| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Session | Specimen |
| Marks | 7 |
| Topic | Second order differential equations |
| Type | Solve via substitution then back-substitute |
| Difficulty | Challenging +1.2 This is a structured second-order DE problem with clear guidance (substitution method specified). The substitution leads to a separable first-order equation, followed by straightforward integration and back-substitution. While it requires multiple steps and careful algebra, the technique is standard for Further Maths students and the path is clearly signposted, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts |
Writing $\frac{dy}{dx} = u$, $\frac{d^2y}{dx^2} = \frac{du}{dx}$, and so the d.e. becomes $\frac{du}{dx} = 2xu^2$: B1
$\Rightarrow \int u^{-2}\,du = \int 2x\,dx \Rightarrow -\frac{1}{u} = x^2 - C$, for separating variables and integrating: M1
$u = \frac{1}{4}$ when $x = 0 \Rightarrow C = 4$, for use of given condition to find $C$: M1
$\Rightarrow u = \frac{dy}{dx} = \frac{1}{4-x^2}$: A1
Then $y = \frac{1}{4}\int\left(\frac{1}{2-x} + \frac{1}{2+x}\right)dx$, method for integrating or use of F.Bk.: M1
$= \frac{1}{4}\ln\left(\frac{2+x}{2-x}\right) + \frac{1}{4}\ln A = \frac{1}{4}\ln\left[A\left(\frac{2+x}{2-x}\right)\right]$
$y = \frac{1}{2}$ when $x = 0 \Rightarrow A = e^2$, for use of given condition to find $A$: M1
$y = \frac{1}{4}\ln\left[e^2\left(\frac{2+x}{2-x}\right)\right]$ or $\frac{1}{4}\ln\left(\frac{2+x}{2-x}\right) + \frac{1}{2}$ **cao**: A1
**Total: 7 marks**5 The variables $y$ and $x$ are related by the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 2 x \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } , \quad - 2 < x < 2 .$$
By writing $u = \frac { \mathrm { d } y } { \mathrm {~d} x }$, determine $y$ explicitly in terms of $x$, given that $y = \frac { 1 } { 2 }$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 4 }$ when $x = 0$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 Q5 [7]}}