Chapter 9 · Differential Equations

Exercise 9.2 · Q7: $xy = \log y + C$ solves $y' = \frac{y^2}{1 - xy}$

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:

7. $xy = \log y + C$ : $y' = \dfrac{y^2}{1 - xy}$ $(xy \ne 1)$

Exam answer

Given $xy = \log y + C$ ⁠. Differentiating both sides with respect to $x$ ⁠,

xy' + y = \frac{1}{y}\,y'

Multiplying by $y$ ⁠,

\begin{aligned} xyy' + y^2 &= y' \\ \Rightarrow\ y' - xyy' &= y^2 \\ \Rightarrow\ y'(1 - xy) &= y^2 \end{aligned}

Dividing by $1 - xy$ (as $xy \ne 1$ ⁠),

y' = \frac{y^2}{1 - xy}

which is the given differential equation.

$\therefore$ $xy = \log y + C$ 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

Here the solution is given implicitly: a relation between $x$ and $y$ ⁠, not a formula $y = \ldots$ ⁠. To verify it, differentiate the relation with respect to $x$ ⁠, treating $y$ as a function of $x$ (so every $y$ -term picks up a factor $y'$ ⁠), then rearrange until the given equation appears. The constant $C$ disappears on differentiation.

Given, to find, and the plan

Given: The relation $xy = \log y + C$ ⁠, with $C$ an arbitrary constant; the equation $y' = \frac{y^2}{1 - xy}$ ⁠, with $xy \ne 1$ ⁠.
To show: That every differentiable $y$ satisfying the relation satisfies the equation.
Plan: (1) Note the domain ( $\log y$ needs $y > 0$ ⁠). (2) Differentiate implicitly. (3) Clear the fraction and collect $y'$ ⁠. (4) Divide by $1 - xy$ ⁠. (5) Conclude, with the role of $C$ ⁠.

Basics you need

1Implicit differentiation

When $y$ is a function of $x$ given by a relation, differentiate both sides with respect to $x$ ⁠; a function of $y$ gives its derivative times $y'$ (Section 5.3.2). Basics: y′ and y″

2Product rule and $\log$

$(xy)' = 1\cdot y + x\,y'$ ⁠, and $\frac{d}{dx}\log y = \frac{1}{y}\,y'$ for $y > 0$ (Sections 5.3.1 and 5.4). Basics: y′ and y″

3Implicit solutions count

Exercise 9.2 asks to verify functions "explicit or implicit": a relation whose $y$ satisfies the equation is a solution. Lesson: general and particular solutions

Solution

The domain

$\log y$ is defined only for $y > 0$ ⁠, so we work where $y > 0$ ⁠. The equation itself needs $xy \ne 1$ ⁠.

Every log, root and division in the working limits where it is valid. Writing the limits first means each later step can be checked against them.

$\log y$ is defined for $y > 0$ (Section 5.4). Lesson: Exercise 9.2 player
Differentiate both sides with respect to $x$
$\begin{aligned} \frac{d}{dx}(xy) &= \frac{d}{dx}(\log y) + \frac{d}{dx}C \\ y + xy' &= \frac{1}{y}\,y' + 0. \end{aligned}$
On the left, $xy$ is a product, so the product rule gives $1 \cdot y + x \cdot y'$ ⁠. On the right, $\log y$ is a function of $y$ ⁠, which is a function of $x$ ⁠, so the chain rule gives $\frac{1}{y}\cdot y'$ ⁠. The constant $C$ gives 0.

Implicit differentiation (Section 5.3.2). Basics: y′ and y″
Clear the fraction and collect $y'$

Multiply both sides by $y$ (allowed, $y > 0$ ⁠):
$\begin{aligned} xyy' + y^2 &= y' \\ y' - xyy' &= y^2 \\ y'(1 - xy) &= y^2. \end{aligned}$
Multiplying by $y$ removes the denominator. Then move the $y'$ terms to one side and take $y'$ out as a common factor.

Rearranging an equation to make $y'$ the subject. Lesson: is it a solution?
Divide by $1 - xy$
$y' = \frac{y^2}{1 - xy}\qquad (xy \ne 1).$
This is exactly the given differential equation, so L.H.S. = R.H.S.

