Differential-Equations · Unit 3 · Video 5 · Interactive Practice

Why the Starting Temperature Stops Mattering: Steady State and Transient in $dT/dt + kT = kT_e$

IKey Formulas

Formula	Name	What it gives
$\dfrac{dT}{dt} + kT = kT_e(t), \quad k > 0$	Conduction equation, standard linear form	Constant coefficient; the right-hand side is $kT_e$ , never $T_e$
$u = e^{\int k\,dt} = e^{kt}$	Integrating factor	$\left(e^{kt}T\right)' = kT_e(t)\,e^{kt}$
$T = e^{-kt}\displaystyle\int_0^t kT_e(t_1)\,e^{kt_1}\,dt_1 + T_0\,e^{-kt}$	Definite-integral solution of $T(0) = T_0$	Every quantity is data of the problem
$T_0\,e^{-kt} \to 0 \quad (t \to \infty)$	The transient	The only place $T_0$ appears; the first term is the steady state

Key Insight: The initial condition enters the solution in exactly one place, the transient. A positive conductivity kills it, so every chamber ends on the same steady state — and steady means the transient is gone, not that the curve is flat.

IIFrom Integrating Factor to the Two Terms

Five moves carry $dT/dt + kT = kT_e$ to a solution split into a part that stays and a part that dies.

Step 1

Step 1 — Standard linear form

\frac{dT}{dt} = k\left(T_e - T\right) \quad\Longrightarrow\quad \frac{dT}{dt} + kT = kT_e(t)

Units force the

k

on the right:

dT/dt

and

kT

are both degrees per unit time, while

T_e

alone is degrees and cannot match.

Step 2 — Integrating factor

\int k\,dt = kt \quad\Longrightarrow\quad u = e^{kt}

e^{kt}\frac{dT}{dt} + k\,e^{kt}T = kT_e(t)\,e^{kt} \quad\Longrightarrow\quad \left(e^{kt}T\right)' = kT_e(t)\,e^{kt}

The prime is now

d/dt

, and the right-hand side was multiplied too.

Step 3 — Integrate, constant outside

e^{kt}T = \int kT_e(t)\,e^{kt}\,dt + C

T = e^{-kt}\int kT_e(t)\,e^{kt}\,dt + C\,e^{-kt}

An indefinite integral is a whole family of functions, so it is not even clear whether a second arbitrary constant is hiding inside it.

Step 4 — Make it definite, from $0$ to $t$

T = e^{-kt}\int_0^t kT_e(t_1)\,e^{kt_1}\,dt_1 + C\,e^{-kt}

T(0) = e^{0}\underbrace{\int_0^0 kT_e(t_1)\,e^{kt_1}\,dt_1}_{=\;0} + C\,e^{0} = C = T_0

The upper limit is the time

t

, so the variable inside is renamed

t_1

. One definite function, and the constant is read off at

t = 0

Step 5 — Steady state and transient

T = \underbrace{e^{-kt}\int_0^t kT_e(t_1)\,e^{kt_1}\,dt_1}_{\text{steady state}} \;+\; \underbrace{T_0\,e^{-kt}}_{\text{transient}}

With

T_e \equiv 20

and

k = 1

the first term is

20\left(1 - e^{-t}\right)

— itself the solution that starts at

0\degree\text{C}

— and the second is

T_0e^{-t}

, which

k > 0

sends to zero.

IIIFour Chambers, One Curve

Chambers starting at $0$ , $10$ , $30$ and $40\degree\text{C}$ in a $20\degree\text{C}$ bath: how long until they agree?

Time 1.5 h

The one-percent mark sits at $t = \ln 100 / k \approx 4.6/k$ : a better-insulated wall (smaller $k$ ) makes the chambers remember their start proportionally longer, but never forever.

IVSteady Does Not Mean Constant

Change what the bath does: the steady state changes with it, while the same transient dies away.

Bath Te(t)

VQuiz Questions

Problem 1 · The Integrating Factor

Given: the constant-coefficient equation $\dfrac{dT}{dt} + kT = kT_e(t)$ , with $k$ constant — find the integrating factor $u$ .

