FP2 June 2013 (R) Q7

EdexcelOld spec13 marksSecond Order Differentials

(a) Find the value of \(\lambda\) for which \(\lambda t^2\mathrm{e}^{3t}\) is a particular integral of the differential equation \[\frac{\mathrm{d}^2y}{\mathrm{d}t^2} - 6\frac{\mathrm{d}y}{\mathrm{d}t} + 9y = 6\mathrm{e}^{3t}, \qquad t \geqslant 0\] (5)

(b) Hence find the general solution of this differential equation. (3)

Given that when \(t = 0\), \(y = 5\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 4\)

(c) find the particular solution of this differential equation, giving your solution in the form \(y = \mathrm{f}(t)\). (5)

Mark scheme (a)

Scheme	Marks
\(y = \lambda t^2\mathrm{e}^{3t}\)
\(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 2\lambda t\mathrm{e}^{3t} + 3\lambda t^2\mathrm{e}^{3t}\)	M1A1
\(\dfrac{\mathrm{d}^2y}{\mathrm{d}t^2} = 2\lambda \mathrm{e}^{3t} + 6\lambda t\mathrm{e}^{3t} + 6\lambda t\mathrm{e}^{3t} + 9\lambda t^2\mathrm{e}^{3t}\)	A1
\(2\lambda \mathrm{e}^{3t} + 6\lambda t\mathrm{e}^{3t} + 6\lambda t\mathrm{e}^{3t} + 9\lambda t^2\mathrm{e}^{3t} - 12\lambda t\mathrm{e}^{3t} - 18\lambda t^2\mathrm{e}^{3t} + 9\lambda t^2\mathrm{e}^{3t} = 6\mathrm{e}^{3t}\)	M1dep
\(\lambda = 3\)	A1cso
NB. Candidates who give \(\lambda = 3\) without all this working get 5/5 provided no erroneous working is seen.
	(5)

Notes

M1 for differentiating \(y = \lambda t^2\mathrm{e}^{3t}\) wrt t. Product rule must be used.

A1 for correct differentiation ie \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 2\lambda t\mathrm{e}^{3t} + 3\lambda t^2\mathrm{e}^{3t}\)

A1 for a correct second differential \(\dfrac{\mathrm{d}^2y}{\mathrm{d}t^2} = 2\lambda \mathrm{e}^{3t} + 6\lambda t\mathrm{e}^{3t} + 6\lambda t\mathrm{e}^{3t} + 9\lambda t^2\mathrm{e}^{3t}\)

M1dep for substituting their differentials in the equation and obtaining a numerical value for \(\lambda \)
Dependent on the first M mark.

A1cso for \(\lambda = 3\) (no incorrect working seen)

NB. Candidates who give \(\lambda = 3\) without all this working get 5/5 provided no erroneous working is seen. Candidates who attempt the differentiation should be marked on that. If they then go straight to \(\lambda = 3\) without showing the substitution, give M1A1 if differentiation correct and M1A0 otherwise, as the solution is incorrect. If \(\lambda \neq 3\) then the M mark is only available if the substitution is shown.

Mark scheme (b)

Scheme	Marks
\(m^2 - 6m + 9 = 0\) \((m - 3)^2 = 0\)
C.F. \((y =)\ (A + Bt)\mathrm{e}^{3t}\)	M1A1
G.S. \(y = (A + Bt)\mathrm{e}^{3t} + 3t^2\mathrm{e}^{3t}\)	A1ft
	(3)

Notes

M1 for solving the 3 term quadratic auxiliary equation to obtain a value or values for \(m\) (usual rules for solving a quadratic equation)

A1 for the CF \((y =)\ (A + Bt)\mathrm{e}^{3t}\)

A1ft for using their CF and their numerical value of \(\lambda \) in the particular integral to obtain the general solution \(y = (A + Bt)\mathrm{e}^{3t} + 3t^2\mathrm{e}^{3t}\) Must have \(y = \ldots\) and rhs must be a function of \(t\).

Mark scheme (c)

Scheme	Marks
\(t = 0\quad y = 5\quad \Rightarrow A = 5\)	B1
\(\dfrac{\mathrm{d}y}{\mathrm{d}t} = B\mathrm{e}^{3t} + 3(A + Bt)\mathrm{e}^{3t} + 6t\mathrm{e}^{3t} + 9t^2\mathrm{e}^{3t}\)	M1
\(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 4\qquad 4 = B + 15\)	M1dep
\(B = -11\)	A1
Solution: \(y = (5 - 11t)\mathrm{e}^{3t} + 3t^2\mathrm{e}^{3t}\)	A1ft
	(5)
	(13 marks)

Notes

B1 for deducing that \(A = 5\)

M1 for differentiating their GS to obtain \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = \ldots\) The product rule must be used.

M1dep for using \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 4\) and their value for \(A\) in their \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) to obtain an equation for \(B\). Dependent on the previous M mark (of (c))

A1cao and cso for \(B = -11\)

A1ft for using their numerical values \(A\) and \(B\) in their GS from (b) to obtain the particular solution. Must have \(y = \ldots\) and rhs must be a function of \(t\).