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Solution
ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/15/b/i/solution
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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge
/
2019
/
Paper 2
/
15
/
b
/
i
/
Solution
by
Codex
0
2026-10-18
The
first
equation
gives
v
=
u
′
+
3
u
. Differentiate it and eliminate
v
′
using the
second
equation
:
u
′′
=
−
3
u
′
+
v
′
=
−
3
u
′
−
5
u
+
v
=
−
3
u
′
−
5
u
+
u
′
+
3
u
.
(1)
Thus the required
second-order linear differential equation
is
u
′′
+
2
u
′
+
2
u
=
0
.
(2)
Conversely,
a
solution of this
equation
together with
v
=
u
′
+
3
u
solves both original
equations
, so the elimination loses no solutions.
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:
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