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Chapter 9 · Differential Equations

Exercise 9.2 · Q8: y−cos⁡y=xy - \cos y = x solves (ysin⁡y+cos⁡y+x)y′=y(y\sin y + \cos y + x)y' = y

Exercise 9.2Verifying an implicit 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:

8. y−cos⁡y=xy - \cos y = x  :  (ysin⁡y+cos⁡y+x) y′=y(y\sin y + \cos y + x)\,y' = y

Exam answer

Given

y−cos⁡y=x…(i)y - \cos y = x\quad\ldots(\text{i})

Differentiating (i) with respect to xx⁠,

y′+sin⁡y⋅y′=1⇒ (1+sin⁡y) y′=1…(ii)\begin{aligned} y' + \sin y \cdot y' &= 1 \\ \Rightarrow\ (1 + \sin y)\,y' &= 1\quad\ldots(\text{ii}) \end{aligned}

From (i), x=y−cos⁡yx = y - \cos y⁠. So

ysin⁡y+cos⁡y+x=ysin⁡y+cos⁡y+(y−cos⁡y)=ysin⁡y+y=y(1+sin⁡y)\begin{aligned} &y\sin y + \cos y + x \\ &= y\sin y + \cos y \\ &\quad + (y - \cos y) \\ &= y\sin y + y \\ &= y(1 + \sin y) \end{aligned} ∴ L.H.S.=y(1+sin⁡y) y′=y⋅1[by (ii)]=y=R.H.S.\begin{aligned} \therefore\ \text{L.H.S.} &= y(1 + \sin y)\,y' \\ &= y \cdot 1\quad[\text{by (ii)}] \\ &= y = \text{R.H.S.} \end{aligned}

∴\therefore y−cos⁡y=xy - \cos y = x 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 equation contains xx⁠, but the solution tells us what xx is: x=y−cos⁡yx = y - \cos y⁠. Replacing xx makes the bracket ysin⁡y+cos⁡y+xy\sin y + \cos y + x collapse to y(1+sin⁡y)y(1 + \sin y)⁠. Differentiating the relation gives (1+sin⁡y)y′=1(1 + \sin y)y' = 1⁠. Multiply the two facts and the left side is just yy⁠.

Given, to find, and the plan
Given
The relation y−cos⁡y=xy - \cos y = x⁠, defining yy implicitly as a function of xx⁠; the equation (ysin⁡y+cos⁡y+x) y′=y(y\sin y + \cos y + x)\,y' = y⁠.
To show
That the function yy defined by the relation satisfies the equation wherever it is differentiable.
Plan
(1) Differentiate the relation implicitly. (2) Use the relation to rewrite the bracket. (3) Substitute and simplify the left side to yy⁠. (4) State where y′y' exists, and conclude.

Basics you need

1Implicit differentiation

Differentiate both sides with respect to xx⁠; a function of yy gives its derivative times y′y' (Section 5.3.2). Basics: y′ and y″

2Derivative of cos⁡y\cos y

ddxcos⁡y=−sin⁡y⋅y′\frac{d}{dx}\cos y = -\sin y \cdot y' by the chain rule (Section 5.3.1). Basics: y′ and y″

3What a solution is

Substituting the function (and y′y'⁠) makes L.H.S. = R.H.S.; an implicit relation counts (Section 9.3). Lesson: general and particular solutions

Solution

  1. Differentiate the relation

    ddx(y−cos⁡y)=ddx(x)y′+sin⁡y⋅y′=1(1+sin⁡y) y′=1.\begin{aligned} \frac{d}{dx}(y - \cos y) &= \frac{d}{dx}(x) \\ y' + \sin y \cdot y' &= 1 \\ (1 + \sin y)\,y' &= 1. \end{aligned}
    yy gives y′y'⁠. −cos⁡y-\cos y gives −(−sin⁡y) y′=sin⁡y⋅y′-(-\sin y)\,y' = \sin y \cdot y' by the chain rule. The right side xx gives 1. Then take y′y' out as a common factor.
    Implicit differentiation and the chain rule (Sections 5.3.1, 5.3.2). Basics: y′ and y″

  2. Rewrite the bracket using the relation

    The relation gives x=y−cos⁡yx = y - \cos y⁠, so

    ysin⁡y+cos⁡y+x=ysin⁡y+cos⁡y+y−cos⁡y=ysin⁡y+y=y(1+sin⁡y).\begin{aligned} &y\sin y + \cos y + x \\ &= y\sin y + \cos y + y - \cos y \\ &= y\sin y + y \\ &= y(1 + \sin y). \end{aligned}
    The cos⁡y\cos y terms cancel, and the rest has the common factor yy⁠. The factor 1+sin⁡y1 + \sin y is the same one that appeared in Step 1, which is what makes the check work.
    Substitute the given relation to remove xx⁠. Lesson: is it a solution?

