Chapter 9 · Differential Equations

Exercise 9.2 · Q4: $y = \sqrt{1 + x^2}$ solves $y' = \frac{xy}{1 + x^2}$

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:

4. $y = \sqrt{1 + x^2}$ : $y' = \dfrac{xy}{1 + x^2}$

Exam answer

Given $y = \sqrt{1 + x^2}$ ⁠. Differentiating with respect to $x$ ⁠,

\begin{aligned} y' &= \frac{2x}{2\sqrt{1 + x^2}} \\ &= \frac{x}{\sqrt{1 + x^2}} \end{aligned}

Substituting $y$ in the R.H.S. and using $1 + x^2 = (\sqrt{1 + x^2})^2$ ⁠,

\begin{aligned} \text{R.H.S.} &= \frac{xy}{1 + x^2} \\ &= \frac{x\sqrt{1 + x^2}}{1 + x^2} \\ &= \frac{x}{\sqrt{1 + x^2}} \\ &= y' = \text{L.H.S.} \end{aligned}

$\therefore$ $y = \sqrt{1 + x^2}$ 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 function satisfies $y^2 = 1 + x^2$ ⁠. So the denominator $1 + x^2$ on the right side is $y^2$ ⁠, and $\frac{xy}{1 + x^2} = \frac{xy}{y^2} = \frac{x}{y}$ ⁠. The chain rule gives $y' = \frac{x}{y}$ as well. Both sides are the same expression written two ways.

Given, to find, and the plan

Given: The function $y = \sqrt{1 + x^2}$ (the positive root), and the equation $y' = \frac{xy}{1 + x^2}$ ⁠.
To show: That the function satisfies the equation, and for which $x$ ⁠.
Plan: (1) Differentiate by the chain rule. (2) Substitute $y$ into the right side and simplify. (3) Compare with $y'$ ⁠. (4) Check the domain of every step and conclude.

Basics you need

1What a solution is

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

2Chain rule for a square root

$\frac{d}{dx}\sqrt{u} = \frac{1}{2\sqrt{u}}\cdot\frac{du}{dx}$ for $u > 0$ (Section 5.3.1). Basics: y′ and y″

3Square of a square root

For $u \ge 0$ ⁠, $(\sqrt{u})^2 = u$ ⁠; so $\frac{\sqrt{u}}{u} = \frac{1}{\sqrt{u}}$ when $u > 0$ ⁠. Lesson: is it a solution?

Solution

Differentiate by the chain rule
$\begin{aligned} y &= (1 + x^2)^{1/2} \\ y' &= \tfrac{1}{2}(1 + x^2)^{-1/2}\cdot 2x \\ &= \frac{x}{\sqrt{1 + x^2}}. \end{aligned}$
The outer function is $\sqrt{u}$ ⁠, with derivative $\frac{1}{2\sqrt{u}}$ ⁠; the inner function is $u = 1 + x^2$ ⁠, with derivative $2x$ ⁠. The chain rule multiplies them, and the 2s cancel.

Chain rule (Section 5.3.1). Basics: y′ and y″
Substitute $y$ into the right side and simplify
$\begin{aligned} \text{R.H.S.} &= \frac{xy}{1 + x^2} = \frac{x\sqrt{1 + x^2}}{1 + x^2} \\ &= \frac{x\sqrt{1 + x^2}}{\sqrt{1 + x^2}\cdot\sqrt{1 + x^2}} \\ &= \frac{x}{\sqrt{1 + x^2}}. \end{aligned}$
Write $1 + x^2$ as $\sqrt{1 + x^2}\cdot\sqrt{1 + x^2}$ (true because $1 + x^2 > 0$ ⁠), then cancel one factor $\sqrt{1 + x^2}$ ⁠, which is never 0.

$(\sqrt{u})^2 = u$ for $u \ge 0$ ⁠. Lesson: is it a solution?
Compare the two sides
$\text{L.H.S.} = y' = \frac{x}{\sqrt{1 + x^2}} = \text{R.H.S.}$
Step 1 gave the left side and Step 2 the right side; they are the same expression. This is the test of a solution.

