| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Solve via substitution then back-substitute |
| Difficulty | Challenging +1.2 This is a structured FP2 differential equation question with clear guidance through substitution. Part (a) is routine differentiation and algebraic verification (given the target form). Part (b) is a standard constant-coefficient second-order DE with particular integral. Part (c) requires only back-substitution. While it involves multiple steps and FP2 content (inherently harder than single maths), the question provides the substitution and target equation, making it a guided exercise rather than requiring problem-solving insight. Slightly above average difficulty due to the algebraic manipulation and FP2 level, but well within standard textbook territory. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dy}{dx} = v + x\frac{dv}{dx}\) | M1 | For attempting to differentiate \(y = xv\) using product rule |
| \(\frac{d^2y}{dx^2} = \frac{dv}{dx} + \frac{dv}{dx} + x\frac{d^2v}{dx^2}\) | M1A1 | For differentiating their \(\frac{dy}{dx}\) using product rule |
| \(4x^2\left(2\frac{dv}{dx} + x\frac{d^2v}{dx^2}\right) - 8x\left(v + x\frac{dv}{dx}\right) + (8+4x^2)\times xv = x^4\) | M1 | For substituting their \(\frac{dy}{dx}\), \(\frac{d^2y}{dx^2}\) and \(y=xv\) into original equation |
| \(4x^3\frac{d^2v}{dx^2} + 4x^3v = x^4\) | M1 | For collecting terms to at most 4 term equation |
| \(4\frac{d^2v}{dx^2} + 4v = x\) ※ | A1 | A1cao and cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(v = \frac{y}{x}\), \(\frac{dv}{dx} = \frac{dy}{dx}\times\frac{1}{x} - y\times\frac{1}{x^2}\) | M1 | For writing \(v=\frac{y}{x}\) and differentiating by quotient/product rule |
| \(\frac{d^2v}{dx^2} = \frac{d^2y}{dx^2}\times\frac{1}{x} - \frac{dy}{dx}\times\frac{1}{x^2} - \frac{dy}{dx}\times\frac{1}{x^2} + 2y\times\frac{1}{x^3}\) | M1A1 | Product or quotient rule must be used |
| \(x^3\frac{d^2v}{dx^2} = x^2\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + 2y\) | M1 | For multiplying their \(\frac{d^2v}{dx^2}\) by \(x^3\) |
| \(4x^3\frac{d^2v}{dx^2} + 4x^3v = 4x^2\frac{d^2y}{dx^2} - 8x\frac{dy}{dx} + 8y + 4x^2y = x^4\) | M1 | For multiplying by 4 and adding \(4x^2y\) to each side |
| \(4\frac{d^2v}{dx^2} + 4v = x\) ※ | A1 | A1cao and cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(4\lambda^2 + 4 = 0\), \(\lambda^2 = -1\) | M1A1 | For forming auxiliary equation and solving |
| \((v =)\, C\cos x + D\sin x\) (or \(Ae^{ix} + Be^{-ix}\)) | A1 | Correct CF in either form; award even if \(\lambda = i\) only shown |
| P.I.: Try \(v = kx\,(+l)\); \(\frac{dv}{dx} = k\), \(\frac{d^2v}{dx^2} = 0\) | M1 | For trying \(v=kx\), \(k\neq 1\) or \(v=kx+l\) or \(v=mx^2+kx+l\) |
| \(4\times 0 + 4(kx\,(+l)) = x\) | M1dep | Substituting into equation; dep on 2nd M; award M0 if original equation used |
| \(k = \frac{1}{4}\), \((l=0)\) | ||
| \(v = C\cos x + D\sin x + \frac{1}{4}x\) (or \(Ae^{ix} + Be^{-ix} + \frac{1}{4}x\)) | A1 | A1cao for correct result in either form |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(y = x\!\left(C\cos x + D\sin x + \frac{1}{4}x\right)\) or \(y = x\!\left(Ae^{ix} + Be^{-ix} + \frac{1}{4}x\right)\) | B1ft | For reversing substitution; follow through their answer to (b) |
