Chapter 9 · Differential Equations

Example 3: $y = a\cos x + b\sin x$ solves $y'' + y = 0$

ExamplesVerifying a solution · Section 9.3

The question

Verify that the function $y = a\cos x + b\sin x$ ⁠, where, $a, b \in \mathbf{R}$ is a solution of the differential equation $\dfrac{d^2y}{dx^2} + y = 0$

Exam answer

Given

y = a\cos x + b\sin x\quad\ldots(\text{i})

Differentiating (i) with respect to $x$ ⁠, successively,

\begin{aligned} \frac{dy}{dx} &= -a\sin x + b\cos x \\ \frac{d^2y}{dx^2} &= -a\cos x - b\sin x \end{aligned}

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

\begin{aligned} \text{L.H.S.} &= (-a\cos x - b\sin x) \\ &\quad + (a\cos x + b\sin x) \\ &= 0 = \text{R.H.S.} \end{aligned}

$\therefore$ $y = a\cos x + b\sin x$ 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

Differentiating $\cos x$ twice gives $-\cos x$ ⁠, and differentiating $\sin x$ twice gives $-\sin x$ ⁠. So for any mix $y = a\cos x + b\sin x$ ⁠, the second derivative is exactly $-y$ ⁠, which is what $y'' + y = 0$ says. The constants $a$ and $b$ just ride along, so the check works for all of them at once.

Given, to find, and the plan

Given: $y = a\cos x + b\sin x$ ⁠, with arbitrary real constants $a$ ⁠, $b$ ⁠; defined for every real $x$ ⁠. The equation $\frac{d^2y}{dx^2} + y = 0$ ⁠.
To show: That the function satisfies the equation for every real $x$ and every choice of $a, b$ ⁠.
Plan: (1) Find $\frac{dy}{dx}$ ⁠. (2) Find $\frac{d^2y}{dx^2}$ ⁠. (3) Substitute into the left side and simplify to 0. (4) Conclude, and count the constants against the order.

Basics you need

1What a solution is

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

2Derivatives of sine and cosine

$\frac{d}{dx}\sin x = \cos x$ and $\frac{d}{dx}\cos x = -\sin x$ ⁠; a constant factor stays in front (Chapter 5). Basics: y′ and y″

3Arbitrary constants and the order

A general solution of an equation of order $n$ carries $n$ arbitrary constants; a particular solution carries none (Section 9.3 and the chapter Summary). Basics: order n gives n constants

Solution

First derivative
$\begin{aligned} \frac{dy}{dx} &= a(-\sin x) + b\cos x \\ &= -a\sin x + b\cos x. \end{aligned}$
Differentiate term by term. The constants $a$ and $b$ stay as factors because the derivative of a constant multiple is the constant times the derivative.

$(\cos x)' = -\sin x$ ⁠, $(\sin x)' = \cos x$ ⁠. Basics: y′ and y″
Second derivative
$\begin{aligned} \frac{d^2y}{dx^2} &= -a\cos x + b(-\sin x) \\ &= -a\cos x - b\sin x \\ &= -(a\cos x + b\sin x) = -y. \end{aligned}$
Differentiating $-a\sin x$ gives $-a\cos x$ ⁠, and differentiating $b\cos x$ gives $-b\sin x$ ⁠. The result is the original function with its sign changed, which is the key fact.

The second derivative is the derivative of the first (Section 5.7). Basics: y′ and y″
Substitute into the left side
$\begin{aligned} \text{L.H.S.} &= \frac{d^2y}{dx^2} + y \\ &= (-a\cos x - b\sin x) \\ &\quad + (a\cos x + b\sin x) \\ &= 0 = \text{R.H.S.} \end{aligned}$
The $a\cos x$ terms cancel and the $b\sin x$ terms cancel. The first derivative is not needed in the substitution, because $\frac{dy}{dx}$ does not appear in the equation; it was only a stage on the way to $\frac{d^2y}{dx^2}$ ⁠.

Section 9.3: substitute and compare the two sides. Lesson: is it a solution?
Conclude, and count the constants

Step 3 holds for every real $x$ and for every choice of $a, b \in \mathbf{R}$ ⁠. So $y = a\cos x + b\sin x$ is a solution of $\frac{d^2y}{dx^2} + y = 0$ ⁠.

It contains two arbitrary constants, $a$ and $b$ ⁠, as many as the order of the equation (2). Giving them values gives particular solutions: for example $a = 1$ ⁠, $b = 0$ gives $y = \cos x$ ⁠, and $a = 0$ ⁠, $b = 1$ gives $y = \sin x$ ⁠.

