Chapter 9 · Differential Equations

Example 2: $y = e^{-3x}$ solves $y'' + y' - 6y = 0$

ExamplesVerifying a solution · Section 9.3

The question

Verify that the function $y = e^{-3x}$ is a solution of the differential equation

\frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0

Exam answer

Given $y = e^{-3x}$ ⁠. Differentiating with respect to $x$ ⁠,

\frac{dy}{dx} = -3e^{-3x}\quad\ldots(\text{i})

Differentiating (i) with respect to $x$ ⁠,

\frac{d^2y}{dx^2} = 9e^{-3x}

Substituting $\frac{d^2y}{dx^2}$ ⁠, $\frac{dy}{dx}$ and $y$ in the given equation,

\begin{aligned} \text{L.H.S.} &= 9e^{-3x} + (-3e^{-3x}) \\ &\quad - 6e^{-3x} \\ &= 9e^{-3x} - 9e^{-3x} \\ &= 0 = \text{R.H.S.} \end{aligned}

$\therefore$ $y = e^{-3x}$ is a solution of the given differential equation.

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

The one idea this problem is built on

A function is a solution of a differential equation when, after you put it and its derivatives into the equation, the left side equals the right side for every $x$ ⁠. So "verify" means: differentiate as often as the equation needs, substitute, simplify, and see that L.H.S. = R.H.S. Here every derivative of $e^{-3x}$ is a number times $e^{-3x}$ ⁠, so everything collapses to $(9 - 3 - 6)e^{-3x} = 0$ ⁠.

Given, to find, and the plan

Given: The function $y = e^{-3x}$ ⁠, defined for every real $x$ ⁠, and the equation $y'' + y' - 6y = 0$ ⁠.
To show: That $y = e^{-3x}$ satisfies the equation for every real $x$ ⁠.
Plan: (1) Find $\frac{dy}{dx}$ ⁠. (2) Find $\frac{d^2y}{dx^2}$ ⁠. (3) Substitute both, and $y$ ⁠, into the left side and simplify to 0. (4) Conclude. (5) See where the number $-3$ comes from.

Basics you need

1What a solution is

A function $y = \varphi(x)$ is a solution when substituting it for $y$ makes L.H.S. = R.H.S. (Section 9.3). Lesson: general and particular solutions

2Derivative of $e^{kx}$

By the chain rule, $\frac{d}{dx}e^{kx} = ke^{kx}$ for any constant $k$ (Sections 5.3.1 and 5.4). Here $k = -3$ ⁠. Basics: e^{kt} is k times itself

3Second derivative

$\frac{d^2y}{dx^2}$ is the derivative of $\frac{dy}{dx}$ (Section 5.7). Basics: y′ and y″

Solution

First derivative, by the chain rule
$\begin{aligned} \frac{dy}{dx} &= e^{-3x}\cdot\frac{d}{dx}(-3x) \\ &= -3e^{-3x}. \end{aligned}$
The outer function is $e^u$ ⁠, whose derivative is $e^u$ ⁠; the inner function is $u = -3x$ ⁠, whose derivative is $-3$ ⁠. The chain rule multiplies the two.

$\frac{d}{dx}e^{kx} = ke^{kx}$ ⁠, here with $k = -3$ ⁠. Basics: e^{kt} is k times itself
Second derivative
$\begin{aligned} \frac{d^2y}{dx^2} &= \frac{d}{dx}\left(-3e^{-3x}\right) \\ &= -3\cdot(-3e^{-3x}) = 9e^{-3x}. \end{aligned}$
Differentiate the result of Step 1 once more: the constant $-3$ stays in front, and $e^{-3x}$ again gives $-3e^{-3x}$ ⁠. Two minus signs make a plus.

The second derivative is the derivative of the first (Section 5.7). Basics: y′ and y″
Substitute into the left side and simplify
$\begin{aligned} \text{L.H.S.} &= \frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y \\ &= 9e^{-3x} + (-3e^{-3x}) - 6e^{-3x} \\ &= (9 - 3 - 6)\,e^{-3x} \\ &= 0 = \text{R.H.S.} \end{aligned}$
Every term is a number times the same factor $e^{-3x}$ ⁠, so the numbers can be collected: $9 - 3 - 6 = 0$ ⁠. The right side of the equation is 0, so the two sides agree.

