Differential-Equations · Unit 11 · Video 3 · Interactive Practice

Normalized Solutions: Why cosh x and sinh x Solve y″ − y = 0

IKey Formulas

Formula	Name	What it says
$Y_1(0) = 1,\ Y_1'(0) = 0;\qquad Y_2(0) = 0,\ Y_2'(0) = 1$	Normalized solutions at $0$	Defined only by their initial values
$y'' + y = 0:\quad Y_1 = \cos x,\ Y_2 = \sin x$	Already normalized	The standard pair needs no change
$y'' - y = 0:\quad Y_1 = \cosh x = \dfrac{e^x + e^{-x}}{2},\ \ Y_2 = \sinh x = \dfrac{e^x - e^{-x}}{2}$	Hyperbolic pair	Built from $e^{x}$ and $e^{-x}$ by fitting $c_1, c_2$
$(\cosh x)' = \sinh x,\quad (\sinh x)' = \cosh x$	Derivatives	Each is the derivative of the other

Key Insight: Any independent pair from the family $c_1y_1 + c_2y_2$ describes the same family; the normalized pair is the one chosen by its starting values.

IIVisualization 1 — Build $Y_1$ and $Y_2$ from $e^x$ and $e^{-x}$

Which constants make $c_1e^x + c_2e^{-x}$ start with the value and slope of $Y_1$ , or of $Y_2$ ?

Target c₁ 1 c₂ 0

IIIVisualization 2 — Standard Pair versus Normalized Pair

The pair found from the characteristic roots and the normalized pair span one family; only the second starts at $(1, 0)$ and $(0, 1)$ .

Equation Pair

IVQuiz Questions

Problem 1 · Normalizing $\sin 2x$

Given: $y'' + 4y = 0$ has solutions $\cos 2x$ and $\sin 2x$ — find the normalized solution $Y_2$ at $0$ .

✅ Correct!

\tfrac{1}{2}\sin 2x

0

0

and its derivative

\cos 2x

1

there.

❌ Check the slope.

(\sin 2x)' = 2\cos 2x

equals

2

0

, not

1

. Unlike

y'' + y = 0

, here the standard pair is not normalized.

❌ Not quite.

2\sin 2x

has slope

4

0

. Divide by the slope of

\sin 2x

instead of multiplying.

❌ Not quite.

\sin x

does not solve

y'' + 4y = 0

(\sin x)'' + 4\sin x = 3\sin x

Show solution

Write $Y_2 = c_1\cos 2x + c_2\sin 2x$ , so $Y_2' = -2c_1\sin 2x + 2c_2\cos 2x$ .

Y_2(0) = c_1 = 0,\qquad Y_2'(0) = 2c_2 = 1 \ \Rightarrow\ c_2 = \tfrac{1}{2}.

So $Y_2 = \tfrac{1}{2}\sin 2x$ (and $Y_1 = \cos 2x$ , which is already normalized).

Problem 2 · The Derivative of cosh

Given: $\cosh x = \frac{e^x + e^{-x}}{2}$ and $\sinh x = \frac{e^x - e^{-x}}{2}$ — which pair of derivative formulas is right?

✅ Correct! Differentiating

e^{-x}

flips its sign, which swaps the two numerators — no minus signs appear.

❌ Not quite. That minus sign belongs to

\cos

and

\sin

. Differentiate

\frac{e^x + e^{-x}}{2}

directly: you get

\frac{e^x - e^{-x}}{2}

❌ Not quite. Only

e^x

is its own derivative. The

e^{-x}

term changes sign, so

\cosh

turns into

\sinh

Show solution

(\cosh x)' = \frac{e^x - e^{-x}}{2} = \sinh x,\qquad (\sinh x)' = \frac{e^x + e^{-x}}{2} = \cosh x.

Hence $(\cosh x)'' = \cosh x$ and $(\sinh x)'' = \sinh x$ , so both solve $y'' - y = 0$ . At $0$ this also gives the initial values: $\cosh 0 = 1$ , $(\cosh)'(0) = \sinh 0 = 0$ , and $\sinh 0 = 0$ , $(\sinh)'(0) = \cosh 0 = 1$ .

Problem 3 · Normalizing $e^x$ and $e^{2x}$

Given: $y'' - 3y' + 2y = 0$ with solutions $e^x$ and $e^{2x}$ — find the normalized solutions $Y_1$ and $Y_2$ at $0$ .

What is $Y_1$ ?

What is $Y_2$ ?

✅ Correct!

Y_1 = 2e^x - e^{2x}

has value

1

and slope

2 - 2 = 0

;

Y_2 = e^{2x} - e^x

has value

0

and slope

2 - 1 = 1

❌ Check $Y_1$ . Solve

c_1 + c_2 = 1

(value) and

c_1 + 2c_2 = 0

(slope) for

y = c_1e^x + c_2e^{2x}

❌ Check $Y_2$ . Solve

c_1 + c_2 = 0

(value) and

c_1 + 2c_2 = 1

(slope).

Show solution

With $y = c_1e^x + c_2e^{2x}$ : $y(0) = c_1 + c_2$ and $y'(0) = c_1 + 2c_2$ .

$Y_1$ : $c_1 + c_2 = 1$ , $c_1 + 2c_2 = 0$ . Subtracting, $c_2 = -1$ , so $c_1 = 2$ : $\ Y_1 = 2e^x - e^{2x}$ .

$Y_2$ : $c_1 + c_2 = 0$ , $c_1 + 2c_2 = 1$ . Subtracting, $c_2 = 1$ , so $c_1 = -1$ : $\ Y_2 = e^{2x} - e^x$ .

Problem 4 · Same Family?

Given: the solutions of $y'' - y = 0$ are $c_1e^x + c_2e^{-x}$ — which pair does NOT describe the same family?

✅ Correct!

3\cosh x

is a constant multiple of

\cosh x

, so their combinations are only the multiples of

\cosh x

—

\sinh x

is missing.

❌ Not quite. These two are members of the family and neither is a constant multiple of the other, so they generate the whole family.

Show solution

A pair $u_1, u_2$ taken from the family describes the whole family exactly when it is independent.

$\cosh x, \sinh x$ : the normalized pair — independent.
$e^x + 2e^{-x}, e^x - e^{-x}$ : coefficient rows $(1, 2)$ and $(1, -1)$ are not proportional — independent.
$2e^x, -e^{-x}$ : multiples of $e^x$ and $e^{-x}$ separately — independent.
$\cosh x, 3\cosh x$ : dependent, so $c_1\cosh x + c_2\cdot 3\cosh x = (c_1 + 3c_2)\cosh x$ never gives $\sinh x$ .

Solved: 0 / 4

Normalized Solutions: Why cosh x and sinh x Solve y″ − y = 0

IKey Formulas

IIVisualization 1 — Build Y1Y_1Y1​ and Y2Y_2Y2​ from exe^xex and e−xe^{-x}e−x

IIIVisualization 2 — Standard Pair versus Normalized Pair

IVQuiz Questions

IIVisualization 1 — Build $Y_1$ and $Y_2$ from $e^x$ and $e^{-x}$