CAIE FP1 2015 November — Question 2 6 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2015
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeStandard non-homogeneous with polynomial RHS
DifficultyStandard +0.8 This is a second-order linear differential equation with constant coefficients and a polynomial forcing term, requiring the auxiliary equation method (which yields a repeated root) plus finding a particular integral by assuming a quadratic form. While systematic, it involves multiple steps (complementary function, particular integral with undetermined coefficients, combining solutions) and is from Further Maths, making it moderately harder than average A-level questions.
Spec4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral
Find the general solution of the differential equation $$\frac{\mathrm{d}^2 x}{\mathrm{d}t^2} + 4\frac{\mathrm{d}x}{\mathrm{d}t} + 4x = 7 - 2t^2.$$ [6]
Question 2:
AnswerMarks
2m2 +4m+4=0⇒(m+2)2 =0⇒m=−2
Ae−2t +Bte−2t
CF: soi
PI : x= pt2 +qt+r⇒x& =2pt+q⇒&x&=2p
⇒2p+8pt+4q+4pt2 +4qt+4r =7−2t2
1
⇒ p=− , q = 1, r = 1
2
1
GS: x= Ae−2t +Bte−2t − t2+t+1
AnswerMarks
2M1
A1
M1
M1
A1
A1
[6]

Total

6
Question 2:
2 | m2 +4m+4=0⇒(m+2)2 =0⇒m=−2
Ae−2t +Bte−2t
CF: soi
PI : x= pt2 +qt+r⇒x& =2pt+q⇒&x&=2p
⇒2p+8pt+4q+4pt2 +4qt+4r =7−2t2
1
⇒ p=− , q = 1, r = 1
2
1
GS: x= Ae−2t +Bte−2t − t2+t+1
2 | M1
A1
M1
M1
A1
A1
[6]
Total
6
Find the general solution of the differential equation
$$\frac{\mathrm{d}^2 x}{\mathrm{d}t^2} + 4\frac{\mathrm{d}x}{\mathrm{d}t} + 4x = 7 - 2t^2.$$ [6]

\hfill \mbox{\textit{CAIE FP1 2015 Q2 [6]}}
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Q1 4 Q2 6 Q3 6 Q4 7 Q5 8 Q6 10 Q7 10 Q8 11 Q9 12 Q10 12 Q11 28