WJEC Further Unit 4 2019 June — Question 6 10 marks

Exam BoardWJEC
ModuleFurther Unit 4 (Further Unit 4)
Year2019
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeParticular solution with initial conditions
DifficultyStandard +0.3 This is a standard second-order linear homogeneous differential equation with constant coefficients. The solution method is routine: find the auxiliary equation (m² - 7m + 10 = 0), solve for roots (m = 2, 5), write general solution (y = Ae^(2x) + Be^(5x)), then apply two initial conditions to find constants. While it's Further Maths content, it's a textbook exercise requiring only algorithmic application of a well-practiced technique with no conceptual challenges or novel insights.
Spec4.10d Second order homogeneous: auxiliary equation method
Solve the differential equation $$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - 7\frac{\mathrm{d}y}{\mathrm{d}x} + 10y = 0,$$ where \(\frac{\mathrm{d}y}{\mathrm{d}x} = 1\) and \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = 8\) when \(x = 0\). [10]
AnswerMarks Guidance
Auxiliary equation \(r^2 - 7r + 10 = 0\)M1
Roots \(r = 2\) and \(r = 5\)A1
General solution \(y = Ae^{5x} + Be^{2x}\)B1
Differentiating,M1
\(\frac{dy}{dx} = 5Ae^{5x} + 2Be^{2x}\)A1
\(\frac{d^2y}{dx^2} = 25Ae^{5x} + 4Be^{2x}\)A1
Substituting.M1
\(1 = 5A + 2B\)A1
\(8 = 25A + 4B\)A1
Solving,
\(A = 2/5\) or \(B = -1/2\) oeA1 1 correct value
\(\therefore y = \frac{2}{5}e^{5x} - \frac{1}{2}e^{2x}\)A1 
Auxiliary equation $r^2 - 7r + 10 = 0$ | M1 |
Roots $r = 2$ and $r = 5$ | A1 |
General solution $y = Ae^{5x} + Be^{2x}$ | B1 |
Differentiating, | M1 |
$\frac{dy}{dx} = 5Ae^{5x} + 2Be^{2x}$ | A1 |
$\frac{d^2y}{dx^2} = 25Ae^{5x} + 4Be^{2x}$ | A1 |
Substituting. | M1 |
$1 = 5A + 2B$ | A1 |
$8 = 25A + 4B$ | A1 |
Solving, | | 
$A = 2/5$ or $B = -1/2$ oe | A1 | 1 correct value
$\therefore y = \frac{2}{5}e^{5x} - \frac{1}{2}e^{2x}$ | A1 |
Solve the differential equation
$$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - 7\frac{\mathrm{d}y}{\mathrm{d}x} + 10y = 0,$$
where $\frac{\mathrm{d}y}{\mathrm{d}x} = 1$ and $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = 8$ when $x = 0$. [10]

\hfill \mbox{\textit{WJEC Further Unit 4 2019 Q6 [10]}}
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