Differential-Equations · Unit 11 · Video 4 · Interactive Practice

Uniqueness: Why c₁y₁ + c₂y₂ Is Every Solution

IKey Formulas

Formula	Name	What it says
$y = y_0\,Y_1 + y_0'\,Y_2$	IVP at $0$ in one line	Solves $y(0) = y_0$ , $y'(0) = y_0'$ : the constants are the initial values
$p, q$ continuous $\Rightarrow$ exactly one $y$ with $y(x_0) = A,\ y'(x_0) = B$	Existence and uniqueness	A value and a slope at one point pin down the solution
$u = u_0\,Y_1 + u_0'\,Y_2$	Every solution is in the family	$u$ and the combination share initial values, so they are equal

Key Insight: Superposition (linearity) puts every $c_1y_1 + c_2y_2$ among the solutions; uniqueness (continuous $p, q$ ) leaves no solution outside them.

IIVisualization 1 — Read the Constants off the Initial Values

For $y'' - y = 0$ the solution with $y(0) = y_0$ , $y'(0) = y_0'$ is $y_0\cosh x + y_0'\sinh x$ , with nothing to solve.

y₀ 2 y₀′ 4

IIIVisualization 2 — Uniqueness Catches Any Solution

A solution $u$ , written any way at all, coincides with $c_1Y_1 + c_2Y_2$ exactly when $c_1 = u(0)$ and $c_2 = u'(0)$ .

Solution u c₁ 0 c₂ 0

IVQuiz Questions

Problem 1 · Writing Down the Answer

Given: $y'' + y = 0$ , $y(0) = 3$ , $y'(0) = -2$ — find the solution.

✅ Correct!

Y_1 = \cos x

Y_2 = \sin x

, so

y = y_0Y_1 + y_0'Y_2 = 3\cos x - 2\sin x

❌ Not quite. The value

y_0

goes with

Y_1

(value

1

0

) and the slope

y_0'

with

Y_2

— you swapped them.

❌ Check the sign. The coefficient of

Y_2 = \sin x

y'(0) = -2

itself.

❌ Not quite.

\cosh

and

\sinh

are the normalized pair of

y'' - y = 0

. For

y'' + y = 0

it is

\cos x

\sin x

Show solution

The normalized solutions of $y'' + y = 0$ at $0$ are $Y_1 = \cos x$ and $Y_2 = \sin x$ , so

y = 3\cos x + (-2)\sin x = 3\cos x - 2\sin x.

Check: $y(0) = 3$ , and $y' = -3\sin x - 2\cos x$ gives $y'(0) = -2$ .

Problem 2 · From cosh and sinh to Exponentials

Given: $y'' - y = 0$ , $y(0) = 1$ , $y'(0) = 3$ — write the solution as $c_1e^x + c_2e^{-x}$ .

✅ Correct!

\cosh x + 3\sinh x = \frac{e^x + e^{-x}}{2} + \frac{3e^x - 3e^{-x}}{2} = 2e^x - e^{-x}

❌ Not quite. The initial values are the coefficients of

\cosh x

and

\sinh x

, not of

e^x

and

e^{-x}

: this one has

y'(0) = 1 - 3 = -2

❌ Not quite. You swapped the coefficients: this has

y(0) = 1

but

y'(0) = -1 - 2 = -3

❌ Not quite.

\cosh

and

\sinh

carry a factor

\frac{1}{2}

: this function has

y(0) = 2

Show solution

y = 1\cdot\cosh x + 3\sinh x = \frac{e^x + e^{-x}}{2} + \frac{3e^x - 3e^{-x}}{2} = \frac{4e^x - 2e^{-x}}{2} = 2e^x - e^{-x}.

Check: $y(0) = 2 - 1 = 1$ and $y'(0) = 2 + 1 = 3$ . Solving $c_1 + c_2 = 1$ , $c_1 - c_2 = 3$ directly gives the same $c_1 = 2$ , $c_2 = -1$ .

Problem 3 · Zero Value, Zero Slope

Given: $u$ solves $y'' + p(x)\,y' + q(x)\,y = 0$ with $p, q$ continuous for all $x$ , and $u(2) = 0$ , $u'(2) = 0$ — what is $u$ ?

✅ Correct! The zero function is a solution with value

0

and slope

0

2

, and uniqueness allows only one such solution.

❌ Not quite.

(x - 2)^2

has the right value and slope at

2

, but uniqueness says only one solution does — and

0

already does.

❌ Not quite.

Y_2

has slope

1

at its normalization point, so a nonzero multiple of it cannot have slope

0

there.

❌ Not quite. Whatever continuous

p

and

q

are,

y = 0

solves the equation, and uniqueness then forces

u = 0

Show solution

The function $y \equiv 0$ solves every homogeneous linear equation and has $y(2) = 0$ , $y'(2) = 0$ . By the existence and uniqueness theorem (with $x_0 = 2$ , $A = B = 0$ ) there is only one such solution, so $u \equiv 0$ .

A solution that touches the axis with zero slope therefore cannot be a nonzero solution — the graph of a nonzero solution crosses the axis with nonzero slope.

Problem 4 · Initial Values at $x_0 = \pi$

Given: $y'' + y = 0$ , $y(\pi) = 2$ , $y'(\pi) = -1$ — find the solution, using the pair normalized at $\pi$ : $\cos(x - \pi)$ and $\sin(x - \pi)$ .

✅ Correct!

2\cos(x - \pi) - \sin(x - \pi) = -2\cos x + \sin x

; check

y(\pi) = 2

y'(\pi) = 2\sin\pi + \cos\pi = -1

❌ Not quite.

\cos x

and

\sin x

are normalized at

0

, not at

\pi

: this function has

y(\pi) = -2

❌ Check a sign.

\sin(x - \pi) = -\sin x

, so

-\sin(x - \pi) = +\sin x

❌ Not quite. The value goes with

\cos(x - \pi)

and the slope with

\sin(x - \pi)

; this function has

y(\pi) = -1

Show solution

$Y_1 = \cos(x - \pi)$ has value $1$ , slope $0$ at $\pi$ ; $Y_2 = \sin(x - \pi)$ has value $0$ , slope $1$ at $\pi$ . So

y = 2\cos(x - \pi) + (-1)\sin(x - \pi) = -2\cos x + \sin x,

using $\cos(x - \pi) = -\cos x$ and $\sin(x - \pi) = -\sin x$ . Check: $y(\pi) = -2(-1) + 0 = 2$ and $y' = 2\sin x + \cos x$ , so $y'(\pi) = -1$ .

Solved: 0 / 4