Differential-Equations · Unit 7 · Video 1 · Interactive Practice

Steady-State and Transient Solutions of y' + ky = q(t)

IKey Formulas

Formula	Name	What it says
$y' + ky = q(t)$ , often $y' + ky = k\,q_e(t)$	Constant-coefficient linear equation	Temperature $T' + kT = kT_e(t)$ , concentration and mixing take the second form; the RC circuit and radioactive decay do not
$y = e^{-kt}\displaystyle\int q(t)\,e^{kt}\,dt + c\,e^{-kt}$	General solution	Integrating factor $e^{kt}$ ; for $q = 1$ the exponentials cancel and leave $\tfrac{1}{k} + c\,e^{-kt}$
$y = \underbrace{e^{-kt}\displaystyle\int_0^t q(u)\,e^{ku}\,du}_{\text{steady state}} + \underbrace{y_0\,e^{-kt}}_{\text{transient}}$	Initial-value solution	For $k > 0$ the transient carries $y_0$ and tends to $0$ ; the steady state does not involve $y_0$
$y_1 - y_2 = (c_1 - c_2)\,e^{-kt} \to 0$	Any two solutions ( $k > 0$ )	Every solution approaches every other, so the steady-state solution is not unique

Key Insight: Any solution could serve as the steady state. The one called the steady-state solution is the simplest, usually the one from integrating with no arbitrary constant, and the terminology only makes sense when $k > 0$ .

IISteady State Plus Transient

For $k > 0$ , every solution is the steady state $S(t)$ plus a transient $c\,e^{-kt}$ that decays to zero.

time t 3

IIIAny Solution Can Be the Steady State

In the bath equation $y' + 0.5y = 10$ , every solution is approached by all the others.

Reference solution

IVWhy $k$ Must Be Positive

Steady state and transient are meaningful only when $e^{-kt}$ decays, that is, when $k > 0$ .

k 0.5

VQuiz Questions

Problem 1 · Split a Solution

Given: $y' + 3y = 6$ with $y(0) = 5$ — find the simplest steady-state solution and the transient that goes with it.

What is the steady-state solution?

What is the transient?

✅ Correct!

y = 2 + 3e^{-3t}

: the constant

2 = 6/3

ignores

y_0

, and

3e^{-3t}

carries the initial offset

5 - 2

and dies away.

❌ Close, but…

6

q

, not the solution. A constant right-hand side

q

produces the constant

q/k

❌ Not quite.

5

is the initial value, and the steady state is exactly the part that does not involve

y_0

❌ Check the steady state. Try a constant

y = A

: then

3A = 6

❌ Close, but…

5e^{-3t}

is the transient that pairs with the definite-integral steady state

2(1 - e^{-3t})

, not with

2

❌ Check the transient. It has the form

c\,e^{-3t}

, and

c

is fixed by

y(0) = 2 + c = 5

Show solution

With the integrating factor $e^{3t}$ and no arbitrary constant in the integral:

y = e^{-3t}\int 6e^{3t}\,dt + c\,e^{-3t} = e^{-3t}\cdot 2e^{3t} + c\,e^{-3t} = 2 + c\,e^{-3t}.

The condition $y(0) = 5$ gives $2 + c = 5$ , so $c = 3$ and

y = \underbrace{2}_{\text{steady state}} + \underbrace{3e^{-3t}}_{\text{transient}}.

Check: $y' + 3y = -9e^{-3t} + 6 + 9e^{-3t} = 6$ . The definite form splits the same function differently, $2(1 - e^{-3t}) + 5e^{-3t}$ ; the constant $2$ is the simpler steady state.

Problem 2 · Which Steady State?

Given: the bath equation $y' + ky = 20k$ with $k > 0$ . The definite integral from $0$ to $t$ gives $20(1 - e^{-kt})$ , and the indefinite integral with no constant gives $20$ — which of these are steady-state solutions?

✅ Correct! Any two solutions differ by a transient, so every solution has the property that all others approach it;

20

is the simplest choice.