✅ Correct! With

p = k

the exponent is

\int k\,dt = kt

, so

u = e^{kt}

and multiplying through gives

\left(e^{kt}T\right)' = kT_e(t)\,e^{kt}

❌ Check the sign of $p$ . Standard form is

y' + p\,y = q

, so here

p = +k

. Reading the coefficient off

dT/dt = -kT + kT_e

instead is exactly the sign trap the standard form exists to prevent.

❌ Only the coefficient of $T$ goes in the exponent.

T_e(t)

lives on the right-hand side; it is

q/k

, not part of

p

❌ Not quite. The integrating factor is

u = e^{\int p\,dt}

, where

p

is the coefficient of

T

once the equation is in standard form.

Show solution

The equation is already in standard linear form $T' + p\,T = q$ with

p = k \quad\text{(a constant)}, \qquad q = kT_e(t)

The integrating factor is $u = e^{\int p\,dt}$ , and no arbitrary constant is needed because only one $u$ is wanted:

u = e^{\int k\,dt} = e^{kt}

Multiplying both sides by $u$ collapses the left-hand side into a single derivative:

e^{kt}\frac{dT}{dt} + k\,e^{kt}T = kT_e(t)\,e^{kt} \quad\Longrightarrow\quad \left(e^{kt}T\right)' = kT_e(t)\,e^{kt}

Check by the product rule: $\left(e^{kt}T\right)' = e^{kt}T' + k\,e^{kt}T$ , which is the left-hand side.

Problem 2 · The Right-Hand Side

Given: Newton's law of cooling, $\dfrac{dT}{dt} = k\left(T_e - T\right)$ with $k > 0$ — write it in standard linear form.

✅ Correct! Expanding

k(T_e - T) = kT_e - kT

and moving

-kT

across gives

kT

on the left and

kT_e

on the right — the

k

appears on both sides, once with the unknown and once with the external value.

❌ The $k$ must appear on the right as well. Units settle it:

dT/dt

and

kT

are degrees per unit time, so

k

carries

1/\text{time}

;

T_e

by itself is degrees and cannot balance the equation.

❌ Both signs are flipped. That equation is

dT/dt = k(T - T_e)

: a warmer bath would now cool the chamber, and the

T_0

term would grow like

e^{+kt}

instead of dying away.

❌ The coefficient of $T$ is $k$ , not $1$ . You cannot drop the conductivity on the left while keeping it on the right.

❌ Not quite. Multiply out the right-hand side first:

k\left(T_e - T\right) = kT_e - kT

Show solution

Expand the right-hand side and collect the $T$ terms on the left:

\frac{dT}{dt} = kT_e - kT \quad\Longrightarrow\quad \frac{dT}{dt} + kT = kT_e

The units check. Both sides of a differential equation must carry the same units:

$dT/dt$ : degrees per unit time
$k$ : $1/\text{time}$ , so $kT$ is degrees per unit time ✅
$T_e$ : degrees only ❌ — it cannot sit alone on the right

The same shape holds for the diffusion twin, $\dfrac{dC}{dt} + k_1C = k_1C_e$ .

Problem 3 · A Chamber in a Fixed Bath

Given: a bath held at $T_e = 20\degree\text{C}$ , conductivity $k = 0.5$ per hour, and $T(0) = 60\degree\text{C}$ — find the temperature and evaluate it after two hours.

What is $T(t)$ ?

What is $T$ after $2$ hours?

✅ Correct! Steady state

20\left(1 - e^{-0.5t}\right)

plus transient

60e^{-0.5t}

collects to

20 + 40e^{-0.5t}

, and at

t = 2

the exponent is

-1

20 + 40(0.3679) = 34.7\degree\text{C}

❌ Build it as steady state plus transient. The steady-state term is

20\left(1 - e^{-0.5t}\right)

and the transient is

T_0e^{-0.5t} = 60e^{-0.5t}

; then collect the exponentials.

❌ That is the bath value plus the transient. The steady-state term is

20\left(1 - e^{-0.5t}\right)

, not

20

— its own

-20e^{-0.5t}

must be collected with the transient, leaving

20 + 40e^{-0.5t}

. Check at

t = 0

: your formula gives

80\degree\text{C}

, not

60

❌ Check the exponent. It is

-kt = -0.5(2) = -1

, so the factor is

e^{-1} = 0.3679

, not

e^{-2}

❌ The $60$ and the $20$ have swapped roles. The bath value

20

is the level the solution approaches;

40 = T_0 - T_e

is the initial gap that decays.

