Chapter 9 · Differential Equations

Exercise 9.3 · Q16: $xy\frac{dy}{dx} = (x + 2)(y + 2)$ through $(1, -1)$

Exercise 9.3Variables separable, particular solution · Sections 9.3 and 9.4.1

The question

16. For the differential equation $xy\dfrac{dy}{dx} = (x + 2)(y + 2)$ ⁠, find the solution curve passing through the point $(1, -1)$ ⁠.

Exam answer

Given $xy\frac{dy}{dx} = (x + 2)(y + 2)$ ⁠. Separating the variables ( $x \ne 0$ ⁠, $y \ne -2$ ⁠),

\frac{y}{y + 2}\,dy = \frac{x + 2}{x}\,dx

\left(1 - \frac{2}{y + 2}\right)dy = \left(1 + \frac{2}{x}\right)dx

Integrating both sides,

\begin{aligned} y - 2\log|y + 2| &= x + 2\log|x| \\ &\quad + C\quad\ldots(\text{i}) \end{aligned}

Putting $x = 1$ ⁠, $y = -1$ in (i): $-1 - 0 = 1 + 0 + C$ ⁠, so $C = -2$ ⁠.

\begin{aligned} y - x + 2 &= 2\log|y + 2| + 2\log|x| \\ &= \log\left(x^2(y + 2)^2\right) \end{aligned}

$\therefore$ The solution curve is $y - x + 2 = \log\left(x^2(y + 2)^2\right)$ ⁠.

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

The one idea this problem is built on

Dividing by $x(y + 2)$ puts all of $y$ on one side and all of $x$ on the other. Each side is an improper fraction, so first write it as "1 plus a simple fraction"; then everything integrates to $y$ ⁠, $x$ and logarithms. The point $(1, -1)$ fixes the constant, and the logs combine into one.

Given, to find, and the plan

Given: $xy\frac{dy}{dx} = (x + 2)(y + 2)$ ⁠, and the curve passes through $(1, -1)$ ⁠.
To find: The solution curve through $(1, -1)$ ⁠: the particular solution.
Plan: (1) Separate, noting what must be nonzero. (2) Split each fraction. (3) Integrate, one constant $C$ ⁠. (4) Use the point. (5) Combine the logs into the textbook's form. (6) Check by implicit differentiation and at the point.

Basics you need

1Variables separable

If $\frac{dy}{dx} = g(x)\,h(y)$ with $h(y) \ne 0$ ⁠, then $\int\frac{dy}{h(y)} = \int g(x)\,dx$ (Section 9.4.1). Here $g(x) = \frac{x + 2}{x}$ and $h(y) = \frac{y + 2}{y}$ ⁠. Lesson: separate, then integrate

2Splitting an improper fraction

$\frac{y}{y + 2} = \frac{(y + 2) - 2}{y + 2} = 1 - \frac{2}{y + 2}$ and $\frac{x + 2}{x} = 1 + \frac{2}{x}$ ⁠.

3Log integral and log laws

$\int\frac{dx}{x + a} = \log|x + a| + C$ (Section 7.2, formula (xii)); $2\log|u| = \log u^2$ ⁠; $\log m + \log n = \log mn$ ⁠. Basics: every method ends with an integral

4Particular solution

Put the given point into the general solution and solve for the constant (Section 9.3). Basics: one constant, one condition

Solution

Separate the variables

The equation has order 1 and degree 1. For $x \ne 0$ and $y \ne -2$ ⁠, divide both sides by $x(y + 2)$ and multiply by $dx$ ⁠:
$\frac{y}{y + 2}\,dy = \frac{x + 2}{x}\,dx.$
This needs $x \ne 0$ and $y \ne -2$ ⁠. The constant function $y = -2$ is itself a solution (both sides are 0), but it does not pass through $(1, -1)$ ⁠. At the given point $x = 1 > 0$ and $y + 2 = 1 > 0$ ⁠, so near it both divisions are safe.

Section 9.4.1: separate, then integrate. Lesson: separate, then integrate
Split each fraction
$\left(1 - \frac{2}{y + 2}\right)dy = \left(1 + \frac{2}{x}\right)dx.$
In $\frac{y}{y + 2}$ the top has the same degree as the bottom, so write $y = (y + 2) - 2$ and divide term by term. Similarly $\frac{x + 2}{x} = \frac{x}{x} + \frac{2}{x}$ ⁠. Each piece now has a standard integral.

Dividing term by term. Basics: every method ends with an integral
Integrate both sides, with one constant
$y - 2\log|y + 2| = x + 2\log|x| + C.$
$\int 1\,dy = y$ ⁠, $\int\frac{2}{y + 2}\,dy = 2\log|y + 2|$ ⁠, $\int 1\,dx = x$ ⁠, $\int\frac{2}{x}\,dx = 2\log|x|$ ⁠. The constants of both sides combine into one $C$ ⁠: one constant for a first order equation. This is the general solution, written as a relation between $x$ and $y$ ⁠.

