Exercise 9.2 · Q6: $y = x\sin x$ solves $xy' = y + x\sqrt{x^2 - y^2}$

Basics you need

Solution

1. Differentiate by the product rule

   $$\begin{aligned} y' &= (x)'\sin x + x(\sin x)' \\ &= \sin x + x\cos x. \end{aligned}$$
   $y$ is a product of $x$ and $\sin x$⁠. The derivative of $x$ is 1 and the derivative of $\sin x$ is $\cos x$⁠.
   Product rule (Chapter 5). Basics: y′ and y″

2. The left side

   $$\text{L.H.S.} = xy' = x\sin x + x^2\cos x.$$
   Multiply the result of Step 1 by $x$⁠, term by term.
   Substitute the derivative into the equation (Section 9.3). Lesson: is it a solution? Examples 2 and 3

3. The right side, keeping the root honest

   Under the root:

   $$\begin{aligned} x^2 - y^2 &= x^2 - x^2\sin^2 x \\ &= x^2(1 - \sin^2 x) = x^2\cos^2 x. \end{aligned}$$

   This is never negative, so the root is defined for every real $x$⁠, and

   $$\sqrt{x^2 - y^2} = \sqrt{(x\cos x)^2} = |x\cos x|.$$

   So

   $$\text{R.H.S.} = x\sin x + x\,|x\cos x|.$$
   $1 - \sin^2 x = \cos^2 x$ turns the expression under the root into a perfect square. The root of a square is the absolute value, because $\sqrt{\ }$ always means the non-negative root.
   $\sin^2 x + \cos^2 x = 1$ and $\sqrt{t^2} = |t|$ (Class XI). Lesson: is it a solution?

4. Compare: where are the two sides equal?

   $$\begin{aligned} &\text{L.H.S.} - \text{R.H.S.} \\ &= x^2\cos x - x\,|x\cos x| \\ &= x\,\big(x\cos x - |x\cos x|\big). \end{aligned}$$

   For $x \ne 0$ this is 0 exactly when $x\cos x = |x\cos x|$⁠, that is, when $x\cos x \ge 0$⁠. When $x\cos x < 0$ the bracket is $2x\cos x \ne 0$ and the sides differ.

   So the equation holds exactly at the $x \ne 0$ with $x\cos x \ge 0$⁠:

   for $x > 0$⁠, where $\cos x \ge 0$⁠, for example $0 < x \le \frac{\pi}{2}$ or $\frac{3\pi}{2} \le x \le \frac{5\pi}{2}$⁠;

   for $x < 0$⁠, where $\cos x \le 0$⁠, for example $-\frac{3\pi}{2} \le x \le -\frac{\pi}{2}$⁠.

   With $t = x\cos x$⁠, the difference is $x(t - |t|)$⁠, and $t - |t|$ is 0 for $t \ge 0$ and $2t$ for $t < 0$⁠. On an interval inside this set, for example $0 < x < \frac{\pi}{2}$⁠, the function is a solution in the full sense: L.H.S. = R.H.S. at every point.
   $|t| = t$ if $t \ge 0$⁠, and $|t| = -t$ if $t < 0$⁠. Lesson: is it a solution?

5. Test the printed condition

   The book prints the condition $x \ne 0$ and $x > y$ or $x < -y$⁠. It does not decide the sign of $x\cos x$⁠:

   At $x = 2$⁠: $y = 2\sin 2 \approx 1.8186 < 2$⁠, so $x > y$ holds; but $\cos 2 < 0$⁠, so $x\cos x < 0$⁠, and L.H.S. $\approx 0.1540$ while R.H.S. $\approx 3.4832$⁠.

   At $x = -1$⁠: $y = \sin 1 \approx 0.8415$⁠, so $x < -y$ holds; but $x\cos x \approx -0.5403 < 0$⁠, and L.H.S. $\approx 1.3818$ while R.H.S. $\approx 0.3012$⁠.

   The condition under which the verification works is $x\cos x \ge 0$ (with $x \ne 0$⁠).

   A single point where both sides differ is enough to show that a condition does not guarantee the equation. The usual working writes $\sqrt{x^2\cos^2 x} = x\cos x$⁠, which silently assumes $x\cos x \ge 0$⁠; that is the condition to state.
   A counterexample disproves a "for all" claim. Lesson: Exercise 9.2 player

6. Conclusion

   Where $x\cos x \ge 0$ (and $x \ne 0$⁠), L.H.S. = R.H.S., so $y = x\sin x$ is a solution of $xy' = y + x\sqrt{x^2 - y^2}$ there, for example on $0 < x < \frac{\pi}{2}$⁠. Hence verified, on that domain.

   The equation has order 1 and degree 1: the root contains $x$ and $y$ but no derivative, so the equation is a polynomial equation in $y'$⁠, with $y'$ to the power 1. The function has no arbitrary constant, so it is a particular solution.

   Stating the domain is part of the answer whenever a root, log or division is involved.
   Order, degree and particular solutions (Sections 9.2 and 9.3). Basics: order n gives n constants

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