Chapter 5 · Continuity and Differentiability

Example 37: $y=3e^{2x}+2e^{3x}$ satisfies $y''-5y'+6y=0$

ExamplesSecond order derivative · Section 5.7

The question

If $y = 3e^{2x} + 2e^{3x}$ ⁠, prove that $\dfrac{d^2y}{dx^2} - 5\dfrac{dy}{dx} + 6y = 0$ ⁠.

Exam answer

Here $y = 3e^{2x} + 2e^{3x}$ ⁠.

Differentiating twice w.r.t. $x$ ⁠,

\begin{aligned} \frac{dy}{dx} &= 6e^{2x} + 6e^{3x} \\ \frac{d^2y}{dx^2} &= 12e^{2x} + 18e^{3x} \end{aligned}

Substituting in the LHS,

\begin{aligned} \text{LHS} &= \frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y \\ &= (12e^{2x} + 18e^{3x}) \\ &\quad - 5(6e^{2x} + 6e^{3x}) \\ &\quad + 6(3e^{2x} + 2e^{3x}) \\ &= (12 - 30 + 18)e^{2x} \\ &\quad + (18 - 30 + 12)e^{3x} \\ &= 0 = \text{RHS} \end{aligned}

$\therefore\ \dfrac{d^2y}{dx^2} - 5\dfrac{dy}{dx} + 6y = 0$ ⁠. Hence proved.

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

The one idea this problem is built on

Differentiating $e^{kx}$ just multiplies it by $k$ ⁠. So every derivative of $y$ is again a combination of $e^{2x}$ and $e^{3x}$ ⁠, and the whole expression $y'' - 5y' + 6y$ collapses to

(\text{number})\,e^{2x} + (\text{number})\,e^{3x}.

We only have to check that both numbers are zero.

Given, to show, and the plan

Given: $y = 3e^{2x} + 2e^{3x}$ ⁠, defined and differentiable for every real $x$ ⁠.
To show: $\dfrac{d^2y}{dx^2} - 5\dfrac{dy}{dx} + 6y = 0$ for every real $x$ ⁠.
Plan: Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ with the chain rule, substitute all three into the left side, and collect the coefficients of $e^{2x}$ and of $e^{3x}$ ⁠.

Basics you need

1Derivative of $e^{kx}$

$\frac{d}{dx}e^x = e^x$ (Theorem 5, Section 5.4). By the chain rule with inner function $u = kx$ ( $k$ a constant):

\frac{d}{dx}e^{kx} = e^{kx}\cdot\frac{d}{dx}(kx) = k\,e^{kx}.

Basics: exponentials and logarithms

2Second order derivative (Section 5.7)

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$ ⁠. Lesson: second order derivative

3Collecting like terms

$p\,e^{2x} + q\,e^{2x} = (p+q)\,e^{2x}$ ⁠. Terms in $e^{2x}$ and terms in $e^{3x}$ are collected separately. Basics: solving for dy/dx

Solution

First derivative

Use Basic 1 with $k = 2$ and $k = 3$ ⁠:
$\begin{aligned} \frac{dy}{dx} &= 3\cdot 2e^{2x} + 2\cdot 3e^{3x} \\ &= 6e^{2x} + 6e^{3x}. \end{aligned}$
Each term is a constant times $e^{kx}$ ⁠. The constant stays in front and the chain rule brings down the factor $k$ ⁠. The book also writes this as $6(e^{2x} + e^{3x})$ ⁠.

$\frac{d}{dx}e^{kx} = k\,e^{kx}$ (Basic 1). Lesson: exponential and logarithmic functions
Second derivative

Differentiate $\frac{dy}{dx}$ again:
$\begin{aligned} \frac{d^2y}{dx^2} &= 6\cdot 2e^{2x} + 6\cdot 3e^{3x} \\ &= 12e^{2x} + 18e^{3x}. \end{aligned}$
The same rule once more: each $e^{2x}$ gains a factor 2 and each $e^{3x}$ a factor 3. (The book writes this as $6(2e^{2x} + 3e^{3x})$ ⁠.)

Section 5.7: the second derivative is the derivative of $\frac{dy}{dx}$ ⁠. Lesson: second order derivative
Substitute into the left side
$\begin{aligned} &\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y \\ &= (12e^{2x} + 18e^{3x}) - 5(6e^{2x} + 6e^{3x}) \\ &\quad + 6(3e^{2x} + 2e^{3x}). \end{aligned}$
The statement to prove is about this particular combination, so we replace each of $\frac{d^2y}{dx^2}$ ⁠, $\frac{dy}{dx}$ ⁠, $y$ by what we found. Brackets keep the multiplication by $-5$ and $6$ correct.

Substitution: an expression may be replaced by anything equal to it.
Collect the coefficients

Expand and collect the $e^{2x}$ terms and the $e^{3x}$ terms separately (Basic 3):
$\begin{aligned} e^{2x}&:\ 12 - 30 + 18 = 0, \\ e^{3x}&:\ 18 - 30 + 12 = 0. \end{aligned}$
So the left side equals $0\cdot e^{2x} + 0\cdot e^{3x} = 0$ for every real $x$ ⁠. Hence $\dfrac{d^2y}{dx^2} - 5\dfrac{dy}{dx} + 6y = 0$ .