Dividing needs $1 - xy \ne 0$ ⁠, which is the condition printed in the question. At points with $xy = 1$ ⁠, Step 3 would read $0 = y^2$ ⁠, impossible for $y > 0$ ⁠, so the relation has no differentiable $y$ through such a point.

Section 9.3: the function satisfies the equation. Lesson: is it a solution?
Conclusion, and the constant $C$

Wherever $y > 0$ and $xy \ne 1$ ⁠, a function $y$ defined by $xy = \log y + C$ satisfies $y' = \frac{y^2}{1 - xy}$ ⁠. Hence verified.

The equation has order 1 and degree 1 (after multiplying by $1 - xy$ it is $(1 - xy)y' = y^2$ ⁠, a polynomial equation in $y'$ of power 1). The relation has one arbitrary constant $C$ ⁠, as many as the order, so it is a general solution. Each value of $C$ picks one curve: for example the curve through $(0.2, 2)$ has $C = 0.4 - \log 2$ ⁠.

The constant $C$ was never used in Steps 2 to 4, so the check works for every member of the family at once.

General solution: as many arbitrary constants as the order (Section 9.3 and the Summary). Basics: order n gives n constants

Result

Differentiating $xy = \log y + C$ gives $y + xy' = \frac{y'}{y}$ ⁠, so $y'(1 - xy) = y^2$ and $y' = \frac{y^2}{1 - xy}$ for $y > 0$ ⁠, $xy \ne 1$ ⁠. Hence $xy = \log y + C$ is a solution (a general solution, one constant) of the given equation. (The answer key gives no answer for this question, since it is a verification.)

Check it with numbers

Take the curve through $x = 0.2$ ⁠, $y = 2$ ⁠: $C = 0.4 - \log 2 \approx -0.293147$ ⁠. There $xy = 0.4 \ne 1$ ⁠, and the equation gives

y' = \frac{2^2}{1 - 0.4} = \frac{4}{0.6} \approx 6.666667.

Solving $xy = \log y + C$ for $y$ numerically near $x = 0.2$ ⁠: $y(0.201) \approx 2.0067077$ and $y(0.199) \approx 1.9933738$ ⁠, so

\frac{2.0067077 - 1.9933738}{0.002} \approx 6.667,

the slope the equation predicts.

Common mistakes

Differentiating $xy$ as $y'$ or as $xy'$ only. It is a product: $y + xy'$ ⁠.
Writing $\frac{d}{dx}\log y = \frac{1}{y}$ without the factor $y'$ ⁠.
Dividing by $1 - xy$ without stating $xy \ne 1$ ⁠.

What to take away

Implicit solution: differentiate the relation, then rearrange to the given equation.
Every $y$ -term picks up $y'$ when differentiated with respect to $x$ ⁠.
Other implicit checks in this exercise: Q8 and Q9.

Every fact used, and where it comes from

Fact	Used in	Source
$\log y$ defined for $y > 0$ ⁠; $\frac{d}{dy}\log y = \frac{1}{y}$	Steps 1, 2	Chapter 5, Section 5.4
Product rule; chain rule; implicit differentiation	Step 2	Chapter 5, Sections 5.3.1 and 5.3.2 (Basics)
A solution makes L.H.S. = R.H.S. when substituted	Step 4	Section 9.3 (Lesson)
Order and degree (polynomial condition)	Step 5	Sections 9.2.1 and 9.2.2 (Lesson)
General solution has as many constants as the order	Step 5	Section 9.3 and the Summary of Chapter 9 (Basics)

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Exercise 9.2 · Q7: xy=log⁡y+Cxy = \log y + Cxy=logy+C solves y′=y21−xyy' = \frac{y^2}{1 - xy}y′=1−xyy2​

Basics you need

Solution

The domain

Differentiate both sides with respect to xxx

Clear the fraction and collect y′y'y′

Divide by 1−xy1 - xy1−xy

Conclusion, and the constant CCC

Every fact used, and where it comes from

Exercise 9.2 · Q7: $xy = \log y + C$ solves $y' = \frac{y^2}{1 - xy}$

Differentiate both sides with respect to $x$

Clear the fraction and collect $y'$

Divide by $1 - xy$

Conclusion, and the constant $C$