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Chapter 9 · Differential Equations

Exercise 9.2 · Q3: y=cos⁡x+Cy = \cos x + C solves y′+sin⁡x=0y' + \sin x = 0

Exercise 9.2Verifying a solution · Section 9.3
The question

In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

3. y=cos⁡x+Cy = \cos x + C  :  y′+sin⁡x=0y' + \sin x = 0

Exam answer

Given y=cos⁡x+Cy = \cos x + C⁠. Differentiating with respect to xx⁠,

y′=−sin⁡xy' = -\sin x

Substituting y′y' in the given equation,

L.H.S.=−sin⁡x+sin⁡x=0=R.H.S.\begin{aligned} \text{L.H.S.} &= -\sin x + \sin x \\ &= 0 = \text{R.H.S.} \end{aligned}

∴\therefore y=cos⁡x+Cy = \cos x + C is a solution of the given differential equation. Hence verified.

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

The one idea this problem is built on

The derivative of cos⁡x\cos x is −sin⁡x-\sin x⁠, and the constant CC contributes nothing. So y′=−sin⁡xy' = -\sin x exactly, and adding sin⁡x\sin x gives 0 for every xx and every CC⁠. The only real danger is the sign.

Given, to find, and the plan
Given
The function y=cos⁡x+Cy = \cos x + C⁠, where CC is an arbitrary constant, defined for every real xx⁠; the equation y′+sin⁡x=0y' + \sin x = 0⁠.
To show
That the function satisfies the equation for every real xx and every CC⁠.
Plan
(1) Find y′y'⁠. (2) Substitute and simplify to 0. (3) State where it holds. (4) Explain the role of CC⁠.

Basics you need

1What a solution is

A function is a solution when substituting it for yy makes L.H.S. = R.H.S. (Section 9.3). Lesson: general and particular solutions

2Derivative of cosine

ddxcos⁡x=−sin⁡x\frac{d}{dx}\cos x = -\sin x⁠, with a minus sign; a constant has derivative 0 (Chapter 5). Basics: y′ and y″

3Arbitrary constants

A general solution carries as many arbitrary constants as the order (Summary, Chapter 9). Basics: order n gives n constants

Solution

  1. Differentiate

    y=cos⁡x+C⇒ y′=−sin⁡x+0=−sin⁡x.\begin{aligned} y &= \cos x + C \\ \Rightarrow\ y' &= -\sin x + 0 \\ &= -\sin x. \end{aligned}
    cos⁡x\cos x gives −sin⁡x-\sin x⁠; the constant CC gives 0. The equation contains only y′y'⁠, so one derivative is enough.
    (cos⁡x)′=−sin⁡x(\cos x)' = -\sin x (Chapter 5). Basics: y′ and y″

  2. Substitute into the left side

    L.H.S.=y′+sin⁡x=−sin⁡x+sin⁡x=0=R.H.S.\begin{aligned} \text{L.H.S.} &= y' + \sin x \\ &= -\sin x + \sin x \\ &= 0 = \text{R.H.S.} \end{aligned}
    Put −sin⁡x-\sin x in place of y′y'⁠. The two sine terms are equal and opposite, so they cancel, leaving the right side 0.
    Section 9.3: substitute and compare the two sides. Lesson: is it a solution? Examples 2 and 3

  3. Where it holds

    No step divides, takes a root or a log, and cos⁡x+C\cos x + C is differentiable for every real xx⁠. The value of CC never entered. So L.H.S. = R.H.S. for every real xx and every CC⁠: hence verified.

    A verification must hold identically in xx⁠; keeping CC as a letter makes it hold for the whole family at once.
    A solution satisfies the equation at every xx of its domain (Section 9.3). Lesson: is it a solution?

  4. The constant CC⁠: a general solution

    The equation y′=−sin⁡xy' = -\sin x has order 1 and degree 1 (a polynomial equation in y′y'⁠, with y′y' to the power 1). The function has one arbitrary constant CC⁠, as many as the order, so it is a general solution. For example C=0C = 0 gives the particular solution y=cos⁡xy = \cos x⁠, and C=2C = 2 gives y=cos⁡x+2y = \cos x + 2⁠.

    Integrating y′=−sin⁡xy' = -\sin x gives y=cos⁡x+Cy = \cos x + C⁠, so every solution has this form.

    Each value of CC shifts the cosine curve up or down without changing any slope, which is why CC disappears from y′y'⁠.
    One integration, one constant (Section 9.3 and the Summary). Basics: order n gives n constants
Result

With y′=−sin⁡xy' = -\sin x⁠, L.H.S. =−sin⁡x+sin⁡x=0== -\sin x + \sin x = 0 = R.H.S. for every real xx and every CC⁠. Hence y=cos⁡x+Cy = \cos x + C is a solution of y′+sin⁡x=0y' + \sin x = 0 (its general solution). (The answer key gives no answer for this question, since it is a verification.)

Check it with numbers

At x=1x = 1⁠: sin⁡1≈0.841471\sin 1 \approx 0.841471⁠, so y′=−sin⁡1≈−0.841471y' = -\sin 1 \approx -0.841471 and y′+sin⁡x≈−0.841471+0.841471=0y' + \sin x \approx -0.841471 + 0.841471 = 0⁠.

From the curve itself, with cos⁡1.001≈0.539461\cos 1.001 \approx 0.539461 and cos⁡0.999≈0.541144\cos 0.999 \approx 0.541144 (any CC cancels in the difference):

0.539461−0.5411440.002≈−0.8415,\frac{0.539461 - 0.541144}{0.002} \approx -0.8415,

which is −sin⁡1-\sin 1 to four decimals.

Common mistakes
  1. Sign of (cos⁡x)′(\cos x)'⁠: it is −sin⁡x-\sin x⁠. Writing +sin⁡x+\sin x gives 2sin⁡x≠02\sin x \ne 0 and the check fails.
  2. Differentiating CC to anything but 0.
  3. Checking at x=0x = 0 only, where sin⁡x=0\sin x = 0 makes almost anything pass.
What to take away
  1. (cos⁡x)′=−sin⁡x(\cos x)' = -\sin x and (sin⁡x)′=cos⁡x(\sin x)' = \cos x⁠: the minus sign belongs to cosine.
  2. Same structure as Q2: one constant added to yy⁠, one order, a general solution.

Every fact used, and where it comes from

FactUsed inSource
(cos⁡x)′=−sin⁡x(\cos x)' = -\sin x⁠; derivative of a constant is 0Step 1Chapter 5 (Basics)
A solution makes L.H.S. = R.H.S. when substitutedSteps 2, 3Section 9.3 (Lesson)
Order and degree (polynomial condition)Step 4Sections 9.2.1 and 9.2.2 (Lesson)
General solution has as many constants as the orderStep 4Section 9.3 and the Summary of Chapter 9 (Basics)
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