Chapter 9 · Differential Equations

Exercise 9.2 · Q2: $y = x^2 + 2x + C$ solves $y' - 2x - 2 = 0$

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:

2. $y = x^2 + 2x + C$ : $y' - 2x - 2 = 0$

Exam answer

Given $y = x^2 + 2x + C$ ⁠. Differentiating with respect to $x$ ⁠,

y' = 2x + 2

Substituting $y'$ in the given equation,

\begin{aligned} \text{L.H.S.} &= (2x + 2) - 2x - 2 \\ &= 0 = \text{R.H.S.} \end{aligned}

$\therefore$ $y = x^2 + 2x + C$ 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

Differentiating $x^2 + 2x + C$ gives $2x + 2$ whatever the constant $C$ is, because a constant has derivative 0. So $y' - 2x - 2$ is 0 for every $x$ and every $C$ ⁠: one calculation verifies a whole family of curves, all shifted copies of one parabola.

Given, to find, and the plan

Given: The function $y = x^2 + 2x + C$ ⁠, where $C$ is an arbitrary constant, defined for every real $x$ ⁠; the equation $y' - 2x - 2 = 0$ ⁠.
To show: That the function satisfies the equation for every real $x$ and every value of $C$ ⁠.
Plan: (1) Find $y'$ ⁠. (2) Substitute into the left side and simplify to 0. (3) State where it holds. (4) Explain the role of $C$ ⁠.

Basics you need

1What a solution is

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

2Derivatives of powers and constants

$\frac{d}{dx}x^n = nx^{n-1}$ ⁠, so $(x^2)' = 2x$ and $(2x)' = 2$ ⁠; a constant has derivative 0 (Chapter 5). Basics: y′ and y″

3Arbitrary constants

A general solution carries as many arbitrary constants as the order of the equation (Summary, Chapter 9). Basics: order n gives n constants

Solution

Differentiate
$\begin{aligned} y &= x^2 + 2x + C \\ \Rightarrow\ y' &= 2x + 2 + 0 \\ &= 2x + 2. \end{aligned}$
Differentiate term by term: $x^2$ gives $2x$ ⁠, $2x$ gives $2$ ⁠, and the constant $C$ gives $0$ ⁠. The equation contains only $y'$ ⁠, so one derivative is enough.

Power rule and derivative of a constant (Chapter 5). Basics: y′ and y″
Substitute into the left side
$\begin{aligned} \text{L.H.S.} &= y' - 2x - 2 \\ &= (2x + 2) - 2x - 2 \\ &= 0 = \text{R.H.S.} \end{aligned}$
Put $2x + 2$ in place of $y'$ ⁠. The $2x$ terms cancel and the $2$ s cancel, leaving 0, the right side.

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

No step divides, takes a root or a log, and $x^2 + 2x + C$ is differentiable for every real $x$ ⁠. The value of $C$ never entered the calculation. So L.H.S. = R.H.S. for every real $x$ and every constant $C$ ⁠: hence verified.

A verification must hold identically. Because $C$ was kept as a letter, the check covers every member of the family at once, not one chosen curve.

A solution must satisfy the equation for all $x$ in its domain (Section 9.3). Lesson: is it a solution?
The constant $C$ ⁠: a general solution

The equation $y' = 2x + 2$ has order 1 (only $y'$ appears) and degree 1 (it is a polynomial equation in $y'$ ⁠, with $y'$ to the power 1). The function contains one arbitrary constant $C$ ⁠, as many as the order, so it is a general solution. Giving $C$ a value gives a particular solution: for example $C = 0$ gives $y = x^2 + 2x$ ⁠, and $C = -1$ gives $y = (x + 1)^2 - 2$ ⁠, the curve through $(0, -1)$ ⁠.

Indeed, integrating $y' = 2x + 2$ gives $y = x^2 + 2x + C$ ⁠: every solution is of this form.

Each value of $C$ shifts the parabola up or down without changing its slope at any $x$ ⁠; that is exactly why $C$ disappears from $y'$ ⁠. One integration produces one constant.

General solution: as many arbitrary constants as the order (Section 9.3 and the Summary). Basics: order n gives n constants

Result

With $y' = 2x + 2$ ⁠, L.H.S. $= (2x + 2) - 2x - 2 = 0 =$ R.H.S. for every real $x$ and every $C$ ⁠. Hence $y = x^2 + 2x + C$ is a solution of $y' - 2x - 2 = 0$ (its general solution, one constant for order 1). (The answer key gives no answer for this question, since it is a verification.)

Check it with numbers

At $x = 1.5$ ⁠: $y' = 2(1.5) + 2 = 5$ ⁠, and $y' - 2x - 2 = 5 - 3 - 2 = 0$ ⁠.

The slope does not depend on $C$ ⁠. With $C = 0$ ⁠: $y(1.501) = 5.255001$ and $y(1.499) = 5.245001$ ⁠, so

\frac{5.255001 - 5.245001}{0.002} = 5.

With $C = 4$ ⁠: $y(1.501) = 9.255001$ and $y(1.499) = 9.245001$ ⁠, so

\frac{9.255001 - 9.245001}{0.002} = 5.

Both give $y' = 5$ ⁠, exactly (for a quadratic this symmetric difference is exact).

Common mistakes

Differentiating $C$ to 1 or keeping it: its derivative is 0.
Choosing a value of $C$ and checking only that curve. The question is about every $C$ ⁠, so keep it as a letter.
Confusing the roles: the equation has no constant; the solution does. Arbitrary constants belong to solutions.

What to take away

An additive constant in $y$ never survives differentiation, so it never affects the check.
One arbitrary constant for an order 1 equation: a general solution. Compare Q3, which has the same structure with $\cos x + C$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$\frac{d}{dx}x^n = nx^{n-1}$ ⁠; derivative of a constant is 0	Step 1	Chapter 5 (Basics)
A solution makes L.H.S. = R.H.S. when substituted	Steps 2, 3	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 · Q2: y=x2+2x+Cy = x^2 + 2x + Cy=x2+2x+C solves y′−2x−2=0y' - 2x - 2 = 0y′−2x−2=0

Basics you need

Solution

Differentiate

Substitute into the left side

Where it holds

The constant CCC⁠: a general solution

Every fact used, and where it comes from

Exercise 9.2 · Q2: $y = x^2 + 2x + C$ solves $y' - 2x - 2 = 0$

The constant $C$ ⁠: a general solution