| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| 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 Further Maths question where part (a) guides students through verifying a given substitution (routine differentiation and algebraic manipulation), and part (b) requires solving a constant-coefficient second-order DE then back-substituting. While it involves multiple steps and is from Further Maths F2, the substitution is provided and the resulting DE is standard, making it moderately above average but not requiring novel insight. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = v + x\frac{dv}{dx}\) or \(\frac{dv}{dx} = x^{-1}\frac{dy}{dx} - x^{-2}y\) | M1A1 | Attempt to find relevant first derivative from \(y = xv\) using product/quotient rule |
| \(\frac{d^2y}{dx^2} = \frac{dv}{dx} + \frac{dv}{dx} + x\frac{d^2v}{dx^2}\) or equivalent | dM1A1 | Attempt to differentiate their \(\frac{dy}{dx}\) using product rule; correct expression |
| \(3\left(2\frac{dv}{dx} + x\frac{d^2v}{dx^2}\right) - \frac{6}{x}\left(v + x\frac{dv}{dx}\right) + \frac{6xv}{x^2} + 3xv = x^2\) | ddM1 | Substitute \(\frac{dy}{dx}\), \(\frac{d^2y}{dx^2}\), and \(y = xv\) into original equation |
| \(3x\frac{d^2v}{dx^2} + 6\frac{dv}{dx} - 6\frac{dv}{dx} - \frac{6}{x}v + \frac{6v}{x} + 3xv = x^2\) | — | — |
| \(3\frac{d^2v}{dx^2} + 3v = x\) * | A1* (6) | Obtain the given equation with no errors; at least one step shown between substitution and result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3\lambda^2 + 3 = 0\) so \(\lambda = \pm i\) | M1 | Forms correct auxiliary equation; attempts to solve (accept \(3m^2+3=0\)) |
| \((v =)\, Ae^{ix} + Be^{-ix}\) or \((v =)\, C\cos x + D\sin x\) | A1 | Correct complementary function |
| P.I.: Try \((v =)\, kx\,(+l)\) | B1 | Suitable form for PI including \(kx\) |
| \(\frac{dv}{dx} = k\), \(\frac{d^2v}{dx^2} = 0\) | — | — |
| \(3(0) + 3(kx + l) = x\) | M1 | Differentiate PI twice and substitute into \(3\frac{d^2v}{dx^2} + 3v = x\) |
| \(k = \frac{1}{3}\), \((l = 0)\) | — | — |
| \(v = Ae^{ix} + Be^{-ix} + \frac{1}{3}x\) or \(v = C\cos x + D\sin x + \frac{1}{3}x\) | A1 | Correct result; must be \(v = \ldots\) |
| \(y = x\!\left(Ae^{ix} + Be^{-ix} + \frac{1}{3}x\right)\) or \(y = x\!\left(C\cos x + D\sin x + \frac{1}{3}x\right)\) | B1ft (6) | Reverse substitution; follow through their previous line; must be \(y = \ldots\) |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = v + x\frac{dv}{dx}$ or $\frac{dv}{dx} = x^{-1}\frac{dy}{dx} - x^{-2}y$ | M1A1 | Attempt to find relevant first derivative from $y = xv$ using product/quotient rule |
| $\frac{d^2y}{dx^2} = \frac{dv}{dx} + \frac{dv}{dx} + x\frac{d^2v}{dx^2}$ or equivalent | dM1A1 | Attempt to differentiate their $\frac{dy}{dx}$ using product rule; correct expression |
| $3\left(2\frac{dv}{dx} + x\frac{d^2v}{dx^2}\right) - \frac{6}{x}\left(v + x\frac{dv}{dx}\right) + \frac{6xv}{x^2} + 3xv = x^2$ | ddM1 | Substitute $\frac{dy}{dx}$, $\frac{d^2y}{dx^2}$, and $y = xv$ into original equation |
| $3x\frac{d^2v}{dx^2} + 6\frac{dv}{dx} - 6\frac{dv}{dx} - \frac{6}{x}v + \frac{6v}{x} + 3xv = x^2$ | — | — |
| $3\frac{d^2v}{dx^2} + 3v = x$ * | A1* (6) | Obtain the given equation with no errors; at least one step shown between substitution and result |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3\lambda^2 + 3 = 0$ so $\lambda = \pm i$ | M1 | Forms correct auxiliary equation; attempts to solve (accept $3m^2+3=0$) |
| $(v =)\, Ae^{ix} + Be^{-ix}$ or $(v =)\, C\cos x + D\sin x$ | A1 | Correct complementary function |
| P.I.: Try $(v =)\, kx\,(+l)$ | B1 | Suitable form for PI including $kx$ |
| $\frac{dv}{dx} = k$, $\frac{d^2v}{dx^2} = 0$ | — | — |
| $3(0) + 3(kx + l) = x$ | M1 | Differentiate PI twice and substitute into $3\frac{d^2v}{dx^2} + 3v = x$ |
| $k = \frac{1}{3}$, $(l = 0)$ | — | — |
| $v = Ae^{ix} + Be^{-ix} + \frac{1}{3}x$ or $v = C\cos x + D\sin x + \frac{1}{3}x$ | A1 | Correct result; must be $v = \ldots$ |
| $y = x\!\left(Ae^{ix} + Be^{-ix} + \frac{1}{3}x\right)$ or $y = x\!\left(C\cos x + D\sin x + \frac{1}{3}x\right)$ | B1ft (6) | Reverse substitution; follow through their previous line; must be $y = \ldots$ |\begin{enumerate}
\item (a) Show that the transformation $y = x v$ transforms the equation
\end{enumerate}
$$3 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \frac { 6 } { x } \frac { \mathrm {~d} y } { \mathrm {~d} x } + \frac { 6 y } { x ^ { 2 } } + 3 y = x ^ { 2 } \quad x \neq 0$$
into the equation
$$3 \frac { \mathrm {~d} ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 3 v = x$$
(b) Hence obtain the general solution of the differential equation (I), giving your answer in the form $y = \mathrm { f } ( x )$
\hfill \mbox{\textit{Edexcel F2 2022 Q7 [12]}}