Differential-Equations · Unit 11 · Video 1 · Interactive Practice

Superposition: Why Linearity Makes c₁y₁ + c₂y₂ a Solution

IKey Formulas

Formula	Name	What it says
$Ly = 0,\quad L = D^2 + p(x)\,D + q(x)$	Operator form	The ODE $y'' + p\,y' + q\,y = 0$ is one linear operator applied to $y$
$L(u_1 + u_2) = L(u_1) + L(u_2),\quad L(cu) = c\,L(u)$	Linearity laws	Sums and constant multiples pass through $L$
$L(c_1y_1 + c_2y_2) = c_1L(y_1) + c_2L(y_2) = 0$	Superposition	Any linear combination of solutions is a solution
$y_2 \neq c\,y_1$ and $y_1 \neq c'\,y_2$	Independence	Both conditions, so $y_1 = 0$ is never half of an independent pair

Key Insight: The proof uses nothing about $p$ , $q$ or the order of the equation — only that $L$ is linear and the right side is $0$ .

IIVisualization 1 — Apply $L$ to a Combination

For a homogeneous equation every combination $c_1y_1 + c_2y_2$ is sent to $0$ ; with a nonzero right side it is not.

Equation c₁ 3/2 c₂ −3/4

IIIVisualization 2 — Which Operators Are Linear?

An operator is linear when both laws hold for every input; a single $y^2$ or $y\,y'$ term breaks them.

Operator Law c 2

IVQuiz Questions

Problem 1 · Feeding $x^2$ into the Box

Given: $L = D^2 + x\,D + 1$ , so $L(y) = y'' + x\,y' + y$ — find $L(x^2)$ .

✅ Correct!

2 + x\cdot 2x + x^2 = 2 + 3x^2

— the video's

2 + 2xp + x^2q

with

p = x

q = 1

❌ Not quite. You dropped the last term:

q = 1

multiplies

y = x^2

itself, adding another

x^2

❌ Not quite. The

D^2

term is applied to

x^2

too:

(x^2)'' = 2

must be added.

❌ Not quite.

p = x

multiplies

y' = 2x

; it is not added to it.

x\,D(x^2) = x\cdot 2x = 2x^2

Show solution

With $y = x^2$ : $y' = 2x$ and $y'' = 2$ . Apply each term of $L$ :

L(x^2) = \underbrace{2}_{D^2y} + \underbrace{x\cdot 2x}_{x\,Dy} + \underbrace{x^2}_{1\cdot y} = 2 + 3x^2.

Computing the output is the easy direction; solving $Ly = 0$ asks which inputs give output $0$ .

Problem 2 · The Zero-Function Trap

Given: four pairs of functions — which pair is independent in the sense of the video (each is not a constant multiple of the other)?

✅ Correct!

x e^x / e^x = x

is not constant, so neither function is a constant multiple of the other.

❌ That is the excluded case.

e^x

is not a multiple of

0

, but

0 = 0\cdot e^x

is a multiple of

e^x

. Both conditions are required, and the second one fails.

❌ Not quite. Here

y_2

is a constant multiple of

y_1

, so every combination

c_1y_1 + c_2y_2

is just a multiple of

y_1

Show solution

Independent means $y_2 \neq c\,y_1$ and $y_1 \neq c'\,y_2$ for all constants.

$(0, e^x)$ : $0 = 0\cdot e^x$ , so $y_1$ is a multiple of $y_2$ — dependent.
$(e^x, 3e^x)$ and $(\sin x, -2\sin x)$ : $y_2 = 3y_1$ and $y_2 = -2y_1$ — dependent.
$(e^x, xe^x)$ : the ratio $x$ is not constant and neither function is $0$ — independent.

Problem 3 · Which Combinations Solve It?

Given: $y_1 = x$ and $y_2 = x^2$ both solve $y'' - \frac{2}{x}\,y' + \frac{2}{x^2}\,y = 0$ for $x > 0$ — for which constants is $c_1x + c_2x^2$ a solution?

✅ Correct!

L

is linear even though

p = -2/x

and

q = 2/x^2

vary, so

L(c_1x + c_2x^2) = c_1\cdot 0 + c_2\cdot 0 = 0

❌ Not quite.

c_1 + c_2 = 1

is the condition for an equation with a nonzero right side

f

, where

L(c_1y_1 + c_2y_2) = (c_1 + c_2)f

. Here the right side is

0

❌ Not quite. The superposition proof never used constant coefficients — only the two linearity laws, which hold for any functions

p(x)

q(x)

❌ Not quite. The sum law lets you add the two solutions as well:

L(c_1x + c_2x^2) = L(c_1x) + L(c_2x^2)

Show solution

Check the two solutions: $L(x) = 0 - \frac{2}{x}\cdot 1 + \frac{2}{x^2}\cdot x = 0$ and $L(x^2) = 2 - \frac{2}{x}\cdot 2x + \frac{2}{x^2}\cdot x^2 = 2 - 4 + 2 = 0$ .

Then by the sum and constant-multiple laws,

L(c_1x + c_2x^2) = c_1L(x) + c_2L(x^2) = 0

for every pair of constants. For instance $3x - 5x^2$ is a solution.

Problem 4 · When the Right Side Is Not Zero

Given: $L$ is linear and $y_1$ , $y_2$ both solve $Ly = f(x)$ with $f \neq 0$ — find $L(2y_1 - 3y_2)$ .

✅ Correct!

2f - 3f = -f

: the combination solves

Ly = f

only when the coefficients add up to

1

❌ Not quite. The output is

0

only for a homogeneous equation. Here

L(y_1) = L(y_2) = f

, not

0

❌ Check the sign. The constant-multiple law brings out

-3

, not

+3

L(-3y_2) = -3\,L(y_2) = -3f

❌ Not quite. Linearity gives

2L(y_1) - 3L(y_2)

, so

f

is multiplied by

2 - 3

, not left as is.

Show solution

L(2y_1 - 3y_2) = 2\,L(y_1) - 3\,L(y_2) = 2f - 3f = -f.

In general $L(c_1y_1 + c_2y_2) = (c_1 + c_2)f$ . Superposition in the form “every combination is a solution” uses the $0$ on the right side.

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Superposition: Why Linearity Makes c₁y₁ + c₂y₂ a Solution

IKey Formulas

IIVisualization 1 — Apply LLL to a Combination

IIIVisualization 2 — Which Operators Are Linear?

IVQuiz Questions

IIVisualization 1 — Apply $L$ to a Combination