Differential-Equations · Unit 7 · Video 2 · Interactive Practice

Input, Response and the Superposition Principle

IKey Formulas

Formula	Name	What it says
$y' + ky = q(t)$ , $q \leadsto y$	Input and response	The right-hand side $q$ is the input; the solution $y$ is the system's response
$y' + ky = k\,q_e(t)$	Physical input $q_e$	$q_e$ has the units of $y$ ; the input $q = kq_e$ has units of $y$ per unit time
$c_1q_1 + c_2q_2 \leadsto c_1y_1 + c_2y_2$	Superposition principle	Holds for any constants $c_1$ , $c_2$ whenever $q_1 \leadsto y_1$ and $q_2 \leadsto y_2$
$(c_1y_1 + c_2y_2)' + k(c_1y_1 + c_2y_2) = c_1q_1 + c_2q_2$	The one-line proof	Differentiate term by term, then regroup into $c_1(y_1' + ky_1) + c_2(y_2' + ky_2)$

Key Insight: Superposition is a consequence of linearity alone: the left-hand side of a combination of functions is the same combination of their left-hand sides. Put $y^2$ in place of $y$ and the cross term $2k\,y_1y_2$ appears, so the sum of two responses is no longer a response to the sum of the inputs.

IIThe Physical Input Has the Units of the Response

For $y' + ky = k\,q_e$ with a constant bath $q_e = 1$ , the response levels off at $q_e$ , not at $q = kq_e$ .

k 2

💡 Many books divide by $k$ to get $\frac{1}{k}\,y' + y = q_e(t)$ and call $q_e$ the input; the coefficient $\frac{1}{k}$ on $y'$ takes it out of standard linear form, so this course keeps $y' + ky = k\,q_e(t)$ .

IIIInputs Combine, Responses Follow

For $y' + 2y = q$ , if $1 \leadsto y_1$ and $e^{-2t} \leadsto y_2$ , then $c_1 + c_2e^{-2t} \leadsto c_1y_1 + c_2y_2$ .

c₁ 1 c₂ 1

💡 Every response here starts from rest, $y(0) = 0$ ; in general the combination $c_1y_1 + c_2y_2$ starts at $c_1y_1(0) + c_2y_2(0)$ .

IVA $y^2$ Breaks Superposition

In $y' + 2y^2 = q$ the sum $y_1 + y_2$ overshoots the response to $q_1 + q_2$ by way of the cross term $2k\,y_1y_2$ .

Equation t 1.00

VQuiz Questions

Problem 1 · Build a Response from Pieces

Given: for $y' + 2y = q(t)$ , the input $q_1 = 1$ has the response $y_1 = \tfrac{1}{2}$ and the input $q_2 = e^{-2t}$ has the response $y_2 = te^{-2t}$ — find a response to the input $q = 3 - 5e^{-2t}$ .

✅ Correct!

q = 3q_1 - 5q_2

, so superposition gives

3y_1 - 5y_2 = \tfrac{3}{2} - 5te^{-2t}

❌ Not quite. The constant

3

multiplies the response

y_1 = \tfrac{1}{2}

, not the input

q_1 = 1

: the first term is

3 \cdot \tfrac{1}{2} = \tfrac{3}{2}

. Substituting

3 - 5te^{-2t}

gives

y' + 2y = 6 - 5e^{-2t}

❌ Not quite. The coefficient

c_2 = -5

carries its sign to the response:

c_2y_2 = -5te^{-2t}

. This option responds to

3 + 5e^{-2t}

❌ Not quite. That is

y_1 + y_2

, the response to

1 + e^{-2t}

. The constants

c_1 = 3

and

c_2 = -5

must multiply the responses too.

❌ Not quite. Write

q = c_1q_1 + c_2q_2

and use

c_1y_1 + c_2y_2

Show solution

Write the input as a combination of the two known inputs:

q = 3 - 5e^{-2t} = 3\cdot q_1 + (-5)\cdot q_2, \qquad c_1 = 3,\ c_2 = -5.

By the superposition principle, $c_1q_1 + c_2q_2 \leadsto c_1y_1 + c_2y_2$ :

y = 3\cdot\tfrac{1}{2} - 5\,te^{-2t} = \tfrac{3}{2} - 5te^{-2t}.

Check: $y' = -5e^{-2t} + 10te^{-2t}$ and $2y = 3 - 10te^{-2t}$ , so

y' + 2y = 3 - 5e^{-2t}\ \checkmark

Problem 2 · Units of the Input

Given: the temperature model $T' + kT = kT_e$ , with $T$ and $T_e$ in degrees Celsius and $t$ in minutes, so that $k$ is in $1/\text{min}$ — find the units of the input $q = kT_e$ .

