| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Topic | Second order differential equations |
| Type | Standard non-homogeneous with polynomial RHS |
| Difficulty | Challenging +1.2 This is a two-part second-order differential equation question requiring standard techniques (complementary function + particular integral for part i) followed by a guided substitution in part ii. While it involves multiple steps and the substitution requires careful differentiation using the product rule, the question is highly structured with clear guidance. The techniques are standard for Further Maths students, though the variable coefficient equation in part ii elevates it slightly above routine A-level questions. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks |
|---|---|
| (ii) | 2i x –2i x |
| Answer | Marks |
|---|---|
| Must have two real arbitrary constants | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Page 6 | Mark Scheme | Syllabus |
| Cambridge Pre-U – May/June 2015 | 9795 | 01 |
Question 9:
--- 9 (i)
(ii) ---
9 (i)
(ii) | 2i x –2i x
Comp. Fn. is u = A cos2x + B sin2x Allow Ae + Be here
For Part. Int. try u = ax + b (u′ = a and u′′ = 0) and substd. into given d.e.
1
a = 2, b =
4
Gen. Soln. is u= Acos2x+Bsin2x+2x+1 ft
4
Must have two arbitrary constants
Condone apparently complex coefficients here
2i x –2i x
Don’t allow Ae + Be here
dy du d2y dy d2u
xy =u ⇒ x + y= and x +2 =
dx dx dx2 dx dx2
d2y dy
⇒ (*) becomes x +2 + 4xy = 8x + 1
dx2 dx
Acos2x Bsin2x 1
⇒ y= + +2+ ft
x x 4x
Accept xy = …
Must have two real arbitrary constants | M1 A1
M1
A1
B1
[5]
M1 A1
A1
B1
[4]
Page 6 | Mark Scheme | Syllabus | Paper
Cambridge Pre-U – May/June 2015 | 9795 | 01The differential equation $(\star)$ is
$$\frac{\text{d}^2 u}{\text{d}x^2} + 4u = 8x + 1.$$
\begin{enumerate}[label=(\roman*)]
\item Find the general solution of $(\star)$. [5]
\item The differential equation $(\star \star)$ is
$$x \frac{\text{d}^2 v}{\text{d}x^2} + 2 \frac{\text{d}v}{\text{d}x} + 4xv = 8x + 1.$$
By using the substitution $u = xv$, show that $(\star)$ becomes $(\star \star)$ and deduce the general solution of $(\star \star)$. [4]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2015 Q9 [9]}}