FP2 June 2007 Q3

EdexcelOld spec14 marksDifferential Equations

3. A scientist is modelling the amount of a chemical in the human bloodstream. The amount \(x\) of the chemical, measured in mg \(l^{-1}\), at time \(t\) hours satisfies the differential equation \[2x\frac{\mathrm{d}^2x}{\mathrm{d}t^2} - 6\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 = x^2 - 3x^4, \qquad x \gt 0.\]

(a) Show that the substitution \(y = \dfrac{1}{x^2}\) transforms this differential equation into \[\frac{\mathrm{d}^2y}{\mathrm{d}t^2} + y = 3. \qquad \boxed{\boldsymbol{I}}\] (5)

(b) Find the general solution of differential equation \(\boxed{\boldsymbol{I}}\). (4)

Given that at time \(t = 0\), \(x = \dfrac{1}{2}\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0\),

(d) write down the maximum value of \(x\) as \(t\) varies. (1)

Mark scheme (a)

Scheme	Marks
\(y = x^{-2} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}t} = -2x^{-3}\dfrac{\mathrm{d}x}{\mathrm{d}t} = -2x^{-3}\dot{x}\) [Use of chain rule; need \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\)]	M1
\(\Rightarrow \dfrac{\mathrm{d}^2y}{\mathrm{d}t^2} = -2x^{-3}\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2},\ \ + 6x^{-4}\left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)^2\)	A1ft, M1A1
(\(\div\) given d.e. by \(x^4\)) \(\ \dfrac{2}{x^3}\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} - \dfrac{6}{x^4}\left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)^2 = \dfrac{1}{x^2} - 3\) becomes \(\left(-\dfrac{\mathrm{d}^2y}{\mathrm{d}t^2} = y - 3\right) \qquad \dfrac{\mathrm{d}^2y}{\mathrm{d}t^2} + y = 3\) AG	A1 cso
	(5)

Notes

Second M1 is for attempt at product rule. (be generous)
Final A1 requires all working correct and sufficient “substitution” work

(corrected from the printed mark scheme: the end of the first line is printed as “\(= -2x - 3\,t\)”, which reads as \(-2x^{-3}\dot{x}\))

Mark scheme (b)

Scheme	Marks
Auxiliary equation: \(m^2 + 1 = 0\) and produce Complementary Function \(y = \ldots\)	M1
\((y) = A\cos t + B\sin t\)	A1cao
Particular integral: \(y = 3\)	B1
\(\therefore\) General solution: \((y) = A\cos t + B\sin t + 3\)	A1ft
	(4)

Notes

Answer can be stated; M1 is implied by correct C.F. stated (allow \(\theta\) for \(t\))
A1 f.t. for candidates CF + PI
Allow \(m^2 + m = 0\) and \(m^2 - 1 = 0\) for M1. Marks for (b) can be gained in (c)

Mark scheme (c)

Scheme	Marks
\(\dfrac{1}{x^2} = A\cos t + B\sin t + 3\)
\(x = \dfrac{1}{2},\ t = 0 \Rightarrow (4 = A + 3)\ A = 1\)	B1
Differentiating (to include \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\)): \(\ -2x^{-3}\dfrac{\mathrm{d}x}{\mathrm{d}t} = -A\sin t + B\cos t\)	M1
\(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0,\ t = 0 \Rightarrow (0 = 0 + B) \qquad B = 0\)	M1
\(\therefore \dfrac{1}{x^2} = 3 + \cos t\) so \(x = \dfrac{1}{\sqrt{3 + \cos t}}\)	A1 cao
	(4)

Notes

Second M : complete method to find other constant (This may involve solving two equations in A and B)

Scheme	Marks
(Max. value of \(x\) when \(\cos t = -1\)) so max \(x = \dfrac{1}{\sqrt{2}}\) or AWRT 0.707	B1
	(1)
	(14 marks)