❌ Not quite. The lower limit

0

is a choice, not a law. Any solution that all others approach qualifies, and

20 - 20(1 - e^{-kt}) = 20e^{-kt} \to 0

❌ Not quite. Steady states need not be constant; in general they follow

q(t)

. Here

20(1 - e^{-kt})

also qualifies, since the two differ by a decaying

20e^{-kt}

❌ Roles reversed. The transient carries

y_0

; the steady state does not involve the initial value at all.

Show solution

Definite integral from $0$ to $t$ :

e^{-kt}\int_0^t 20k\,e^{ku}\,du = e^{-kt}\cdot 20\left(e^{kt} - 1\right) = 20\left(1 - e^{-kt}\right).

Indefinite integral with no arbitrary constant:

e^{-kt}\int 20k\,e^{kt}\,dt = e^{-kt}\cdot 20e^{kt} = 20.

Both satisfy $y' + ky = 20k$ , and their difference is

20 - 20\left(1 - e^{-kt}\right) = 20e^{-kt} \to 0 \quad (k > 0),

a transient. Every solution approaches both, so both are steady-state solutions; the one usually meant is the simplest, $y = 20$ .

Problem 3 · Starting at 127

Given: $y' + 0.5y = 10$ (a bath held at $20$ , $k = 0.5$ ). One solution starts at $y(0) = 127$ , another at $y(0) = 20$ — when are they exactly $1$ apart?

✅ Correct! The gap is the transient difference

107e^{-0.5t}

, and it falls to

1

t = 2\ln 107

❌ Close, but… The gap between the two solutions is

(127 - 20)e^{-0.5t}

: the shared steady state

20

cancels.

❌ Check the last step. From

0.5t = \ln 107

you divide by

0.5

, which doubles

\ln 107

❌ Not quite. The exponent is

-0.5t

, not

-t

; the factor

k = 0.5

cannot be dropped.

Show solution

Every solution is $y = 20 + c\,e^{-0.5t}$ with $c = y(0) - 20$ :

y_1 = 20 + 107e^{-0.5t}, \qquad y_2 = 20.

Their difference is a transient: $y_1 - y_2 = 107e^{-0.5t}$ . Setting it equal to $1$ :

e^{-0.5t} = \frac{1}{107} \;\Rightarrow\; 0.5t = \ln 107 \;\Rightarrow\; t = 2\ln 107 \approx 9.35.

The solution that starts at $127$ comes down and stays above $20$ , closing in on it as $t \to \infty$ .

Problem 4 · When k Is Negative

Given: $y' - 0.2y = 3\sin t$ . Two solutions start $1$ apart at $t = 0$ — how far apart are they at $t = 10$ , and does a steady state make sense?

✅ Correct! Here

k = -0.2

, so the difference

(c_1 - c_2)e^{0.2t}

grows:

e^{2}

t = 10

❌ Watch the sign of $k$ . Matching

y' + ky

with

y' - 0.2y

gives

k = -0.2

, so

e^{-kt} = e^{0.2t}

grows.

❌ Not quite. The difference is

(c_1 - c_2)e^{-kt}

, which stays constant only when

k = 0

❌ Check the exponent.

e^{0.2}

is the factor at

t = 1

; at

t = 10

the exponent is

0.2 \cdot 10 = 2

Show solution

In the form $y' + ky = q(t)$ this equation has $k = -0.2$ and $q(t) = 3\sin t$ . Any two solutions share the first term and differ only in their constants:

y_1 - y_2 = (c_1 - c_2)\,e^{-kt} = (c_1 - c_2)\,e^{0.2t}.

At $t = 0$ the difference is $c_1 - c_2 = 1$ , so at $t = 10$ it is

e^{0.2 \cdot 10} = e^{2} \approx 7.39.

The term carrying the initial value grows instead of dying away, the solutions fan apart, and the words steady state and transient are inappropriate.

Solved: 0 / 4

Steady-State and Transient Solutions of y' + ky = q(t)

IKey Formulas

IISteady State Plus Transient

IIIAny Solution Can Be the Steady State

IVWhy kkk Must Be Positive

VQuiz Questions

IVWhy $k$ Must Be Positive