Question 12 - A-Level Maths

WJEC Further Unit 4 2022 June — Question 12 12 marks

Exam Board	WJEC
Module	Further Unit 4 (Further Unit 4)
Year	2022
Session	June
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Particular solution with initial conditions
Difficulty	Challenging +1.2 This is a second-order linear differential equation with constant coefficients and a polynomial particular integral. While it requires multiple steps (auxiliary equation, complementary function, particular integral, applying initial conditions), each step follows standard algorithmic procedures taught in Further Maths. The polynomial right-hand side makes finding the particular integral straightforward, and the 12 marks reflect length rather than conceptual difficulty. It's harder than average A-level due to being Further Maths content, but routine within that context.
Spec	4.10e Second order non-homogeneous: complementary + particular integral

Find the solution of the differential equation $$3\frac{d^2y}{dx^2} + 5\frac{dy}{dx} - 2y = 8 + 6x - 2x^2,$$ where $y = 6$ and $\frac{dy}{dx} = 5$ when $x = 0$. [12]

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Answer	Marks	Guidance
Solve auxiliary $3t^2 + 5t - 2 = 0$	M1	M0A0 no working
$(3t - 1)(t + 2) = 0$
$t = \frac{1}{3}$ or $t = -2$	A1	Both values
Complementary function: $y = Ae^{3x} + Be^{-2x}$
Use particular integral of the form $Cx^2 + Dx + E$	M1
$\frac{dy}{dx} = 2Cx + D$	A1	Both $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$
$\frac{d^2y}{dx^2} = 2C$
Therefore,	A1	Substitution
$6c + 5(2Cx + D) - 2(Cx^2 + Dx + E) = 8 + 6x - 2x^2$	A1	All values
$-2C = -2 \rightarrow C = 1$	A1
$10C - 2D = 6 \rightarrow D = 2$
$6C + 5D - 2E = 8 \rightarrow E = 4$
General Solution:
$y = Ae^{3x} + Be^{-2x} + x^2 + 2x + 4$	M1	FT C,D,E for M1A1M1A1. Sub and differentiate
$\frac{dy}{dx} = \frac{1}{3}Ae^{3x} - 2Be^{-2x} + 2x + 2$	A1
When $x = 0$, $y = A + B + 4 = 6$	M1	Substitution
$\frac{dy}{dx} = \frac{1}{3}A - 2B + 2 = 5$	A1	Both $y$ and $\frac{dy}{dx}$
Solving,
$A = 3$ and $B = -1$	B1	cao
Therefore,
$y = 3e^{\frac{1}{3}x} - e^{-2x} + x^2 + 2x + 4$	B1	cao

Solve auxiliary $3t^2 + 5t - 2 = 0$ | M1 | M0A0 no working
$(3t - 1)(t + 2) = 0$ | | 
$t = \frac{1}{3}$ or $t = -2$ | A1 | Both values

Complementary function: $y = Ae^{3x} + Be^{-2x}$ | | 

Use particular integral of the form $Cx^2 + Dx + E$ | M1 | 
$\frac{dy}{dx} = 2Cx + D$ | A1 | Both $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$
$\frac{d^2y}{dx^2} = 2C$ | | 

Therefore, | A1 | Substitution
$6c + 5(2Cx + D) - 2(Cx^2 + Dx + E) = 8 + 6x - 2x^2$ | A1 | All values

$-2C = -2 \rightarrow C = 1$ | A1 | 
$10C - 2D = 6 \rightarrow D = 2$ | | 
$6C + 5D - 2E = 8 \rightarrow E = 4$ | | 

General Solution: | | 
$y = Ae^{3x} + Be^{-2x} + x^2 + 2x + 4$ | M1 | FT C,D,E for M1A1M1A1. Sub and differentiate

$\frac{dy}{dx} = \frac{1}{3}Ae^{3x} - 2Be^{-2x} + 2x + 2$ | A1 | 

When $x = 0$, $y = A + B + 4 = 6$ | M1 | Substitution

$\frac{dy}{dx} = \frac{1}{3}A - 2B + 2 = 5$ | A1 | Both $y$ and $\frac{dy}{dx}$

Solving, | | 
$A = 3$ and $B = -1$ | B1 | cao

Therefore, | | 
$y = 3e^{\frac{1}{3}x} - e^{-2x} + x^2 + 2x + 4$ | B1 | cao

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Find the solution of the differential equation
$$3\frac{d^2y}{dx^2} + 5\frac{dy}{dx} - 2y = 8 + 6x - 2x^2,$$
where $y = 6$ and $\frac{dy}{dx} = 5$ when $x = 0$. [12]

\hfill \mbox{\textit{WJEC Further Unit 4 2022 Q12 [12]}}

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Q1 8 Q2 4 Q3 9 Q4 5 Q5 5 Q6 6 Q7 8 Q8 6 Q9 12 Q10 9 Q11 15 Q12 12 Q13 11 Q14 10

Answer	Marks	Guidance
Solve auxiliary \(3t^2 + 5t - 2 = 0\)	M1	M0A0 no working
\((3t - 1)(t + 2) = 0\)
\(t = \frac{1}{3}\) or \(t = -2\)	A1	Both values
Complementary function: \(y = Ae^{3x} + Be^{-2x}\)
Use particular integral of the form \(Cx^2 + Dx + E\)	M1
\(\frac{dy}{dx} = 2Cx + D\)	A1	Both \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\)
\(\frac{d^2y}{dx^2} = 2C\)
Therefore,	A1	Substitution
\(6c + 5(2Cx + D) - 2(Cx^2 + Dx + E) = 8 + 6x - 2x^2\)	A1	All values
\(-2C = -2 \rightarrow C = 1\)	A1
\(10C - 2D = 6 \rightarrow D = 2\)
\(6C + 5D - 2E = 8 \rightarrow E = 4\)
General Solution:
\(y = Ae^{3x} + Be^{-2x} + x^2 + 2x + 4\)	M1	FT C,D,E for M1A1M1A1. Sub and differentiate
\(\frac{dy}{dx} = \frac{1}{3}Ae^{3x} - 2Be^{-2x} + 2x + 2\)	A1
When \(x = 0\), \(y = A + B + 4 = 6\)	M1	Substitution
\(\frac{dy}{dx} = \frac{1}{3}A - 2B + 2 = 5\)	A1	Both \(y\) and \(\frac{dy}{dx}\)
Solving,
\(A = 3\) and \(B = -1\)	B1	cao
Therefore,
\(y = 3e^{\frac{1}{3}x} - e^{-2x} + x^2 + 2x + 4\)	B1	cao