Section 7.2, formula (xii). Basics: every method ends with an integral
Use the point to find $C$

Put $x = 1$ ⁠, $y = -1$ ⁠: $|y + 2| = 1$ and $|x| = 1$ ⁠, so both logs are $\log 1 = 0$ ⁠:
$-1 - 0 = 1 + 0 + C \;\Rightarrow\; C = -2.$
One condition fixes the one constant. $\log 1 = 0$ makes the arithmetic short.

Section 9.3: a particular solution gives the constant a value. Basics: one constant, one condition
Combine the logs

With $C = -2$ ⁠, move $x$ and $-2$ to the left and the log to the right:
$\begin{aligned} y - x + 2 &= 2\log|x| + 2\log|y + 2| \\ &= \log x^2 + \log(y + 2)^2 \\ &= \log\left(x^2(y + 2)^2\right). \end{aligned}$
$2\log|u| = \log|u|^2 = \log u^2$ ⁠, since $|u|^2 = u^2$ ⁠; then $\log m + \log n = \log mn$ ⁠. Squares are positive (for $x \ne 0$ ⁠, $y \ne -2$ ⁠), so no modulus is needed. This is the textbook's form.

Laws of logarithms. Lesson: general and particular solutions
Check: differentiate implicitly, and test the point

Differentiate $y - 2\log|y + 2| = x + 2\log|x| - 2$ with respect to $x$ ⁠:
$\begin{aligned} y' - \frac{2y'}{y + 2} &= 1 + \frac{2}{x} \\ y'\cdot\frac{y}{y + 2} &= \frac{x + 2}{x}, \end{aligned}$
and multiplying by $x(y + 2)$ gives $xy\,y' = (x + 2)(y + 2)$ ⁠. At $(1, -1)$ ⁠: $-1 - 1 + 2 = 0$ and $\log(1\cdot 1) = 0$ ⁠.

An implicit answer is checked by implicit differentiation: it returns the equation, so every curve of the relation is a solution, and the point lies on ours. At $(1, -1)$ the equation gives the slope $\frac{3\cdot 1}{1\cdot(-1)} = -3$ ⁠.

Implicit differentiation (Chapter 5). Basics: y′ is the slope

Result

The solution curve through $(1, -1)$ is $y - x + 2 = \log\left(x^2(y + 2)^2\right)$ ( $x \ne 0$ ⁠, $y \ne -2$ ⁠; near the point, $x > 0$ and $y > -2$ ⁠). (Textbook answer: $y - x + 2 = \log\left(x^2(y + 2)^2\right)$ ⁠.)

Check it with numbers

Solving the relation numerically near the point: at $x = 1.2$ ⁠, $y \approx -1.3748344$ ⁠, and the equation gives the slope $\frac{(x + 2)(y + 2)}{xy} \approx -1.2125883$ ⁠.

The curve itself: $y(1.201) \approx -1.3760450$ and $y(1.199) \approx -1.3736198$ ⁠, so

\begin{aligned} &\frac{-1.3760450 + 1.3736198}{0.002} \\ &\approx -1.2126. \end{aligned}

Common mistakes

Integrating $\frac{y}{y + 2}$ as $\log|y + 2|$ ⁠. That needs the derivative of the bottom on top; first split it as $1 - \frac{2}{y + 2}$ ⁠.
Losing the 2s: $\int\frac{2}{x}\,dx = 2\log|x|$ ⁠, and $2\log|x| = \log x^2$ ⁠.
Sign of $C$ ⁠: $-1 = 1 + C$ gives $C = -2$ ⁠, not $2$ ⁠.

What to take away

When a separated side is an improper fraction, divide first.
An implicit answer is fine; check it by implicit differentiation.
$\log 1 = 0$ often makes the point condition quick.

Every fact used, and where it comes from

Fact	Used in	Source
Variables separable: separate, then integrate	Step 1	Section 9.4.1 (Lesson)
$\int\frac{dx}{x + a} = \log\|x + a\| + C$	Step 3	Section 7.2, formula (xii) (Basics)
A particular solution gives the constant a value	Step 4	Section 9.3 (Basics)
$2\log\|u\| = \log u^2$ ⁠; $\log m + \log n = \log mn$	Step 5	Laws of logarithms
Implicit differentiation; a solution makes L.H.S. = R.H.S.	Step 6	Chapter 5; Section 9.3 (Basics)

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Exercise 9.3 · Q16: xydydx=(x+2)(y+2)xy\frac{dy}{dx} = (x + 2)(y + 2)xydxdy​=(x+2)(y+2) through (1,−1)(1, -1)(1,−1)

Basics you need

Solution

Separate the variables

Split each fraction

Integrate both sides, with one constant

Use the point to find CCC

Combine the logs

Check: differentiate implicitly, and test the point

Every fact used, and where it comes from

Exercise 9.3 · Q16: $xy\frac{dy}{dx} = (x + 2)(y + 2)$ through $(1, -1)$

Use the point to find $C$