Chapter 5 · Continuity and Differentiability

Exercise 5.7 · Q15: $y=500e^{7x}+600e^{-7x}$ satisfies $y''=49y$

Exercise 5.7Second order derivative · Section 5.7

The question

15. If $y = 500e^{7x} + 600e^{-7x}$ ⁠, show that $\dfrac{d^2y}{dx^2} = 49y$ ⁠.

Exam answer

Here $y = 500e^{7x} + 600e^{-7x}$ ⁠.

\begin{aligned} \frac{dy}{dx} &= 500\cdot 7e^{7x} + 600\cdot(-7)e^{-7x} \\ &= 3500e^{7x} - 4200e^{-7x} \\ \frac{d^2y}{dx^2} &= 24500e^{7x} + 29400e^{-7x} \\ &= 49(500e^{7x} + 600e^{-7x}) \end{aligned}

$\therefore\ \dfrac{d^2y}{dx^2} = 49y$ ⁠. Hence proved.

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

The one idea this problem is built on

Differentiating $e^{kx}$ twice multiplies it by $k^2$ ⁠. Here the two exponents are $7x$ and $-7x$ ⁠, and

7^2 = (-7)^2 = 49,

so both terms of $y$ get multiplied by the same number 49, and therefore so does $y$ ⁠.

Given, to show, and the plan

Given: $y = 500e^{7x} + 600e^{-7x}$ ⁠, defined and differentiable for every real $x$ ⁠.
To show: $\dfrac{d^2y}{dx^2} = 49y$ for every real $x$ ⁠.
Plan: (1) $\frac{dy}{dx}$ by the chain rule. (2) $\frac{d^2y}{dx^2}$ the same way. (3) Compute the right side $49y$ in full and compare the two sides term by term.

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 $u = kx$ ⁠, $\frac{du}{dx} = k$ ⁠:

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

For $k = -7$ this gives $\frac{d}{dx}e^{-7x} = -7e^{-7x}$ ⁠. Basics: exponentials and logarithms

2Sum rule and constant multiples

$(u + v)' = u' + v'$ (Section 5.3, rule (1)) and $(cu)' = cu'$ for a constant $c$ (Class XI, Limits and Derivatives). Basics: derivative rules

3Second order derivative (Section 5.7)

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

Solution

First derivative

Term by term, with Basic 1 ( $k = 7$ ⁠, then $k = -7$ ⁠):
$\begin{aligned} \frac{dy}{dx} &= 500\cdot 7e^{7x} + 600\cdot(-7)e^{-7x} \\ &= 3500e^{7x} - 4200e^{-7x}. \end{aligned}$
The chain rule brings down the constant in each exponent. For the second term that constant is $-7$ ⁠, which makes the term negative: $600 \times (-7) = -4200$ ⁠.

Basic 1; Section 5.3, rule (1). Lesson: chain rule
Second derivative
$\begin{aligned} \frac{d^2y}{dx^2} &= 3500\cdot 7e^{7x} - 4200\cdot(-7)e^{-7x} \\ &= 24500e^{7x} + 29400e^{-7x}. \end{aligned}$
Again each exponent's constant comes down. In the second term $-4200 \times (-7) = +29400$ ⁠: the two minus signs cancel, so after two differentiations both terms are positive again, like the original $y$ ⁠.

Section 5.7. Lesson: second order derivative
Compute the right side and compare

The right side is
$\begin{aligned} 49y &= 49\cdot 500e^{7x} + 49\cdot 600e^{-7x} \\ &= 24500e^{7x} + 29400e^{-7x}. \end{aligned}$
This is exactly the expression for $\frac{d^2y}{dx^2}$ in Step 2, term for term. Hence
$\frac{d^2y}{dx^2} = 49y \quad\text{for every real } x.$
To show two expressions are equal, compute both fully and compare. Both are combinations of $e^{7x}$ and $e^{-7x}$ with the same coefficients ( $24500 = 49 \times 500$ and $29400 = 49 \times 600$ ⁠), so they agree at every $x$ ⁠. Equivalently, $\frac{d^2y}{dx^2} = 49(500e^{7x} + 600e^{-7x})$ ⁠.

Result

$\dfrac{dy}{dx} = 3500e^{7x} - 4200e^{-7x}$ and $\dfrac{d^2y}{dx^2} = 24500e^{7x} + 29400e^{-7x}$ $= 49y$ ⁠, so

\frac{d^2y}{dx^2} = 49y \quad\text{for every real } x.

(The answer key gives no entry for this "show that" question.)

Check it with numbers

At $x = 0$ : $y = 500 + 600 = 1100$ and $\frac{d^2y}{dx^2} = 24500 + 29400 = 53900$ ⁠, which is $49 \times 1100$ ⁠.

At $x = 0.1$ : $e^{0.7} \approx 2.0137527$ and $e^{-0.7} \approx 0.4965853$ ⁠, so

\begin{aligned} y &\approx 1006.8763537 + 297.9511823 \\ &\approx 1304.8275360, \\ \frac{d^2y}{dx^2} &\approx 49336.9413330 + 14599.6079315 \\ &\approx 63936.5492645, \end{aligned}

and $49 \times 1304.8275360$ $\approx 63936.5492645$ ⁠, the same.

Common mistakes

Dropping the minus from the exponent. $\frac{d}{dx}e^{-7x} = -7e^{-7x}$ ⁠, not $7e^{-7x}$ ⁠.
Wrong sign after the second differentiation. $\frac{d}{dx}\left(-4200e^{-7x}\right) = +29400e^{-7x}$ ⁠.
Multiplying by 7 instead of 49. Each differentiation brings a factor $\pm 7$ ⁠; two of them give $49$ ⁠.

What to take away

$\frac{d^2}{dx^2}e^{kx} = k^2e^{kx}$ ⁠, so any combination $Ae^{kx} + Be^{-kx}$ satisfies $y'' = k^2y$ ⁠.
This is Question 14 with $A = 500$ ⁠, $B = 600$ ⁠, $m = 7$ ⁠, $n = -7$ ⁠: there $m + n = 0$ and $mn = -49$ ⁠, so $y'' - (m+n)y' + mny = 0$ becomes $y'' - 49y = 0$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$\frac{d}{dx}e^x = e^x$	Steps 1, 2	Theorem 5, Section 5.4 (Lesson: exponential and logarithmic functions)
Chain rule: $\frac{d}{dx}e^{kx} = ke^{kx}$	Steps 1, 2	Theorem 4, Section 5.3.1 (Lesson: chain rule)
Sum rule and constant multiples	Steps 1–3	Section 5.3, rule (1) (sum); Class XI, Limits and Derivatives (constant multiple) (Basics: derivative rules)
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$	Step 2	Section 5.7 (Lesson: second order derivative)

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Exercise 5.7 · Q15: y=500e7x+600e−7xy=500e^{7x}+600e^{-7x}y=500e7x+600e−7x satisfies y′′=49yy''=49yy′′=49y

Basics you need

Solution

First derivative

Second derivative

Compute the right side and compare

Every fact used, and where it comes from

Exercise 5.7 · Q15: $y=500e^{7x}+600e^{-7x}$ satisfies $y''=49y$