Chapter 9 · Differential Equations

Miscellaneous · Q7: $(1 + e^{2x})\,dy + (1 + y^2)e^x\,dx = 0$ ⁠, $y(0) = 1$

Miscellaneous ExerciseVariables separable, particular solution · Section 9.4.1

The question

7. Find the particular solution of the differential equation $(1 + e^{2x})\,dy + (1 + y^2)\,e^x\,dx = 0$ ⁠, given that $y = 1$ when $x = 0$ ⁠.

Exam answer

Since $1 + e^{2x} > 0$ and $1 + y^2 > 0$ ⁠, dividing by their product, the variables separate:

\frac{dy}{1 + y^2} + \frac{e^x\,dx}{1 + e^{2x}} = 0

Integrating both sides,

\int\frac{dy}{1 + y^2} + \int\frac{e^x\,dx}{1 + e^{2x}} = C

In the second integral, put $e^x = t$ ⁠. Then $e^x\,dx = dt$ ⁠.

\begin{aligned} \tan^{-1}y + \int\frac{dt}{1 + t^2} &= C \\ \tan^{-1}y + \tan^{-1}t &= C \\ \tan^{-1}y + \tan^{-1}(e^x) &= C\ \ldots(\text{i}) \end{aligned}

Putting $x = 0$ ⁠, $y = 1$ in (i): $\tan^{-1}1 + \tan^{-1}1 = C$ ⁠, so $C = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$ ⁠.

$\therefore$ The particular solution is $\tan^{-1}y + \tan^{-1}(e^x) = \dfrac{\pi}{2}$ ⁠.

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

The one idea this problem is built on

The $dy$ term carries a factor in $x$ and the $dx$ term a factor in $y$ ⁠. Dividing by both factors, $(1 + e^{2x})(1 + y^2)$ ⁠, separates the variables. The $y$ -integral is a standard inverse tangent, and the $x$ -integral becomes one too after $e^x = t$ ⁠, because $e^{2x} = t^2$ ⁠. The condition $y = 1$ at $x = 0$ then fixes the constant; the answer even simplifies to $y = e^{-x}$ ⁠.

Given, to find, and the plan

Given: $(1 + e^{2x})\,dy + (1 + y^2)e^x\,dx = 0$ ⁠, and $y = 1$ when $x = 0$ ⁠.
To find: The particular solution satisfying the condition.
Plan: (1) Separate the variables. (2) Integrate, with $e^x = t$ for the $x$ -part and one constant. (3) Put $x = 0$ ⁠, $y = 1$ to find $C$ ⁠. (4) Simplify the particular solution. (5) Check by substituting back, and see the curve.

Basics you need

1Variables separable

Arrange the equation as (function of $y$ ⁠) $dy$ $+$ (function of $x$ ⁠) $dx = 0$ and integrate each part (Section 9.4.1). Lesson: variables separable

2 $\int\frac{dt}{1 + t^2}$ and substitution

$\int\frac{dt}{1 + t^2} = \tan^{-1}t + C$ (Section 7.2, formula (x)); with $e^x = t$ ⁠, $e^x\,dx = dt$ (Section 7.3.1). Basics: every method ends with an integral

3Particular solution from a condition

Put the given values into the general solution to fix the constant (Section 9.3). Lesson: general and particular solutions

4Inverse tangent values

$\tan^{-1}1 = \frac{\pi}{4}$ ⁠; $\tan^{-1}$ takes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (Chapter 2, Section 2.2); $\tan\left(\frac{\pi}{2} - \theta\right) = \cot\theta$ (Class XI). Basics: e^{kt} and its rate

Solution

Separate the variables

Both $1 + e^{2x}$ and $1 + y^2$ are always positive. Divide every term by $(1 + e^{2x})(1 + y^2)$ ⁠:
$\frac{dy}{1 + y^2} + \frac{e^x\,dx}{1 + e^{2x}} = 0.$
In the first term $1 + e^{2x}$ cancels, in the second $1 + y^2$ cancels, so each term has one variable. Because the divisor is never 0, no solution is lost and the equation is defined for all real $x$ and $y$ ⁠.

