Chapter 9 · Differential Equations

Exercise 9.3 · Q14: Particular solution of $\frac{dy}{dx} = y\tan x$ ⁠, $y(0) = 1$

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

The question

For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition:

14. $\dfrac{dy}{dx} = y\tan x$ ⁠; $y = 1$ when $x = 0$

Exam answer

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

\frac{dy}{y} = \tan x\,dx

Integrating both sides,

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

Putting $x = 0$ ⁠, $y = 1$ in (i): $1 = C\sec 0 = C$ ⁠, so $C = 1$ ⁠.

$\therefore$ $y = \sec x$ $\left(-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}\right)$ is the particular solution.

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

The one idea this problem is built on

The right side is a function of $x$ times a function of $y$ ⁠, so dividing by $y$ separates the variables. Both sides then integrate to logarithms, and writing the constant as $\log|C|$ lets the logs combine into the simple family $y = C\sec x$ ⁠. The condition at $x = 0$ gives $C = 1$ ⁠.

Given, to find, and the plan

Given: $\frac{dy}{dx} = y\tan x$ ⁠, and $y = 1$ when $x = 0$ ⁠.
To find: The particular solution: no arbitrary constant, passing through $(0, 1)$ ⁠.
Plan: (1) Separate, noting where this is allowed. (2) Integrate, with the constant written as $\log|C|$ ⁠. (3) Remove the logs to get $y = C\sec x$ ⁠. (4) Use the condition. (5) Check by substituting back, and draw the family.

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 $g(x) = \tan x$ and $h(y) = y$ ⁠. Lesson: separate, then integrate

2Two standard integrals

$\int\frac{dy}{y} = \log|y| + C$ (Section 7.2, formula (xii)) and $\int\tan x\,dx = \log|\sec x| + C$ (Section 7.3.1). Basics: every method ends with an integral

3The constant as a logarithm

Any real constant can be written as $\log|C|$ with $C \ne 0$ ⁠; then $\log|y| = \log|C\sec x|$ gives $|y| = |C\sec x|$ ⁠. Basics: one constant, one condition

4Particular solution

Put the given $x$ and $y$ into the general solution and solve for the constant (Section 9.3). Lesson: general and particular solutions

Solution

Separate the variables

This is a first order equation of degree 1 (a polynomial of degree 1 in $\frac{dy}{dx}$ ⁠). For $y \ne 0$ ⁠, divide by $y$ and multiply by $dx$ ⁠:
$\frac{dy}{y} = \tan x\,dx.$
The equation has the form $\frac{dy}{dx} = g(x)\,h(y)$ with $g(x) = \tan x$ ⁠, $h(y) = y$ ⁠. Dividing by $h(y)$ needs $y \ne 0$ ⁠; the constant function $y = 0$ also solves the equation, but it does not pass through $(0, 1)$ ⁠, so we lose nothing. $\tan x$ is defined except at odd multiples of $\frac{\pi}{2}$ ⁠; the condition is at $x = 0$ ⁠, so we work on $-\frac{\pi}{2} < x < \frac{\pi}{2}$ ⁠.

Section 9.4.1: separate, then integrate. Lesson: separate, then integrate
Integrate both sides, with one constant
$\begin{aligned} \int\frac{dy}{y} &= \int\tan x\,dx \\ \log|y| &= \log|\sec x| + \log|C|. \end{aligned}$
Both integrals are standard. The constants of the two sides combine into one, and we write it as $\log|C|$ ( $C \ne 0$ ⁠): as $C$ runs over the nonzero numbers, $\log|C|$ takes every real value, so nothing is lost, and the next step becomes a one-liner. One constant for a first order equation.

Section 7.2, formula (xii); Section 7.3.1, $\int\tan x\,dx$ ⁠. Basics: every method ends with an integral
Remove the logarithms
$\begin{aligned} \log|y| &= \log|C\sec x| \\ \Rightarrow\ |y| &= |C\sec x| \\ \Rightarrow\ y &= C\sec x. \end{aligned}$
$\log m + \log n = \log mn$ ⁠, and $\log$ is one to one, so the moduli are equal. Then $y = \pm C\sec x$ ⁠; the sign can be absorbed into $C$ ⁠, which may be any nonzero number. This is the general solution: the family $y = C\sec x$ ⁠.