  3. Substitute into the left side

    L.H.S.=(ysin⁡y+cos⁡y+x) y′=y [(1+sin⁡y) y′]=y⋅1=y=R.H.S.\begin{aligned} \text{L.H.S.} &= (y\sin y + \cos y + x)\,y' \\ &= y\,\big[(1 + \sin y)\,y'\big] \\ &= y \cdot 1 = y = \text{R.H.S.} \end{aligned}
    Group the factor (1+sin⁡y) y′(1 + \sin y)\,y'⁠, which equals 1 by Step 1. No division was needed.
    Section 9.3: substitute and compare the two sides. Lesson: is it a solution?

  4. Where y′y' exists, and conclusion

    As yy runs over all reals, x=y−cos⁡yx = y - \cos y increases (its rate dxdy=1+sin⁡y\frac{dx}{dy} = 1 + \sin y is never negative and is zero only at isolated points), and takes every real value. So the relation defines yy as a function of xx for every real xx⁠. By Step 1, y′=11+sin⁡yy' = \frac{1}{1 + \sin y} exists wherever 1+sin⁡y≠01 + \sin y \ne 0⁠. At the points where sin⁡y=−1\sin y = -1⁠, that is y=2kπ−π2y = 2k\pi - \frac{\pi}{2} and so x=2kπ−π2x = 2k\pi - \frac{\pi}{2} (kk an integer), the tangent is vertical and y′y' does not exist.

    So L.H.S. = R.H.S. at every xx where y′y' exists, that is, x≠2kπ−π2x \ne 2k\pi - \frac{\pi}{2}⁠. Hence verified.

    The equation has order 1 and degree 1 (a polynomial equation in y′y'⁠, power 1). The relation has no arbitrary constant, so it is a particular solution.

    The verification needs y′y' to exist; the derivative of the relation shows exactly where it does. Stating it completes the answer.
    A solution satisfies the equation where it is defined and differentiable (Section 9.3). Basics: order n gives n constants
Result

From y−cos⁡y=xy - \cos y = x⁠: (1+sin⁡y) y′=1(1 + \sin y)\,y' = 1 and ysin⁡y+cos⁡y+x=y(1+sin⁡y)y\sin y + \cos y + x = y(1 + \sin y)⁠, so (ysin⁡y+cos⁡y+x) y′=y(y\sin y + \cos y + x)\,y' = y⁠. Hence y−cos⁡y=xy - \cos y = x is a solution, wherever y′y' exists (1+sin⁡y≠01 + \sin y \ne 0⁠). (The answer key gives no answer for this question, since it is a verification.)

Check it with numbers

Take y=2y = 2⁠: then x=2−cos⁡2≈2.416147x = 2 - \cos 2 \approx 2.416147⁠, with sin⁡2≈0.909297\sin 2 \approx 0.909297 and cos⁡2≈−0.416147\cos 2 \approx -0.416147⁠.

The bracket:

2sin⁡2+cos⁡2+x≈1.818595−0.416147+2.416147=3.818595,\begin{aligned} &2\sin 2 + \cos 2 + x \\ &\approx 1.818595 - 0.416147 \\ &\quad + 2.416147 \\ &= 3.818595, \end{aligned}

which is 2(1+sin⁡2)≈2×1.9092972(1 + \sin 2) \approx 2 \times 1.909297 (to rounding).

From Step 1, y′=11.909297≈0.523753y' = \frac{1}{1.909297} \approx 0.523753⁠, and the left side is 3.818595×0.523753≈2.00000=y3.818595 \times 0.523753 \approx 2.00000 = y⁠.

Solving y−cos⁡y=xy - \cos y = x numerically: y(x+0.001)≈2.0005238y(x + 0.001) \approx 2.0005238 and y(x−0.001)≈1.9994763y(x - 0.001) \approx 1.9994763⁠, so

2.0005238−1.99947630.002≈0.5238,\begin{aligned} &\frac{2.0005238 - 1.9994763}{0.002} \\ &\approx 0.5238, \end{aligned}

matching y′y'⁠.

Common mistakes
  1. Sign of ddxcos⁡y\frac{d}{dx}\cos y⁠: it is −sin⁡y⋅y′-\sin y \cdot y'⁠, so −cos⁡y-\cos y gives +sin⁡y⋅y′+\sin y \cdot y'⁠.
  2. Forgetting the factor y′y' when differentiating cos⁡y\cos y⁠.
  3. Leaving xx in the bracket. Replace it by y−cos⁡yy - \cos y from the relation.
What to take away
  1. In an implicit check, use the relation itself to remove xx (or yy⁠) from the equation.
  2. Look for the factor from the derivative (here 1+sin⁡y1 + \sin y⁠) in the expression you are simplifying.
  3. Compare Q7 and Q9.

Every fact used, and where it comes from

FactUsed inSource
Chain rule; (cos⁡u)′=−sin⁡u⋅u′(\cos u)' = -\sin u \cdot u'⁠; implicit differentiationStep 1Chapter 5, Sections 5.3.1 and 5.3.2 (Basics)
A solution makes L.H.S. = R.H.S. when substitutedSteps 2, 3Section 9.3 (Lesson)
sin⁡y=−1\sin y = -1 exactly at y=2kπ−π2y = 2k\pi - \frac{\pi}{2}Step 4Class XI trigonometry
Order and degree (polynomial condition); particular solutionStep 4Sections 9.2.1, 9.2.2 and 9.3 (Lesson)
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