## Question 7:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = v + x\frac{dv}{dx}$ | M1 | For attempting to differentiate $y = xv$ using product rule |
| $\frac{d^2y}{dx^2} = \frac{dv}{dx} + \frac{dv}{dx} + x\frac{d^2v}{dx^2}$ | M1A1 | For differentiating their $\frac{dy}{dx}$ using product rule |
| $4x^2\left(2\frac{dv}{dx} + x\frac{d^2v}{dx^2}\right) - 8x\left(v + x\frac{dv}{dx}\right) + (8+4x^2)\times xv = x^4$ | M1 | For substituting their $\frac{dy}{dx}$, $\frac{d^2y}{dx^2}$ and $y=xv$ into original equation |
| $4x^3\frac{d^2v}{dx^2} + 4x^3v = x^4$ | M1 | For collecting terms to at most 4 term equation |
| $4\frac{d^2v}{dx^2} + 4v = x$ ※ | A1 | A1cao and cso |
**Alternative for (a):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $v = \frac{y}{x}$, $\frac{dv}{dx} = \frac{dy}{dx}\times\frac{1}{x} - y\times\frac{1}{x^2}$ | M1 | For writing $v=\frac{y}{x}$ and differentiating by quotient/product rule |
| $\frac{d^2v}{dx^2} = \frac{d^2y}{dx^2}\times\frac{1}{x} - \frac{dy}{dx}\times\frac{1}{x^2} - \frac{dy}{dx}\times\frac{1}{x^2} + 2y\times\frac{1}{x^3}$ | M1A1 | Product or quotient rule must be used |
| $x^3\frac{d^2v}{dx^2} = x^2\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + 2y$ | M1 | For multiplying their $\frac{d^2v}{dx^2}$ by $x^3$ |
| $4x^3\frac{d^2v}{dx^2} + 4x^3v = 4x^2\frac{d^2y}{dx^2} - 8x\frac{dy}{dx} + 8y + 4x^2y = x^4$ | M1 | For multiplying by 4 and adding $4x^2y$ to each side |
| $4\frac{d^2v}{dx^2} + 4v = x$ ※ | A1 | A1cao and cso |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $4\lambda^2 + 4 = 0$, $\lambda^2 = -1$ | M1A1 | For forming auxiliary equation and solving |
| $(v =)\, C\cos x + D\sin x$ (or $Ae^{ix} + Be^{-ix}$) | A1 | Correct CF in either form; award even if $\lambda = i$ only shown |
| P.I.: Try $v = kx\,(+l)$; $\frac{dv}{dx} = k$, $\frac{d^2v}{dx^2} = 0$ | M1 | For trying $v=kx$, $k\neq 1$ or $v=kx+l$ or $v=mx^2+kx+l$ |
| $4\times 0 + 4(kx\,(+l)) = x$ | M1dep | Substituting into equation; dep on 2nd M; award M0 if original equation used |
| $k = \frac{1}{4}$, $(l=0)$ | | |
| $v = C\cos x + D\sin x + \frac{1}{4}x$ (or $Ae^{ix} + Be^{-ix} + \frac{1}{4}x$) | A1 | A1cao for correct result in either form |
### Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $y = x\!\left(C\cos x + D\sin x + \frac{1}{4}x\right)$ or $y = x\!\left(Ae^{ix} + Be^{-ix} + \frac{1}{4}x\right)$ | B1ft | For reversing substitution; follow through their answer to (b) |
---\begin{enumerate}
\item (a) Show that the transformation $y = x v$ transforms the equation
\end{enumerate}
$$4 x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 8 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + \left( 8 + 4 x ^ { 2 } \right) y = x ^ { 4 }$$
into the equation
$$4 \frac { \mathrm {~d} ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 4 v = x$$
(b) Solve the differential equation (II) to find $v$ as a function of $x$.\\
(c) Hence state the general solution of the differential equation (I).\\
\hfill \mbox{\textit{Edexcel FP2 2013 Q7 [13]}}