✅ Correct!

q = kT_e

has units

(1/\text{min})\cdot{}^\circ\text{C}

, a rate of temperature change, just like

T'

❌ Not quite. Degrees Celsius are the units of the physical input

T_e

and of the response

T

. The input

q

carries the extra factor

k

, in

1/\text{min}

❌ Not quite. Per minute is the unit of

k

alone;

q

k

times a temperature.

❌ Not quite. Minutes are the units of

t

and of

1/k

;

q

k

times a temperature.

Show solution

Every term of an equation has the same units. $T'$ is a temperature per unit time, in ${}^\circ\text{C}/\text{min}$ , so $kT$ and $kT_e$ are too. Directly:

[q] = [k]\,[T_e] = \frac{1}{\text{min}}\cdot{}^\circ\text{C} = {}^\circ\text{C}/\text{min}.

The physical input $T_e$ is in ${}^\circ\text{C}$ , the same as the response $T$ . That shared unit is why the course calls $q_e$ the physical input: it can be compared with $y$ directly, while $q$ cannot.

Problem 3 · Where the Cross Term Shows Up

Given: for the nonlinear equation $y' + y^2 = q(t)$ (so $k = 1$ ), $y_1 = t$ is the response to $q_1 = 1 + t^2$ and $y_2 = 1$ is the response to $q_2 = 1$ — find the input whose response is $y_1 + y_2 = t + 1$ .

✅ Correct!

(t+1)' + (t+1)^2 = t^2 + 2t + 2

, which exceeds

q_1 + q_2 = t^2 + 2

by the cross term

2k\,y_1y_2 = 2t

❌ Not quite.

t^2 + 2

q_1 + q_2

, the answer superposition would give, and superposition fails here: the square of

y_1 + y_2

contains the cross term

2y_1y_2 = 2t

❌ Not quite.

(y_1 + y_2)^2 = y_1^2 + 2y_1y_2 + y_2^2

: the cross term is

2t

, not

t

❌ Not quite.

t^2 + 2t + 1

(t+1)^2

alone; the input also includes the derivative

(t+1)' = 1

❌ Not quite. Substitute

y = t + 1

into the left-hand side

y' + y^2

Show solution

The input that produces a given function is its left-hand side. Substitute $y = t + 1$ :

y' + y^2 = 1 + (t+1)^2 = 1 + t^2 + 2t + 1 = t^2 + 2t + 2.

Compare with the sum of the inputs, $q_1 + q_2 = (1 + t^2) + 1 = t^2 + 2$ . The difference is

(t^2 + 2t + 2) - (t^2 + 2) = 2t = 2k\,y_1y_2 \quad (k = 1,\ y_1 = t,\ y_2 = 1),

the cross term from $(y_1 + y_2)^2 = y_1^2 + 2y_1y_2 + y_2^2$ . So $q_1 + q_2$ does not produce $y_1 + y_2$ : superposition fails for this nonlinear equation.

Problem 4 · Scaling Fails Too

Given: for the nonlinear equation $y' + 2y^2 = q$ , the constant input $q = 2$ has the constant response $y = 1$ — find the positive constant response to the input $q = 8$ , four times as large.

✅ Correct! A constant

y

has

y' = 0

, so

2y^2 = 8

and

y = 2

: four times the input gives only twice the response.

❌ Not quite. "Four times the input gives four times the response" is the special case

c_2 = 0

of superposition, which needs a linear equation. Check:

2\cdot 4^2 = 32 \ne 8

❌ Not quite.

2\sqrt{2}

solves

y^2 = 8

; the equation is

2y^2 = 8

, with

k = 2

❌ Not quite.

\sqrt{2} = \sqrt{8}/2

divides by

k

outside the square root. Check:

2(\sqrt{2})^2 = 4 \ne 8

Show solution

A constant response has $y' = 0$ , so the equation reduces to $2y^2 = q$ .

For $q = 2$ : $2y^2 = 2$ gives $y = 1$ , as stated. For $q = 8$ :

2y^2 = 8 \;\Longrightarrow\; y^2 = 4 \;\Longrightarrow\; y = 2 \quad (y > 0).

The input was multiplied by $4$ , but the response only by $2 = \sqrt{4}$ . For a linear equation the scaling rule $c\,q \leadsto c\,y$ would give $y = 4$ ; the $y^2$ term breaks it, just as it breaks the sum rule.

Solved: 0 / 4

Input, Response and the Superposition Principle

IKey Formulas

IIThe Physical Input Has the Units of the Response

IIIInputs Combine, Responses Follow

IVA y2y^2y2 Breaks Superposition

VQuiz Questions

IVA $y^2$ Breaks Superposition