx^2} = 1 - 2y\frac{dy}{dx} \Rightarrow \frac{d^3y}{dx^3} = -2y\frac{d^2y}{dx^2} - 2\left(\frac{dy}{dx}\right)^2$ | M1, A1 | M1: Attempts second and third derivatives, requiring $\frac{d^2y}{dx^2} = 1 \pm 2y\frac{dy}{dx}$ followed by $\frac{d^3y}{dx^3} = \pm 2y\frac{d^2y}{dx^2} \pm \ldots$. A1: Correct second **and** third derivatives. |
| $\frac{d^4y}{dx^4} = -2\frac{dy}{dx}\frac{d^2y}{dx^2} - 2y\frac{d^3y}{dx^3} - 4\frac{dy}{dx}\frac{d^2y}{dx^2} = -6\frac{dy}{dx}\frac{d^2y}{dx^2} - 2y\frac{d^3y}{dx^3}$ | dM1 | Continues to differentiate to reach 5th derivative. Dependent on first M mark. No need to check detail as long as 5th derivative is reached. |
| $\frac{d^5y}{dx^5} = -6\frac{dy}{dx}\frac{d^3y}{dx^3} - 6\left(\frac{d^2y}{dx^2}\right)^2 - 2y\frac{d^4y}{dx^4} - 2\frac{dy}{dx}\frac{d^3y}{dx^3}$ $= -2y\frac{d^4y}{dx^4} - 8\frac{dy}{dx}\frac{d^3y}{dx^3} - 6\left(\frac{d^2y}{dx^2}\right)^2$ | A1 | Correct expression. NB $a = -2,\ b = -8,\ c = -6$ |
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# Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=0,\ y=1 \Rightarrow \left(\frac{dy}{dx}\right)_0 = 0 - 1^2 = -1$ | B1 | Deduces correct value for $y'(0)$ |
| $\left(\frac{d^2y}{dx^2}\right)_0 = 1-2(1)(-1) = 3$ $\left(\frac{d^3y}{dx^3}\right)_0 = -2(1)(3)-2(-1)^2 = -8$ $\left(\frac{d^4y}{dx^4}\right)_0 = -6(-1)(3)-2(1)(-8) = 34$ $\left(\frac{d^5y}{dx^5}\right)_0 = -2(1)(34)-8(-1)(-8)-6(3)^2 = -186$ | M1, A1 | M1: Finds values of all derivatives at $x=0$ up to 5th. A1: All values correct as single simplified values. |
| $y = y(0) + x\left(\frac{dy}{dx}\right)_0 + \frac{x^2}{2!}\left(\frac{d^2y}{dx^2}\right)_0 + \frac{x^3}{3!}\left(\frac{d^3y}{dx^3}\right)_0 + \frac{x^4}{4!}\left(\frac{d^4y}{dx^4}\right)_0 + \frac{x^5}{5!}\left(\frac{d^5y}{dx^5}\right)_0 + \ldots$ | M1 | Applies correct Maclaurin series with factorials up to term in $x^5$ |
| $y = 1 - x + \frac{3}{2}x^2 - \frac{4}{3}x^3 + \frac{17}{12}x^4 - \frac{31}{20}x^5 + \ldots$ | A1ft | Correct expansion, follow through on their derivative values. Factorials must be evaluated. Once correct expression seen apply isw. |3.
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = x - y ^ { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } ^ { 5 } y } { \mathrm {~d} x ^ { 5 } } = a y \frac { \mathrm {~d} ^ { 4 } y } { \mathrm {~d} x ^ { 4 } } + b \frac { \mathrm {~d} y } { \mathrm {~d} x } \frac { \mathrm {~d} ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } + c \left( \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } \right) ^ { 2 }$$
where $a$, $b$ and $c$ are integers to be determined.
\item Hence find a series solution, in ascending powers of $x$ as far as the term in $x ^ { 5 }$, of the differential equation (I), given that $y = 1$ at $x = 0$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2019 Q3 [9]}}