Both coefficients vanish, so the identity holds at every $x$ ⁠, not just at chosen points. The book reaches the same zero by writing the three pieces as $6(2e^{2x}+3e^{3x})$ ⁠, $-30(e^{2x}+e^{3x})$ and $6(3e^{2x}+2e^{3x})$ ⁠.

$-5 \times 6 = -30$ and $6 \times 3 = 18$ ⁠, $6 \times 2 = 12$ ⁠.
Why the numbers 2 and 3 make this work

For a single term $y = e^{kx}$ we have $y' = k e^{kx}$ and $y'' = k^2 e^{kx}$ ⁠, so
$\begin{aligned} y'' - 5y' + 6y &= (k^2 - 5k + 6)\,e^{kx} \\ &= (k-2)(k-3)\,e^{kx}. \end{aligned}$
This is zero exactly when $k = 2$ or $k = 3$ ⁠, because $e^{kx}$ is never zero. And since the left side is built from $y$ and its derivatives only by adding and multiplying by constants, any combination $C_1 e^{2x} + C_2 e^{3x}$ also gives zero; the example is the case $C_1 = 3$ ⁠, $C_2 = 2$ ⁠.

This explains the "coincidence" in Step 4: the coefficient of $e^{2x}$ was $3(2^2 - 5\cdot 2 + 6) = 3\cdot 0$ and that of $e^{3x}$ was $2(3^2 - 5\cdot 3 + 6) = 2\cdot 0$ ⁠. It is a first glimpse of how such differential equations are solved (Chapter 9).

$e^{kx} \gt 0$ for every real $x$ ⁠. Basics: exponentials and logarithms

Result

$\dfrac{dy}{dx} = 6e^{2x} + 6e^{3x}$ ⁠, $\dfrac{d^2y}{dx^2} = 12e^{2x} + 18e^{3x}$ ⁠, and

\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = 0

for every real $x$ ⁠.

Check it with numbers

At $x = 0$ ⁠: $y = 3 + 2 = 5$ ⁠, $\frac{dy}{dx} = 6 + 6 = 12$ ⁠, $\frac{d^2y}{dx^2} = 12 + 18 = 30$ ⁠. Then $30 - 5(12) + 6(5) = 30 - 60 + 30 = 0$ ⁠.

At $x = 0.5$ ⁠: $y \approx 17.1182236$ ⁠, $\frac{dy}{dx} \approx 43.1998254$ ⁠, $\frac{d^2y}{dx^2} \approx 113.2897852$ ⁠. Then

\begin{aligned} &113.2897852 - 215.9991270 \\ &\quad + 102.7093418 = 0.0000000. \end{aligned}

The derivative values themselves can be checked from $y$ alone: with $h = 0.001$ ⁠, $\frac{y(0.501) - y(0.499)}{0.002} \approx 43.19988$ ⁠, and the second difference $\frac{y(0.501) - 2y(0.5) + y(0.499)}{0.000001}$ is about $113.290$ ⁠.

Common mistakes

Forgetting the chain-rule factor. $\frac{d}{dx}e^{2x} = 2e^{2x}$ ⁠, not $e^{2x}$ ⁠.
Sign slips when multiplying by $-5$ ⁠. Keep $\frac{dy}{dx}$ in brackets: $-5(6e^{2x} + 6e^{3x}) = -30e^{2x} - 30e^{3x}$ ⁠.
Mixing the two exponentials. $e^{2x}$ and $e^{3x}$ are different functions; their coefficients must be collected separately.

What to take away

Derivatives of $e^{kx}$ are multiples of $e^{kx}$ ⁠: $y' = k e^{kx}$ ⁠, $y'' = k^2 e^{kx}$ ⁠.
To prove a relation such as $y'' - 5y' + 6y = 0$ ⁠, compute each derivative, substitute, and collect coefficients.
$e^{kx}$ satisfies $y'' - 5y' + 6y = 0$ exactly when $k^2 - 5k + 6 = 0$ ⁠, that is $k = 2$ or $3$ ⁠. Exercise 5.7, Question 14 ( $y = Ae^{mx} + Be^{nx}$ ⁠) is the general version of this example.

Every fact used, and where it comes from

Fact	Used in	Source
$\frac{d}{dx}e^x = e^x$	Steps 1, 2, 5	Theorem 5, Section 5.4 (Lesson: exponential and logarithmic functions)
Chain rule, giving $\frac{d}{dx}e^{kx} = k e^{kx}$	Steps 1, 2, 5	Theorem 4, Section 5.3.1 (as in Example 26(i))
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$	Step 2	Section 5.7 (Lesson: second order derivative)
$k^2 - 5k + 6 = (k-2)(k-3)$ ⁠; $e^{kx} \neq 0$	Step 5	Factorisation; Section 5.4

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Example 37: y=3e2x+2e3xy=3e^{2x}+2e^{3x}y=3e2x+2e3x satisfies y′′−5y′+6y=0y''-5y'+6y=0y′′−5y′+6y=0

Basics you need

Solution

First derivative

Second derivative

Substitute into the left side

Collect the coefficients

Why the numbers 2 and 3 make this work

Every fact used, and where it comes from

Example 37: $y=3e^{2x}+2e^{3x}$ satisfies $y''-5y'+6y=0$