Section 9.4.1: separate, then integrate. Lesson: variables separable
Integrate, with one constant

The first integral is standard: $\int\frac{dy}{1 + y^2} = \tan^{-1}y$ ⁠. For the second, put $e^x = t$ ⁠, so $e^x\,dx = dt$ and $e^{2x} = t^2$ ⁠:
$\begin{aligned} \int\frac{e^x\,dx}{1 + e^{2x}} &= \int\frac{dt}{1 + t^2} \\ &= \tan^{-1}t = \tan^{-1}(e^x). \end{aligned}$
So the general solution is
$\tan^{-1}y + \tan^{-1}(e^x) = C,\quad\ldots(1)$
where $C$ is an arbitrary constant.

$e^x\,dx$ is exactly the differential of $e^x$ ⁠, and $e^{2x} = (e^x)^2$ ⁠, so the substitution turns the $x$ -part into the same standard form as the $y$ -part. One constant suffices for a first order equation.

Section 7.2, formula (x); substitution, Section 7.3.1. Basics: every method ends with an integral
Use the condition $y = 1$ when $x = 0$

Putting $x = 0$ ⁠, $y = 1$ in (1), with $e^0 = 1$ ⁠:
$\begin{aligned} \tan^{-1}1 + \tan^{-1}1 &= C, \\ C = \frac{\pi}{4} + \frac{\pi}{4} &= \frac{\pi}{2}. \end{aligned}$
So the particular solution is
$\tan^{-1}y + \tan^{-1}(e^x) = \frac{\pi}{2}.$
Of all the curves (1), the condition picks the one through $(0, 1)$ ⁠. $\tan^{-1}1 = \frac{\pi}{4}$ because $\tan\frac{\pi}{4} = 1$ and $\frac{\pi}{4}$ lies in the principal range.

Section 9.3: a particular solution from the general one. Lesson: Examples 4 to 9
Simplify: the particular solution is $y = e^{-x}$

Let $\theta = \tan^{-1}(e^x)$ ⁠. Since $e^x > 0$ ⁠, $0 < \theta < \frac{\pi}{2}$ ⁠, so $\frac{\pi}{2} - \theta$ also lies in $\left(0, \frac{\pi}{2}\right)$ ⁠, inside the principal range. The particular solution says $\tan^{-1}y = \frac{\pi}{2} - \theta$ ⁠, hence
$y = \tan\left(\frac{\pi}{2} - \theta\right) = \cot\theta = \frac{1}{e^x} = e^{-x},$
for every real $x$ ⁠.

The key form with inverse tangents and the explicit form $y = e^{-x}$ are the same curve. The explicit form makes the check in Step 5 short and shows that the solution exists for all $x$ ⁠.

$\tan\left(\frac{\pi}{2} - \theta\right) = \cot\theta$ (Class XI); principal range of $\tan^{-1}$ (Chapter 2, Section 2.2). Lesson: Miscellaneous exercise player
Check by substituting back, and see the curve

For $y = e^{-x}$ ⁠, $dy = -e^{-x}\,dx$ and $1 + y^2 = 1 + e^{-2x}$ ⁠. The left side of the equation becomes
$\begin{aligned} &(1 + e^{2x})(-e^{-x})\,dx + (1 + e^{-2x})e^x\,dx \\ &= (-e^{-x} - e^x + e^x + e^{-x})\,dx = 0. \end{aligned}$
And $y(0) = e^0 = 1$ ⁠. So the particular solution satisfies both the equation and the condition.

Slope field of $\frac{dy}{dx} = -\frac{(1 + y^2)e^x}{1 + e^{2x}}$ ⁠, with equal scales. The bold curve is the particular solution $y = e^{-x}$ through $(0, 1)$ ⁠; the thin curves are the members $C = 1.2$ and $C = 0.6$ of $\tan^{-1}y + \tan^{-1}(e^x) = C$ ⁠.

