Question 5 - A-Level Maths

OCR Further Pure Core 1 2023 June — Question 5 6 marks

Exam Board	OCR
Module	Further Pure Core 1 (Further Pure Core 1)
Year	2023
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Combined polynomial and exponential RHS
Difficulty	Standard +0.8 This is a standard Further Maths second-order differential equation question requiring the auxiliary equation method for complex roots (part a) and particular integral for polynomial RHS (part b). While methodical, it involves multiple techniques (complex roots, trial polynomial PI) and careful algebraic manipulation, placing it moderately above average difficulty but still within standard FM curriculum expectations.
Spec	4.10d Second order homogeneous: auxiliary equation method 4.10e Second order non-homogeneous: complementary + particular integral

Find the general solution of the differential equation \(\frac { d ^ { 2 } y } { d x ^ { 2 } } - 2 \frac { d y } { d x } + 5 y = 0\).
Hence find the general solution of the differential equation \(\frac { d ^ { 2 } y } { d x ^ { 2 } } - 2 \frac { d y } { d x } + 5 y = x ( 4 - 5 x )\).

Show mark scheme Show mark scheme source

Question 5:

Answer	Marks	Guidance
5	(a)	A u x i l i a r y e q u a t i o n : n 2 − 2 n + 5 = 0

 n =  1 2 i

Answer	Marks
 y = x e ( A c o s 2 x + B s i n 2 x ) o e	B1
B1	Correct roots of auxiliary equation.

ft complex k only.

Or 𝑦 = 𝑅e𝑥cos(2𝑥+𝜑) or 𝑦 = 𝑅e𝑥sin(2𝑥+𝜑)

Or 𝑦 = 𝐴e(2i+1)𝑥 +𝐵e(−2i+1)𝑥

Final equation must be y = f(x)

[2]

Answer	Marks
(b)	T r i a l f u n c t i o n : y = a x 2 + b x + c

 y'=2ax+b, y''=2a

2a−2(2ax+b)+5(ax2 +bx+c)4x−5x2

5ax2 +x(−4a+5b)+5c−2b+2a4x−5x2

2  a = − 1 , b = 0 , c =

5

2 GS: y =ex(Acos2x+Bsin2x)−x2 +

Answer	Marks
5	B1

M1

A1

Answer	Marks
A1ft	Allow any extraneous terms in the trial function (eg.dx3) as long as d shown

to be zero, a,b,c0

𝑑𝑦 𝑑2𝑦

Differentiates their trial function to find and and substitutes

𝑑𝑥 𝑑𝑥2

ft their particular integral, and their CF from (a) (dependent on CF

containing exactly two arbitrary constants)

[4]

Question 5:
5 | (a) | A u x i l i a r y e q u a t i o n : n 2 − 2 n + 5 = 0
 n =  1 2 i
 y = x e ( A c o s 2 x + B s i n 2 x ) o e | B1
B1 | Correct roots of auxiliary equation.
ft complex k only.
Or 𝑦 = 𝑅e𝑥cos(2𝑥+𝜑) or 𝑦 = 𝑅e𝑥sin(2𝑥+𝜑)
Or 𝑦 = 𝐴e(2i+1)𝑥 +𝐵e(−2i+1)𝑥
Final equation must be y = f(x)
[2]
(b) | T r i a l f u n c t i o n : y = a x 2 + b x + c
 y'=2ax+b, y''=2a
2a−2(2ax+b)+5(ax2 +bx+c)4x−5x2
5ax2 +x(−4a+5b)+5c−2b+2a4x−5x2
2
 a = − 1 , b = 0 , c =
5
2
GS: y =ex(Acos2x+Bsin2x)−x2 +
5 | B1
M1
A1
A1ft | Allow any extraneous terms in the trial function (eg.dx3) as long as d shown
to be zero, a,b,c0
𝑑𝑦 𝑑2𝑦
Differentiates their trial function to find and and substitutes
𝑑𝑥 𝑑𝑥2
ft their particular integral, and their CF from (a) (dependent on CF
containing exactly two arbitrary constants)
[4]

Show LaTeX source

5
\begin{enumerate}[label=(\alph*)]
\item Find the general solution of the differential equation $\frac { d ^ { 2 } y } { d x ^ { 2 } } - 2 \frac { d y } { d x } + 5 y = 0$.
\item Hence find the general solution of the differential equation $\frac { d ^ { 2 } y } { d x ^ { 2 } } - 2 \frac { d y } { d x } + 5 y = x ( 4 - 5 x )$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 1 2023 Q5 [6]}}

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