Laws of logarithms; renaming the constant. Basics: one constant, one condition
Use the condition to find $C$

Put $x = 0$ ⁠, $y = 1$ ⁠: $1 = C\sec 0 = C\cdot 1$ ⁠, so $C = 1$ and
$y = \sec x,\qquad -\frac{\pi}{2} < x < \frac{\pi}{2}.$
$\sec 0 = \frac{1}{\cos 0} = 1$ ⁠. One condition fixes the one constant. The solution lives on the interval through $x = 0$ on which $\sec x$ is defined; it cannot be continued past $x = \pm\frac{\pi}{2}$ ⁠, where $\sec x$ becomes infinite.

Section 9.3: a particular solution gives the constant a value. Lesson: general and particular solutions
Check: substitute back, and see the family

$\frac{d}{dx}\sec x = \sec x\tan x = y\tan x$ ⁠, so L.H.S. = R.H.S. At $x = 0$ ⁠: $y = \sec 0 = 1$ ⁠.

The particular solution $y = \sec x$ (bold) through $(0, 1)$ ⁠, with two other members of the family $y = C\sec x$ ⁠: $C = 2$ and $C = -1$ (thin). Drawn with equal scales on $-\frac{\pi}{2} < x < \frac{\pi}{2}$ ⁠; the dashed lines $x = \pm\frac{\pi}{2}$ are where every member becomes infinite.

Substituting back is the definition of a solution (Section 9.3). The picture shows the whole family; only the bold curve passes through $(0, 1)$ ⁠, and its lowest point is exactly there, where the slope $\sec 0\tan 0$ is 0.

$\frac{d}{dx}\sec x = \sec x\tan x$ (Chapter 5). Basics: y′ is the slope

Result

The particular solution is $y = \sec x$ ⁠, for $-\frac{\pi}{2} < x < \frac{\pi}{2}$ ⁠. (Textbook answer: $y = \sec x$ ⁠.)

Check it with numbers

At $x = 0.5$ ⁠: $y = \sec 0.5 \approx 1.1394939$ and $\tan 0.5 \approx 0.5463025$ ⁠, so the equation asks for the slope $y\tan x \approx 0.6225084$ ⁠.

The curve itself: $y(0.501) \approx 1.1401173$ and $y(0.499) \approx 1.1388723$ ⁠, so

\begin{aligned} &\frac{1.1401173 - 1.1388723}{0.002} \\ &\approx 0.6225. \end{aligned}

Common mistakes

$\int\tan x\,dx = \sec^2x$ ⁠. That is the derivative of $\tan x$ ⁠; the integral is $\log|\sec x|$ ⁠.
Adding $C$ after removing the logs: $y = \sec x + C$ is wrong. A constant added to logs becomes a factor.
Forgetting the interval. $\sec x$ is not defined at $x = \pm\frac{\pi}{2}$ ⁠, so the solution through $(0, 1)$ stays between them.

What to take away

When both sides integrate to logs, write the constant as $\log|C|$ and combine.
$\frac{dy}{dx} = y\,g(x)$ always gives $y = Ce^{\int g\,dx}$ ⁠; here $e^{\log|\sec x|} = |\sec x|$ ⁠.
Check the particular solution in the equation and at the given point.

Every fact used, and where it comes from

Fact	Used in	Source
Variables separable: separate, then integrate	Step 1	Section 9.4.1 (Lesson)
$\int\frac{dy}{y} = \log\|y\| + C$ ⁠; $\int\tan x\,dx = \log\|\sec x\| + C$	Step 2	Section 7.2, formula (xii); Section 7.3.1 (Basics)
Laws of logarithms; a constant written as $\log\|C\|$	Steps 2, 3	Laws of logarithms (Basics)
A particular solution gives the constant a value	Step 4	Section 9.3 (Lesson)
$\frac{d}{dx}\sec x = \sec x\tan x$	Step 5	Chapter 5 (Basics)

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Exercise 9.3 · Q14: Particular solution of dydx=ytan⁡x\frac{dy}{dx} = y\tan xdxdy​=ytanx⁠, y(0)=1y(0) = 1y(0)=1

Basics you need

Solution

Separate the variables

Integrate both sides, with one constant

Remove the logarithms

Use the condition to find CCC

Check: substitute back, and see the family

Every fact used, and where it comes from

Exercise 9.3 · Q14: Particular solution of $\frac{dy}{dx} = y\tan x$ ⁠, $y(0) = 1$

Use the condition to find $C$