Substituting the function and its derivatives is the test of Section 9.3. Lesson: is it a solution?
Conclude, for every real $x$

The calculation of Step 3 used no special value of $x$ ⁠; it holds for every real $x$ ⁠, where $e^{-3x}$ is defined and differentiable. So $y = e^{-3x}$ is a solution of the given differential equation.

It contains no arbitrary constant, so it is a particular solution. The equation has order 2, so a general solution would carry two arbitrary constants.

A "verify" question is answered only by L.H.S. = R.H.S. holding identically, not at one point. Noting the absence of constants connects the answer to the definitions of general and particular solutions.

Section 9.3: a solution free from arbitrary constants is a particular solution. Basics: order n gives n constants
Where the $-3$ comes from

Try $y = e^{mx}$ for a constant $m$ ⁠. Then $y' = me^{mx}$ and $y'' = m^2e^{mx}$ ⁠, and
$\begin{aligned} y'' + y' - 6y &= (m^2 + m - 6)\,e^{mx} \\ &= (m + 3)(m - 2)\,e^{mx}. \end{aligned}$
Since $e^{mx} \ne 0$ ⁠, this is 0 for every $x$ exactly when $m = -3$ or $m = 2$ ⁠. So $e^{-3x}$ works, and so does $e^{2x}$ ⁠.

This explains why the check in Step 3 came out to exactly 0: $-3$ is a root of $m^2 + m - 6 = 0$ ⁠. It also shows that the equation has other solutions, so verifying one function says nothing about it being the only one.

Derivative of $e^{mx}$ by the chain rule. Basics: e^{kt} is k times itself

Result

Substituting $y = e^{-3x}$ ⁠, $y' = -3e^{-3x}$ and $y'' = 9e^{-3x}$ gives L.H.S. $= 9e^{-3x} - 3e^{-3x} - 6e^{-3x} = 0 =$ R.H.S. for every real $x$ ⁠, so $y = e^{-3x}$ is a solution of $\frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0$ ⁠. (Textbook answer: the same, L.H.S. = 0 = R.H.S.)

Check it with numbers

At $x = 0.5$ ⁠: $y = e^{-1.5} \approx 0.22313016$ ⁠, $y' = -3y \approx -0.66939048$ ⁠, $y'' = 9y \approx 2.00817144$ and $6y \approx 1.33878096$ ⁠. Then

\begin{aligned} &2.00817144 - 0.66939048 \\ &\quad - 1.33878096 = 0.00000000. \end{aligned}

The slope $y'$ also agrees with the curve itself: with $e^{-1.5003} \approx 0.2230632311$ and $e^{-1.4997} \approx 0.2231971092$ ⁠,

\begin{aligned} &\frac{0.2230632311 - 0.2231971092}{0.0002} \\ &\approx -0.66939, \end{aligned}

the value of $y'$ found above.

Common mistakes

Forgetting the chain rule: writing $\frac{d}{dx}e^{-3x} = e^{-3x}$ ⁠. The factor $-3$ is essential.
Sign slip in $y''$ ⁠: $-3 \times -3 = +9$ ⁠, not $-9$ ⁠.
Checking at one point only. A single value of $x$ cannot prove a solution; the simplification must give 0 for every $x$ ⁠.
Stopping at "= 0" without the sentence "L.H.S. = R.H.S., so it is a solution". The conclusion carries a mark.

What to take away

To verify a solution: differentiate as many times as the order, substitute, simplify to L.H.S. = R.H.S., and say so.
For $y = e^{mx}$ ⁠, each derivative just multiplies by $m$ ⁠, so the check becomes a quadratic in $m$ ⁠.
A function with no arbitrary constant is a particular solution.

Every fact used, and where it comes from

Fact	Used in	Source
$\frac{d}{dx}e^{kx} = ke^{kx}$ (chain rule)	Steps 1, 2, 5	Chapter 5, Sections 5.3.1 and 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)
General solution (arbitrary constants) and particular solution (none)	Step 4	Section 9.3 (Basics)
$e^{mx} \ne 0$ for every real $x$	Step 5	Chapter 5, Section 5.4

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Example 2: y=e−3xy = e^{-3x}y=e−3x solves y′′+y′−6y=0y'' + y' - 6y = 0y′′+y′−6y=0

Basics you need

Solution

First derivative, by the chain rule

Second derivative

Substitute into the left side and simplify

Conclude, for every real xxx

Where the −3-3−3 comes from

Every fact used, and where it comes from

Example 2: $y = e^{-3x}$ solves $y'' + y' - 6y = 0$

Conclude, for every real $x$

Where the $-3$ comes from