Chapter 9 · Differential Equations

Exercise 9.2 · Q1: $y = e^x + 1$ solves $y'' - y' = 0$

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:

1. $y = e^x + 1$ : $y'' - y' = 0$

Exam answer

Given $y = e^x + 1$ ⁠. Differentiating with respect to $x$ ⁠,

y' = e^x\quad\ldots(\text{i})

Differentiating (i) with respect to $x$ ⁠,

y'' = e^x

Substituting $y''$ and $y'$ in the given equation,

\begin{aligned} \text{L.H.S.} &= e^x - e^x \\ &= 0 = \text{R.H.S.} \end{aligned}

$\therefore$ $y = e^x + 1$ 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

To verify a solution, differentiate as often as the equation needs, substitute, and show L.H.S. = R.H.S. for every $x$ ⁠. Here $e^x$ is its own derivative and the constant $1$ disappears on the first differentiation, so $y'$ and $y''$ are both $e^x$ ⁠, and their difference is 0.

Given, to find, and the plan

Given: The function $y = e^x + 1$ ⁠, defined and differentiable for every real $x$ ⁠, and the equation $y'' - y' = 0$ ⁠.
To show: That $y = e^x + 1$ satisfies the equation for every real $x$ ⁠.
Plan: (1) Find $y'$ ⁠. (2) Find $y''$ ⁠. (3) Substitute into the left side and simplify to 0. (4) State where this holds, and what kind of solution it is.

Basics you need

1What a solution is

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

2The exponential is its own derivative

$\frac{d}{dx}e^x = e^x$ (Section 5.4), and the derivative of a constant is 0. Basics: e^{kt} is k times itself

3 $y'$ and $y''$

$y'$ is the first derivative of $y$ ⁠; $y''$ is the derivative of $y'$ (Section 5.7). Basics: y′ and y″

Solution

First derivative
$y = e^x + 1 \;\Rightarrow\; y' = e^x + 0 = e^x.$
Differentiate term by term. The term $e^x$ gives $e^x$ ⁠, and the constant $1$ gives $0$ because a constant does not change.

$\frac{d}{dx}e^x = e^x$ and $\frac{d}{dx}(\text{constant}) = 0$ ⁠. Basics: e^{kt} is k times itself
Second derivative
$y'' = \frac{d}{dx}\left(e^x\right) = e^x.$
The equation contains $y''$ ⁠, so we differentiate the result of Step 1 once more. Again $e^x$ gives $e^x$ ⁠.

The second derivative is the derivative of the first (Section 5.7). Basics: y′ and y″
Substitute into the left side
$\begin{aligned} \text{L.H.S.} &= y'' - y' \\ &= e^x - e^x \\ &= 0 = \text{R.H.S.} \end{aligned}$
Put the expressions from Steps 2 and 1 in place of $y''$ and $y'$ ⁠. They are equal, so their difference is 0, which is the right side of the equation. The function $y$ itself does not appear in the equation, so it is not substituted.

Section 9.3: substitute and compare the two sides. Lesson: is it a solution? Examples 2 and 3
Where it holds, and what kind of solution it is

The working used no special value of $x$ and no division or root, and $e^x + 1$ is differentiable for every real $x$ ⁠. So L.H.S. = R.H.S. for every real $x$ ⁠: hence verified, $y = e^x + 1$ is a solution of $y'' - y' = 0$ ⁠.

The equation has order 2 (highest derivative $y''$ ⁠) and degree 1 (it is a polynomial equation in $y''$ and $y'$ ⁠, with $y''$ to the power 1). The function has no arbitrary constant, so it is a particular solution. A general solution of this order 2 equation carries two constants, for example $y = Ae^x + B$ ⁠: then $y' = y'' = Ae^x$ and $y'' - y' = 0$ again. Our function is the case $A = 1$ ⁠, $B = 1$ ⁠.

A verification must hold identically, not at one point; checking the steps for divisions, roots or logs is how we know the domain. Counting constants against the order connects the answer to the definitions in Section 9.3.

General solution: as many arbitrary constants as the order; particular: none (Summary, Chapter 9). Basics: order n gives n constants

Result

With $y' = e^x$ and $y'' = e^x$ ⁠, L.H.S. $= y'' - y' = e^x - e^x = 0 =$ R.H.S. for every real $x$ ⁠. Hence $y = e^x + 1$ is a solution of $y'' - y' = 0$ (a particular solution). (The answer key gives no answer for this question, since it is a verification.)

Check it with numbers

At $x = 0.5$ ⁠: $e^{0.5} \approx 1.648721$ ⁠, so $y \approx 2.648721$ ⁠, and $y' = y'' = e^{0.5} \approx 1.648721$ ⁠. Then $y'' - y' = 1.648721 - 1.648721 = 0$ ⁠.

The slope also agrees with the curve itself. With $y(0.501) \approx 2.650371$ and $y(0.499) \approx 2.647073$ ⁠,

\frac{2.650371 - 2.647073}{0.002} \approx 1.649,

which is $y' = e^{0.5} \approx 1.649$ to three decimals.

Common mistakes

Keeping the constant: writing $y' = e^x + 1$ ⁠. The derivative of the constant 1 is 0.
Treating $e^x$ like a power: $\frac{d}{dx}e^x$ is $e^x$ ⁠, not $xe^{x-1}$ ⁠.
Checking at one value of $x$ only. The simplification must give 0 for every $x$ ⁠.
Stopping at "= 0" without the sentence that L.H.S. = R.H.S., so the function is a solution.

What to take away

Verify = differentiate, substitute, simplify to L.H.S. = R.H.S., and say so.
An added constant vanishes on differentiation, so $y = e^x + B$ passes this test for every $B$ ⁠.
Same method as Example 2, where $y = e^{-3x}$ is checked in a second order equation.

Every fact used, and where it comes from

Fact	Used in	Source
$\frac{d}{dx}e^x = e^x$ ⁠; derivative of a constant is 0	Steps 1, 2	Chapter 5, Section 5.4 (Basics)
Second order derivative	Step 2	Chapter 5, Section 5.7 (Basics)
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)
General solution has as many constants as the order; particular has none	Step 4	Section 9.3 and the Summary of Chapter 9 (Basics)

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Exercise 9.2 · Q1: y=ex+1y = e^x + 1y=ex+1 solves y′′−y′=0y'' - y' = 0y′′−y′=0

Basics you need

Solution

First derivative

Second derivative

Substitute into the left side

Where it holds, and what kind of solution it is

Every fact used, and where it comes from

Exercise 9.2 · Q1: $y = e^x + 1$ solves $y'' - y' = 0$