How do you find the particular solution to dT+k(T−70)dt=0 that satisfies T(0)=140?
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___k off
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1 Answer
Steve M
Nov 8, 2016
Answer:
T
(
t
)
=
70
+
70
e
−
k
t
Explanation:
We have
d
T
+
k
(
T
−
70
)
d
t
=
0
which is a First Order separable DE
We can rearrange as follows:
d
T
=
−
k
(
T
−
70
)
d
t
⇒
d
T
k
(
T
−
70
)
=
−
d
t
∴
1
k
∫
1
T
−
70
d
T
=
−
∫
d
t
Integrating gives:
∴
1
k
ln
(
T
−
70
)
=
−
t
+
C
I always prefer to aptly the initial conditions as soon as possible to minimise the change of an algebraic slip:
T
=
140
when
t
=
0
∴
1
k
ln
(
140
−
70
)
=
C
⇒
1
k
ln
(
70
)
Substituting back into our DE solution gives us:
1
k
ln
(
T
−
70
)
=
−
t
+
1
k
ln
(
70
)
∴
ln
(
T
−
70
)
=
−
k
t
+
ln
(
70
)
∴
T
−
70
=
e
−
k
t
+
ln
(
70
)
∴
T
−
70
=
e
−
k
t
e
ln
(
70
)
∴
T
−
70
=
e
−
k
t
(
70
)
∴
T
=
70
+
70
e
−
k
t
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