| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Topic | Second order differential equations |
| Type | Solve via substitution then back-substitute |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring multiple sophisticated steps: computing second derivatives with the product rule, substituting into a complex differential equation, recognizing the resulting simpler form, solving it, and applying initial conditions. The substitution method for variable-coefficient ODEs is non-standard and requires significant technical facility beyond typical A-level content. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
$x = t^2u \Rightarrow \frac{dx}{dt} = t^2\frac{du}{dt}+2tu$ **B1** and $\frac{d^2x}{dt^2} = t^2\frac{d^2u}{dt^2}+4t\frac{du}{dt}+2u$ **B1**
Substitution into given d.e. and eliminating all $x$'s **M1** $\frac{d^2u}{dt^2}-4u = 0$ **A1** (assuming $t\neq 0$)
Gen. Soln. $u = Ae^{2t}+Be^{-2t}$ **B1**
$x = t^2(Ae^{2t}+Be^{-2t})$ **M1** Re-writing in terms of $x$ (possibly later)
$\frac{dx}{dt} = t^2(2Ae^{2t}-2Be^{-2t})+2t(Ae^{2t}+Be^{-2t})$ **M1** Differentiating this or $\frac{du}{dt}$
**M1** for substitution of at least one initial condition: $t=1$, $x = e-1$, $\frac{dx}{dt} = 4e$ to find $A$, $B$
$e-1 = Ae^2+\frac{B}{e^2}$ and $4e = 2Ae^2-\frac{2B}{e^2}+2Ae^2+\frac{2B}{e^2} \Rightarrow A = \frac{1}{e}$ **A1** $B = -e^2$ **A1**
i.e. $x = t^2(e^{2t-1}-e^{2-2t})$
**[10]**8 For the differential equation $t ^ { 2 } \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 4 t \frac { \mathrm {~d} x } { \mathrm {~d} t } + \left( 6 - 4 t ^ { 2 } \right) x = 0$, use the substitution $x = t ^ { 2 } u$ to find a differential equation involving $t$ and $u$ only. Hence solve the above differential equation, given that $x = \mathrm { e } - 1$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = 4 \mathrm { e }$ when $t = 1$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2010 Q8 [10]}}