Because $a$ and $b$ were never given values, one calculation proves the claim for the whole family at once. The count of constants matches the order, which is what the chapter Summary asks of a general solution.

Section 9.3: general and particular solutions; the chapter Summary: as many constants as the order. Basics: order n gives n constants
The same family as the book's $a\sin(x + b)$

Section 9.3 uses $y = A\sin(x + B)$ as a solution of the same equation. By the addition formula,
$\begin{aligned} A\sin(x + B) &= (A\sin B)\cos x \\ &\quad + (A\cos B)\sin x, \end{aligned}$
which is $a\cos x + b\sin x$ with $a = A\sin B$ and $b = A\cos B$ ⁠. Conversely, any $a, b$ arise this way: take $A = \sqrt{a^2 + b^2}$ and an angle $B$ with $\sin B = \frac{a}{A}$ ⁠, $\cos B = \frac{b}{A}$ (when $A \ne 0$ ⁠).

The two forms describe the same functions, so a general solution can look different in different books and still be correct. This matters when you compare your answer with a key.

$\sin(x + B) = \sin x\cos B + \cos x\sin B$ (Class XI). Lesson: general and particular solutions

Result

With $y'' = -a\cos x - b\sin x$ ⁠, L.H.S. $= y'' + y = 0 =$ R.H.S. for every real $x$ and all $a, b \in \mathbf{R}$ ⁠, so $y = a\cos x + b\sin x$ is a solution of $\frac{d^2y}{dx^2} + y = 0$ ⁠. (Textbook answer: the same, L.H.S. = 0 = R.H.S.)

Check it with numbers

Take $a = 2$ ⁠, $b = 3$ and $x = 1$ ⁠. Then $y = 2\cos 1 + 3\sin 1$ ⁠, with $2\cos 1 \approx 1.080604612$ and $3\sin 1 \approx 2.524412954$ ⁠, so $y \approx 3.605017566$ ⁠, and the formula of Step 2 gives $y'' \approx -3.605017566$ ⁠, and $y'' + y = 0$ ⁠.

The curve itself agrees. With $y(0.99) \approx 3.6054576570$ ⁠, $y(1) \approx 3.6050175662$ and $y(1.01) \approx 3.6042169766$ ⁠, the second difference

\begin{aligned} &\frac{y(1.01) - 2y(1) + y(0.99)}{0.01^2} \\ &\approx -3.6050 \end{aligned}

matches $-y(1) \approx -3.6050$ to four decimals (a step of 0.01 makes it an approximation).

Common mistakes

Sign of $(\cos x)'$ ⁠: it is $-\sin x$ ⁠. Getting it wrong gives $y'' = a\cos x - b\sin x$ ⁠, and the check fails.
Choosing values for $a$ and $b$ and checking only those. The question is about all real $a, b$ ⁠, so keep them as letters.
Substituting $\frac{dy}{dx}$ into an equation that does not contain it.

What to take away

$\sin x$ and $\cos x$ each satisfy $y'' = -y$ ⁠, and so does any combination of them.
Keep arbitrary constants as letters: one calculation then proves the whole family.
Two constants for an order 2 equation: the count matches, as a general solution should.

Every fact used, and where it comes from

Fact	Used in	Source
$(\sin x)' = \cos x$ ⁠, $(\cos x)' = -\sin x$ ⁠, constant multiple rule	Steps 1, 2	Chapter 5 (Basics)
Second order derivative	Step 2	Chapter 5, Section 5.7
A solution makes L.H.S. = R.H.S. when substituted	Step 3	Section 9.3 (Lesson)
General solution: as many arbitrary constants as the order; particular: none	Step 4	Section 9.3; chapter Summary (Basics)
$y = a\sin(x + b)$ solves $y'' + y = 0$	Step 5	Section 9.3, equations (3) and (4)
$\sin(x + B) = \sin x\cos B + \cos x\sin B$	Step 5	Class XI trigonometry

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Example 3: y=acos⁡x+bsin⁡xy = a\cos x + b\sin xy=acosx+bsinx solves y′′+y=0y'' + y = 0y′′+y=0

Basics you need

Solution

First derivative

Second derivative

Substitute into the left side

Conclude, and count the constants

The same family as the book's asin⁡(x+b)a\sin(x + b)asin(x+b)

Every fact used, and where it comes from

Example 3: $y = a\cos x + b\sin x$ solves $y'' + y = 0$

The same family as the book's $a\sin(x + b)$