Chapter 5 · Continuity and Differentiability

Exercise 5.7 · Q11: $y=5\cos x-3\sin x$ satisfies $y''+y=0$

Exercise 5.7Second order derivative · Section 5.7

The question

11. If $y = 5\cos x - 3\sin x$ ⁠, prove that $\dfrac{d^2y}{dx^2} + y = 0$ ⁠.

Exam answer

Here $y = 5\cos x - 3\sin x$ ⁠.

\begin{aligned} \frac{dy}{dx} &= -5\sin x - 3\cos x \\ \frac{d^2y}{dx^2} &= -5\cos x + 3\sin x \\ &= -(5\cos x - 3\sin x) \\ &= -y \end{aligned}

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

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

The one idea this problem is built on

Differentiating $\sin$ or $\cos$ twice brings back the same function with a minus sign:

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

So any combination of $\cos x$ and $\sin x$ satisfies $\frac{d^2y}{dx^2} = -y$ ⁠. This is Example 36 of the book with $A = -3$ and $B = 5$ ⁠.

Given, to show, and the plan

Given: $y = 5\cos x - 3\sin x$ ⁠, defined and differentiable (any number of times) for every real $x$ ⁠.
To show: $\dfrac{d^2y}{dx^2} + y = 0$ for every real $x$ ⁠.
Plan: (1) Find $\frac{dy}{dx}$ ⁠. (2) Find $\frac{d^2y}{dx^2}$ ⁠. (3) Compute the left side $\frac{d^2y}{dx^2} + y$ in full and see that it is $0$ ⁠, the right side.

Basics you need

1Derivatives of $\sin x$ and $\cos x$

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

2Sum rule and constant multiples

$(u \pm v)' = u' \pm v'$ (Section 5.3, rule (1)) and $(cu)' = c\,u'$ for a constant $c$ (Class XI, Limits and Derivatives). So we differentiate term by term and keep the numbers 5 and 3 in front.

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

Term by term, using Basic 1:
$\begin{aligned} \frac{dy}{dx} &= 5(-\sin x) - 3\cos x \\ &= -5\sin x - 3\cos x. \end{aligned}$
The constants 5 and 3 simply stay as factors. The derivative of $\cos x$ carries a minus sign, so the first term becomes $-5\sin x$ ⁠; the second term, $-3\sin x$ ⁠, becomes $-3\cos x$ ⁠.

Table 5.3; Section 5.3, rule (1). Basics: derivative rules
Second derivative

Differentiate $\frac{dy}{dx}$ again, term by term:
$\begin{aligned} \frac{d^2y}{dx^2} &= -5\cos x - 3(-\sin x) \\ &= -5\cos x + 3\sin x. \end{aligned}$
$-5\sin x$ gives $-5\cos x$ ⁠. $-3\cos x$ gives $-3\cdot(-\sin x) = +3\sin x$ ⁠: two minus signs make a plus.

Section 5.7: the second derivative is the derivative of $\frac{dy}{dx}$ ⁠. Lesson: second order derivative
Compare with $y$

Take out a factor $-1$ ⁠:
$\frac{d^2y}{dx^2} = -(5\cos x - 3\sin x) = -y.$
The bracket is exactly the given $y$ ⁠. Each term has turned into its own negative after two differentiations, which is the idea of this problem.
Compute the left side in full
$\begin{aligned} \frac{d^2y}{dx^2} + y &= (-5\cos x + 3\sin x) \\ &\quad + (5\cos x - 3\sin x) \\ &= (-5 + 5)\cos x + (3 - 3)\sin x \\ &= 0. \end{aligned}$
The right side of the required relation is $0$ ⁠, so $\dfrac{d^2y}{dx^2} + y = 0$ for every real $x$ ⁠.

To prove an identity we work out the left side completely and reach the right side. Collecting the $\cos x$ terms and the $\sin x$ terms separately shows each coefficient is exactly zero, so the result holds at every $x$ ⁠, not just at some.

Result

$\dfrac{dy}{dx} = -5\sin x - 3\cos x$ and $\dfrac{d^2y}{dx^2} = -5\cos x + 3\sin x = -y$ ⁠, so

\frac{d^2y}{dx^2} + y = 0 \quad\text{for every real } x.

(The answer key gives no entry for this "prove that" question.)

Check it with numbers

Take $x = 1$ (radian): $\cos 1 \approx 0.5403023$ ⁠, $\sin 1 \approx 0.8414710$ ⁠. Then

\begin{aligned} y &\approx 2.701512 - 2.524413 \approx 0.177099, \\ \frac{d^2y}{dx^2} &\approx -2.701512 + 2.524413 \\ &\approx -0.177099, \end{aligned}

(using $5\cos 1 \approx 2.701512$ and $3\sin 1 \approx 2.524413$ ⁠), and the sum is $0$ ⁠. From values of $y$ alone, with $h = 0.001$ ⁠, the second difference $\frac{y(1.001) - 2y(1) + y(0.999)}{h^2}$ is about $-0.177099$ ⁠, matching.

Common mistakes

Sign of $\cos$ 's derivative. $\frac{d}{dx}\cos x = -\sin x$ ⁠, not $\sin x$ ⁠. With that slip the second derivative comes out as $+y$ and the proof "fails".
Losing the minus in front of $3\sin x$ ⁠. It is part of the term: $\frac{d}{dx}(-3\sin x) = -3\cos x$ ⁠.
Checking at one point only. A numerical check is not a proof; the algebra in Step 4 shows the coefficients cancel for all $x$ ⁠.

What to take away

For every pair of constants $A, B$ ⁠, $y = A\sin x + B\cos x$ satisfies $y'' + y = 0$ (Example 36); this question is $A = -3$ ⁠, $B = 5$ ⁠.
To "prove that" a relation holds, compute each derivative, substitute, and collect like terms until the left side visibly equals the right side.

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 (Basics: derivative rules)
$(u \pm v)' = u' \pm v'$ ⁠; $(cu)' = cu'$	Steps 1, 2	Section 5.3, rule (1) (sum); Class XI, Limits and Derivatives (constant multiple)
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$	Step 2	Section 5.7 (Lesson: second order derivative)
$A\sin x + B\cos x$ satisfies $y'' + y = 0$	Idea, take-away	Example 36, Section 5.7

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Exercise 5.7 · Q11: y=5cos⁡x−3sin⁡xy=5\cos x-3\sin xy=5cosx−3sinx satisfies y′′+y=0y''+y=0y′′+y=0

Basics you need

Solution

First derivative

Second derivative

Compare with yyy

Compute the left side in full

Every fact used, and where it comes from

Exercise 5.7 · Q11: $y=5\cos x-3\sin x$ satisfies $y''+y=0$

Compare with $y$