Every term cancels, so the curve really solves the equation. In the picture the slope marks are tangent to every member of the family, and only the bold one passes through $(0, 1)$ ⁠.

Section 9.3: check a solution by substituting it. Basics: a rule for slopes

Result

The general solution is $\tan^{-1}y + \tan^{-1}(e^x) = C$ ⁠, and the particular solution with $y = 1$ when $x = 0$ is

\tan^{-1}y + \tan^{-1}(e^x) = \frac{\pi}{2},

that is, $y = e^{-x}$ ⁠.

(Textbook answer: $\tan^{-1}y + \tan^{-1}(e^x) = \dfrac{\pi}{2}$ ⁠.)

Check it with numbers

At $x = 1$ ⁠: $y = e^{-1} \approx 0.367879$ ⁠. Then $\tan^{-1}(e) \approx 1.218283$ and $\tan^{-1}(e^{-1}) \approx 0.352513$ ⁠, whose sum $1.570796$ is $\frac{\pi}{2}$ ⁠.

The equation gives the slope

\begin{aligned} &-\frac{(1 + e^{-2})e}{1 + e^2} \\ &= -\frac{e + e^{-1}}{1 + e^2} \\ &= -e^{-1} \approx -0.367879, \end{aligned}

and the curve $y = e^{-x}$ has slope $-e^{-1}$ there. They agree.

Common mistakes

Integrating $\frac{e^x}{1 + e^{2x}}$ as a logarithm. The top is the derivative of $e^x$ ⁠, not of $1 + e^{2x}$ ⁠; the substitution $e^x = t$ gives $\tan^{-1}(e^x)$ ⁠.
Taking $C = \frac{\pi}{4}$ by putting in only one of the two inverse tangents. Both terms give $\frac{\pi}{4}$ ⁠.
Dividing by $1 + y^2$ with a worry about losing solutions: it is never 0, so nothing is lost.

What to take away

Divide by the "wrong" factors to separate: here $(1 + e^{2x})(1 + y^2)$ ⁠.
$\int\frac{e^x}{1 + e^{2x}}\,dx = \tan^{-1}(e^x) + C$ with $e^x = t$ ⁠.
An inverse-tangent answer can sometimes be simplified: here $\tan^{-1}y = \frac{\pi}{2} - \tan^{-1}(e^x)$ gives $y = e^{-x}$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Separating the variables; integrate both sides	Steps 1, 2	Section 9.4.1 (Lesson)
$\int\frac{dt}{1 + t^2} = \tan^{-1}t + C$	Step 2	Section 7.2, formula (x) (Basics)
Substitution $e^x = t$ ⁠, $e^x\,dx = dt$	Step 2	Section 7.3.1 (Basics)
Particular solution from a condition	Step 3	Section 9.3 (Lesson)
$\tan^{-1}1 = \frac{\pi}{4}$ ⁠; principal range of $\tan^{-1}$	Steps 3, 4	Chapter 2, Section 2.2
$\tan\left(\frac{\pi}{2} - \theta\right) = \cot\theta$	Step 4	Class XI trigonometry
$(e^{-x})' = -e^{-x}$	Step 5	Chapter 5, Section 5.4 (Basics)

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Miscellaneous · Q7: (1+e2x) dy+(1+y2)ex dx=0(1 + e^{2x})\,dy + (1 + y^2)e^x\,dx = 0(1+e2x)dy+(1+y2)exdx=0⁠, y(0)=1y(0) = 1y(0)=1

Basics you need

Solution

Separate the variables

Integrate, with one constant

Use the condition y=1y = 1y=1 when x=0x = 0x=0

Simplify: the particular solution is y=e−xy = e^{-x}y=e−x

Check by substituting back, and see the curve

Every fact used, and where it comes from

Miscellaneous · Q7: $(1 + e^{2x})\,dy + (1 + y^2)e^x\,dx = 0$ ⁠, $y(0) = 1$

Use the condition $y = 1$ when $x = 0$

Simplify: the particular solution is $y = e^{-x}$