| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Substitution changing independent variable |
| Difficulty | Challenging +1.2 This is a structured multi-part question with clear guidance at each step. Part (i) is routine chain rule application, part (ii) is algebraic substitution following given results, and part (iii) requires solving a standard constant-coefficient second-order DE with particular integral. While it involves multiple techniques and careful algebra, the path is well-signposted and uses standard Further Maths methods without requiring novel insight. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = 2t^{\frac{1}{2}}\frac{dy}{dt}\) (AG) | B1 | |
| \(\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right)\times\frac{dt}{dx} = \left\{t^{-\frac{1}{2}}\frac{dy}{dt} + 2t^{\frac{1}{2}}\frac{d^2y}{dt^2}\right\}2t^{\frac{1}{2}} = 2\frac{dy}{dt} + 4t\frac{d^2y}{dt^2}\) (AG) | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitute in (*): \(2\frac{dy}{dt} + 4t\frac{d^2y}{dt^2} - 16t\frac{dy}{dt} - 2\frac{dy}{dt} + 12ty = 4te^{-t}\) | B1 | |
| \(\Rightarrow \frac{d^2y}{dt^2} - 4\frac{dy}{dt} + 3y = e^{-t}\) (AG) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| CF: \((m-1)(m-3) = 0\) | M1 | |
| \(y = Ae^t + Be^{3t}\) | A1 | |
| PI: \(y = ke^{-t} \Rightarrow y' = -ke^{-t} \Rightarrow y'' = ke^{-t}\) | M1 | |
| \(ke^{-t} + 4ke^{-t} + 3ke^{-t} = e^{-t}\) | M1 | |
| \(\Rightarrow k = \frac{1}{8}\) | A1 | |
| GS: \(y = Ae^t + Be^{3t} + \frac{1}{8}e^{-t}\) | A1 FT | |
| \(y = Ae^{x^2} + Be^{3x^2} + \frac{1}{8}e^{-x^2}\) | A1 FT |
## Question 10(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = 2t^{\frac{1}{2}}\frac{dy}{dt}$ (AG) | B1 | |
| $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right)\times\frac{dt}{dx} = \left\{t^{-\frac{1}{2}}\frac{dy}{dt} + 2t^{\frac{1}{2}}\frac{d^2y}{dt^2}\right\}2t^{\frac{1}{2}} = 2\frac{dy}{dt} + 4t\frac{d^2y}{dt^2}$ (AG) | M1A1 | |
## Question 10(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute in (*): $2\frac{dy}{dt} + 4t\frac{d^2y}{dt^2} - 16t\frac{dy}{dt} - 2\frac{dy}{dt} + 12ty = 4te^{-t}$ | B1 | |
| $\Rightarrow \frac{d^2y}{dt^2} - 4\frac{dy}{dt} + 3y = e^{-t}$ (AG) | | |
## Question 10(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| CF: $(m-1)(m-3) = 0$ | M1 | |
| $y = Ae^t + Be^{3t}$ | A1 | |
| PI: $y = ke^{-t} \Rightarrow y' = -ke^{-t} \Rightarrow y'' = ke^{-t}$ | M1 | |
| $ke^{-t} + 4ke^{-t} + 3ke^{-t} = e^{-t}$ | M1 | |
| $\Rightarrow k = \frac{1}{8}$ | A1 | |
| GS: $y = Ae^t + Be^{3t} + \frac{1}{8}e^{-t}$ | A1 FT | |
| $y = Ae^{x^2} + Be^{3x^2} + \frac{1}{8}e^{-x^2}$ | A1 FT | |10 It is given that $x = t ^ { \frac { 1 } { 2 } }$, where $x > 0$ and $t > 0$, and $y$ is a function of $x$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 t ^ { \frac { 1 } { 2 } } \frac { \mathrm {~d} y } { \mathrm {~d} t }$ and $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 4 t \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }$.\\
(ii) Hence show that the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 8 x + \frac { 1 } { x } \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 12 x ^ { 2 } y = 4 x ^ { 2 } \mathrm { e } ^ { - x ^ { 2 } }$$
reduces to the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 y = \mathrm { e } ^ { - t }$$
(iii) Find the general solution of ( $*$ ), giving $y$ in terms of $x$.\\
\hfill \mbox{\textit{CAIE FP1 2017 Q10 [11]}}