Chapter 5 · Continuity and Differentiability

Example 36: $y=A\sin x+B\cos x$ satisfies $y''+y=0$

ExamplesSecond order derivative · Section 5.7

The question

If $y = A\sin x + B\cos x$ ⁠, then prove that $\dfrac{d^2y}{dx^2} + y = 0$ ⁠.

Exam answer

Here $y = A\sin x + B\cos x$ ⁠.

Differentiating w.r.t. $x$ ⁠,

\frac{dy}{dx} = A\cos x - B\sin x

Differentiating again,

\begin{aligned} \frac{d^2y}{dx^2} &= -A\sin x - B\cos x \\ &= -(A\sin x + B\cos x) = -y \end{aligned}

$\therefore\ \dfrac{d^2y}{dx^2} + y = 0$ ⁠. Hence proved.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Differentiating sine and cosine twice brings each one back with a minus sign:

\begin{aligned} \frac{d^2}{dx^2}\sin x &= -\sin x, \\ \frac{d^2}{dx^2}\cos x &= -\cos x. \end{aligned}

So the second derivative of any combination $A\sin x + B\cos x$ is minus that same combination, $-y$ ⁠, and adding $y$ gives zero.

Given, to show, and the plan

Given: $y = A\sin x + B\cos x$ ⁠, where $A$ and $B$ are constants (any real numbers) and $x$ is any real number.
To show: $\dfrac{d^2y}{dx^2} + y = 0$ for every real $x$ ⁠.
Plan: Differentiate twice, then recognise the second derivative as $-y$ ⁠.

Basics you need

1Derivatives of sine and cosine

$\frac{d}{dx}\sin x = \cos x$ and $\frac{d}{dx}\cos x = -\sin x$ (Table 5.3), for every real $x$ ⁠. Basics: derivative rules

2Constants and sums

$\frac{d}{dx}\big(A\,u + B\,v\big) = A\,u' + B\,v'$ when $A$ ⁠, $B$ are constants. A letter like $A$ is a constant because it does not change when $x$ changes. Basics: derivative rules

3Second order derivative (Section 5.7)

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$ ⁠. Lesson: second order derivative

Solution

First derivative
$\begin{aligned} \frac{dy}{dx} &= A\cos x + B(-\sin x) \\ &= A\cos x - B\sin x. \end{aligned}$
Each term is a constant times a standard function, so the constants stay in front (Basic 2) and we use Basic 1 for $\sin x$ and $\cos x$ ⁠.

$\frac{d}{dx}\cos x = -\sin x$ ⁠: this minus sign produces the $-B\sin x$ ⁠. Basics: derivative rules
Second derivative

Differentiate $\frac{dy}{dx}$ again:
$\begin{aligned} \frac{d^2y}{dx^2} &= \frac{d}{dx}\left(A\cos x - B\sin x\right) \\ &= A(-\sin x) - B\cos x \\ &= -A\sin x - B\cos x. \end{aligned}$
By definition the second derivative is the derivative of the first derivative (Basic 3). Again the constants $A$ and $B$ stay in front, $\cos x$ becomes $-\sin x$ ⁠, and $\sin x$ becomes $\cos x$ ⁠.

Section 5.7. Lesson: second order derivative
Recognise $-y$

Take out the common factor $-1$ ⁠:
$\frac{d^2y}{dx^2} = -\left(A\sin x + B\cos x\right) = -y.$
The bracket is exactly the given expression for $y$ ⁠. Seeing this is the point of the problem: the second derivative reproduces the original function, up to sign.

Factorising: $-p - q = -(p + q)$ ⁠.
Conclude

Add $y$ to both sides:
$\frac{d^2y}{dx^2} + y = -y + y = 0.$
Every step holds for every real $x$ and for any constants $A$ ⁠, $B$ ⁠, so $\dfrac{d^2y}{dx^2} + y = 0$ for all $x$ ⁠.

The statement to prove is an equation involving $\frac{d^2y}{dx^2}$ and $y$ ⁠. Once we know $\frac{d^2y}{dx^2} = -y$ ⁠, substituting it makes the left side zero. No special values of $x$ ⁠, $A$ or $B$ were used, so the result is general.

An equation linking a function and its derivatives is called a differential equation; you will meet them in Chapter 9. Here every function of the form $A\sin x + B\cos x$ is a solution of $y'' + y = 0$ ⁠.

Result

For $y = A\sin x + B\cos x$ ⁠: $\dfrac{dy}{dx} = A\cos x - B\sin x$ and $\dfrac{d^2y}{dx^2} = -A\sin x - B\cos x = -y$ ⁠. Hence $\dfrac{d^2y}{dx^2} + y = 0$ for all real $x$ ⁠, whatever the constants $A$ and $B$ ⁠.

Check it with numbers

Take $A = 2$ ⁠, $B = -3$ ⁠, $x = 1$ ⁠. Then $y = 2\sin 1 - 3\cos 1$ ⁠, which is about $1.6829420 - 1.6209069 = 0.0620351$ ⁠.

Estimate $\frac{d^2y}{dx^2}$ from three values with $h = 0.001$ ⁠: $y(0.999) \approx 0.0584300040285$ ⁠, $y(1) \approx 0.0620350520114$ ⁠, $y(1.001) \approx 0.0656400379592$ ⁠, so

\begin{aligned} &y(1.001) - 2y(1) + y(0.999) \\ &\approx -0.000000062035, \end{aligned}

and dividing by $h^2 = 0.000001$ gives $\frac{d^2y}{dx^2} \approx -0.062035$ ⁠.

So $\frac{d^2y}{dx^2} + y \approx -0.062035 + 0.062035 = 0$ to six decimal places, as proved.

Common mistakes

Sign errors. $\frac{d}{dx}\cos x = -\sin x$ but $\frac{d}{dx}\sin x = +\cos x$ ⁠. Mixing these up gives $\frac{d^2y}{dx^2} = y$ or other wrong forms.
Treating $A$ and $B$ as variables. They are constants; their derivatives are zero, and they simply stay in front.
Checking only special values. Showing the equation for $A = 1$ ⁠, $B = 0$ is not a proof. The algebra must keep $A$ and $B$ general.

What to take away

$\sin x$ and $\cos x$ each satisfy $y'' = -y$ ⁠, and so does every combination $A\sin x + B\cos x$ ⁠.
To "prove" a relation between $y$ and its derivatives: compute the derivatives, then substitute and simplify.
Exercise 5.7, Question 11 ( $y = 5\cos x - 3\sin x$ ⁠) is this example with $A = -3$ ⁠, $B = 5$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$\frac{d}{dx}\sin x = \cos x$ ⁠, $\frac{d}{dx}\cos x = -\sin x$	Steps 1, 2	Table 5.3, Section 5.3
Constant multiple and sum rules	Steps 1, 2	Algebra of derivatives, Section 5.3
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$	Step 2	Section 5.7 (Lesson: second order derivative)
Differential equations	Step 4	Chapter 9 (preview)

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Example 36: y=Asin⁡x+Bcos⁡xy=A\sin x+B\cos xy=Asinx+Bcosx satisfies y′′+y=0y''+y=0y′′+y=0

Basics you need

Solution

First derivative

Second derivative

Recognise −y-y−y

Conclude

Every fact used, and where it comes from

Example 36: $y=A\sin x+B\cos x$ satisfies $y''+y=0$

Recognise $-y$