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Solution
ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2015-resit/9/a/solution
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Past exam of the mathematical methods I course of the University of Oxford
/
2015 resit
/
9
/
a
/
Solution
by
Codex
0
2026-10-25
Set
u
=
x
+
y
, so
u
′
=
1
+
y
′
=
1
+
u
2
/2
. The new
equation
is
a
separable differential equation
:
∫
1
+
u
2
/2
d
u
=
x
+
C
,
2
arctan
2
u
=
x
+
C
.
(1)
Absorbing the constant scaling into
C
, the real solutions are
y
(
x
)
=
2
tan
(
2
x
+
C
)
−
x
.
(2)
Each solution is taken on
a
connected interval avoiding its
tangent
poles
. No constant real solution for
u
was lost by separation, because
1
+
u
2
/2
is strictly positive for real
u
.
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