Show solution

Step 1 — the two terms. For a constant bath the steady-state integral can be done:

e^{-kt}\int_0^t kT_e\,e^{kt_1}\,dt_1 = T_e\,e^{-kt}\left[e^{kt_1}\right]_0^t = T_e\left(1 - e^{-kt}\right) = 20\left(1 - e^{-0.5t}\right)

The transient is $T_0e^{-kt} = 60e^{-0.5t}$ .

Step 2 — add and collect.

T = 20 - 20e^{-0.5t} + 60e^{-0.5t} = 20 + 40e^{-0.5t}

Check the start: $T(0) = 20 + 40 = 60\degree\text{C}$ ✅, and $T \to 20\degree\text{C}$ as $t \to \infty$ because $k > 0$ .

Step 3 — evaluate at $t = 2$ . The exponent is $-0.5(2) = -1$ :

T(2) = 20 + 40e^{-1} = 20 + 40(0.36788) = 34.7\degree\text{C}

Note the regrouping: $20 + 40e^{-0.5t}$ is not the steady-state/transient split. The steady state is $20 - 20e^{-0.5t}$ and the transient is $60e^{-0.5t}$ ; only the transient carries $T_0$ .

Problem 4 · A Bath That Never Settles

Given: $k = 1$ per hour and a swinging bath $T_e(t) = 20 + 8\sin t$ , so that $T = 20 + 4\left(\sin t - \cos t\right) - 16e^{-t} + T_0e^{-t}$ .

Which part of the solution carries $T_0$ ?

What happens as $t \to \infty$ ?

✅ Correct!

T_0

multiplies

e^{-t}

and nothing else, and

k > 0

kills it. What survives is

20 + 4\sqrt{2}\sin\!\left(t - \tfrac{\pi}{4}\right)

— steady, because the transient is gone, but far from constant.

❌ Look at where $T_0$ appears. The steady-state term

e^{-kt}\int_0^t kT_e(t_1)e^{kt_1}dt_1

is built from the bath alone; the initial condition enters only through

C = T_0

, multiplying

e^{-kt}

❌ Drop only what decays. Both

-16e^{-t}

and

T_0e^{-t}

vanish, leaving

20 + 4\sin t - 4\cos t

; now combine that into a single sine.

❌ Steady state does not mean constant. Only the exponentials die. The oscillation the bath drives is part of the steady-state term and lasts forever.

❌ The chamber cannot follow the whole swing. Conduction through the wall takes time:

4\sin t - 4\cos t

has amplitude

\sqrt{4^2 + 4^2} = 4\sqrt{2} \approx 5.66

, not

8

, and it peaks after the bath does.

Show solution

Where $T_0$ lives. The definite-integral solution is

T = \underbrace{e^{-kt}\int_0^t kT_e(t_1)\,e^{kt_1}\,dt_1}_{\text{steady state — the bath's doing}} + \underbrace{T_0\,e^{-kt}}_{\text{transient — the start's doing}}

The first term is the solution that starts at $0\degree\text{C}$ ; the second is the only place the initial condition appears. Here $k = 1 > 0$ , so $T_0e^{-t} \to 0$ .

The long run. Dropping every $e^{-t}$ term leaves

T \longrightarrow 20 + 4\sin t - 4\cos t

Write $4\sin t - 4\cos t = R\sin(t - \varphi)$ with $R = \sqrt{4^2 + 4^2} = 4\sqrt{2} \approx 5.657$ and $\tan\varphi = 4/4 = 1$ , so $\varphi = \pi/4$ :

T \longrightarrow 20 + 4\sqrt{2}\,\sin\!\left(t - \frac{\pi}{4}\right)

The chamber oscillates about the bath's mean with a smaller amplitude and a phase lag — a steady state that is never constant.

Solved: 0 / 4

Why the Starting Temperature Stops Mattering: Steady State and Transient in dT/dt+kT=kTedT/dt + kT = kT_edT/dt+kT=kTe​

IKey Formulas

IIFrom Integrating Factor to the Two Terms

IIIFour Chambers, One Curve

IVSteady Does Not Mean Constant

VQuiz Questions

Why the Starting Temperature Stops Mattering: Steady State and Transient in $dT/dt + kT = kT_e$