Section 9.3: substitute and compare the two sides. Lesson: is it a solution?
Where every step is valid, and conclusion

For every real $x$ ⁠, $1 + x^2 \ge 1 > 0$ ⁠. So the square root is defined and differentiable, the denominators $1 + x^2$ and $\sqrt{1 + x^2}$ are never 0, and the cancellation in Step 2 is allowed. Hence verified: $y = \sqrt{1 + x^2}$ is a solution for every real $x$ ⁠.

The equation has order 1 and degree 1 (it is a polynomial equation in $y'$ ⁠, with $y'$ to the power 1). The function has no arbitrary constant, so it is a particular solution. In fact $y = A\sqrt{1 + x^2}$ works for every constant $A$ (both sides just gain the factor $A$ ⁠); our function is $A = 1$ ⁠.

A root or a denominator can fail for some $x$ ⁠; here neither does, which is why the answer needs no restriction on $x$ ⁠.

Particular solution: no arbitrary constants (Section 9.3). Basics: order n gives n constants

Result

$y' = \frac{x}{\sqrt{1 + x^2}}$ and $\frac{xy}{1 + x^2} = \frac{x}{\sqrt{1 + x^2}}$ ⁠, so L.H.S. = R.H.S. for every real $x$ ⁠. Hence $y = \sqrt{1 + x^2}$ is a solution of $y' = \frac{xy}{1 + x^2}$ ⁠. (The answer key gives no answer for this question, since it is a verification.)

Check it with numbers

At $x = 0.75$ ⁠: $1 + x^2 = 1.5625$ and $y = \sqrt{1.5625} = 1.25$ ⁠. Then $y' = \frac{0.75}{1.25} = 0.6$ ⁠, and the right side is

\frac{0.75 \times 1.25}{1.5625} = \frac{0.9375}{1.5625} = 0.6.

From the curve itself, with $y(0.751) \approx 1.2506003$ and $y(0.749) \approx 1.2494003$ ⁠,

\begin{aligned} &\frac{1.2506003 - 1.2494003}{0.002} \\ &\approx 0.6000. \end{aligned}

Common mistakes

Forgetting the inner derivative $2x$ and writing $y' = \frac{1}{2\sqrt{1 + x^2}}$ ⁠.
Substituting $y'$ on the right side. The right side contains $y$ ⁠, not $y'$ ⁠: put $y = \sqrt{1 + x^2}$ there.
Leaving the two sides in different forms ( $\frac{x\sqrt{1 + x^2}}{1 + x^2}$ against $\frac{x}{\sqrt{1 + x^2}}$ ⁠) without showing they are equal.

What to take away

When the right side contains $y$ ⁠, substitute the function there and simplify until it matches $y'$ ⁠.
Use the defining relation (here $y^2 = 1 + x^2$ ⁠) to see why the two sides agree.
For a root in the function, check that what is under it stays positive on the domain; compare Q10, where it does only on $(-a, a)$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Chain rule; $\frac{d}{dx}\sqrt{u} = \frac{u'}{2\sqrt{u}}$	Step 1	Chapter 5, Section 5.3.1 (Basics)
$(\sqrt{u})^2 = u$ for $u \ge 0$	Step 2	Class XI algebra
A solution makes L.H.S. = R.H.S. when substituted	Steps 3, 4	Section 9.3 (Lesson)
Order and degree (polynomial condition)	Step 4	Sections 9.2.1 and 9.2.2 (Lesson)
Particular solution: no arbitrary constants	Step 4	Section 9.3 (Basics)

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Exercise 9.2 · Q4: y=1+x2y = \sqrt{1 + x^2}y=1+x2​ solves y′=xy1+x2y' = \frac{xy}{1 + x^2}y′=1+x2xy​

Basics you need

Solution

Differentiate by the chain rule

Substitute yyy into the right side and simplify

Compare the two sides

Where every step is valid, and conclusion

Every fact used, and where it comes from

Exercise 9.2 · Q4: $y = \sqrt{1 + x^2}$ solves $y' = \frac{xy}{1 + x^2}$

Substitute $y$ into the right side and simplify