Particular Solutions from Initial…

Part 1 · Hook

Why this matters

Separation of variables gives a family of curves, one for each constant. A real problem wants one curve: the temperature of this coffee, which was 90 °C at noon. That one fact, an initial condition, picks out a single member of the family. Use it the moment the constant appears, and watch the domain: some solutions shoot off to infinity and simply stop.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. The general solution of $\frac{dy}{dx} = xy$ is

$y = Ae^{x^2/2}$
$y = e^{x^2/2} + C$
$y = \frac{x^2}{2} + C$
$y = Ae^{x}$

Show the answer

$\ln|y| = \frac{x^2}{2} + C$ , so $y = Ae^{x^2/2}$ .

Correct: $y = Ae^{x^2/2}$ :
$y = e^{x^2/2} + C$ :
$y = \frac{x^2}{2} + C$ :
$y = Ae^{x}$ :

2. $\int 3x^2\,dx =$

$x^3 + C$
$6x + C$
$x^3$
$3x^3 + C$

Show the answer

Raise the power by one and divide by the new power; add C.

Correct: $x^3 + C$ :
$6x + C$ :
$x^3$ :
$3x^3 + C$ :

Part 4 · See it

See it first

The slope field of dy/dx = y² with the curve y = 1/(1 − x). The crimson branch through the closed point (0, 1) rises without bound as x approaches the dashed vertical asymptote x = 1; it is the solution, with domain x < 1. The gray dashed branch for x > 1, below the x-axis, also fits the equation but is not connected to (0, 1), so it is not part of this solution. — $\frac{dy}{dx} = y^2$ with $y(0) = 1$ gives $y = \frac{1}{1 - x}$ . The particular solution is the crimson branch, on $x < 1$ ; the other branch is not connected to $(0, 1)$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Separate and integratean equation with one constant $C$
Substitute the initial condition right awaya number for $C$ , before any rearranging
Solve for $y$ , choosing the sign or branch that passes through the given pointone function, the particular solution
The formula breaks somewhere (division by 0, a negative under a root, a log of 0)the domain is the largest open interval around the initial x where the solution is defined and differentiable

Part 6 · Key ideas

Key ideas

An initial condition $y(x_0) = y_0$ says the solution passes through $(x_0, y_0)$ . Put $x_0$ in for $x$ and $y_0$ in for $y$ .
Use it as soon as $C$ appears. Solving for $y$ first invites algebra slips and wrong constants.
After a square root, keep the sign that matches $y_0$ : $y(1) = -3$ needs the negative root.
State the domain: an open interval containing $x_0$ on which the formula is defined and its derivative exists.

Part 7 · Misconception

A common mistake

The wrong idea: From $\ln|y| = \sin x + C$ : $y = e^{\sin x} + C$ , and $y(0) = 2$ gives $C = 1$ , so $y = e^{\sin x} + 1$ .

What actually happens: Exponentiating gives $y = Ae^{\sin x}$ , not $e^{\sin x} + C$ . Better, use the condition first: $\ln 2 = \sin 0 + C$ , so $C = \ln 2$ and $y$ $= e^{\sin x + \ln 2}$ $= 2e^{\sin x}$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

A student solves an initial value problem

A student's four steps for $\frac{dy}{dx} = \frac{2x}{y}$ with $y(1) = -3$
Step	1	2	3	4
Work	$y\,dy = 2x\,dx$	$\frac{y^2}{2} = x^2 + C$	$\frac{9}{2} = 1 + C$ , so $C = \frac{7}{2}$	$y = \sqrt{2x^2 + 7}$

1. Which step contains an error?

Step 4: since $y(1) = -3$ , the root must be negative.
Step 3: $(-3)^2$ should be $-9$ .
Step 2: the constant belongs on the left side.
Step 1: $y$ should be in the denominator.

Show the answer

$y^2 = 2x^2 + 7$ has two roots. The initial condition $y(1) = -3$ picks the negative one: $y = -\sqrt{2x^2 + 7}$ .

Correct: Step 4: since $y(1) = -3$ , the root must be negative.: $y = \sqrt{2x^2 + 7}$ gives $y(1) = 3$ , not $-3$ .
Step 3: $(-3)^2$ should be $-9$ .: $(-3)^2 = 9$ ; step 3 is right.
Step 2: the constant belongs on the left side.: One constant on either side is fine.
Step 1: $y$ should be in the denominator.: Multiplying by $y\,dx$ gives $y\,dy = 2x\,dx$ ; step 1 is right.

2. Write the correct particular solution for $\frac{dy}{dx} = \frac{2x}{y}$ with $y(1) = -3$ .

Type an expression for $y =$ .

Show the answer

$\frac{y^2}{2} = x^2 + \frac{7}{2}$ , so $y^2 = 2x^2 + 7$ and, since $y(1) < 0$ , $y = -\sqrt{2x^2 + 7}$ .

Answer: $y =$ $-\sqrt{2x^{2} + 7}$

3. What is the domain of the particular solution $y = -\sqrt{2x^2 + 7}$ ?

$x \ge 1$
All real numbers
$x > -\frac{7}{2}$
$|x| \le \sqrt{\frac{7}{2}}$

Show the answer

$2x^2 + 7$ is never below 7, so $y$ is defined and never 0, and $\frac{dy}{dx} = \frac{2x}{y}$ exists for every real $x$ .

$x \ge 1$ : The initial point is inside the domain; the domain is not cut off there.
Correct: All real numbers: $2x^2 + 7 \ge 7 > 0$ , so the root is defined and $y \ne 0$ for every $x$ .
$x > -\frac{7}{2}$ : $2x^2 + 7$ is positive for every $x$ , so nothing is cut off.
$|x| \le \sqrt{\frac{7}{2}}$ : That would need $7 - 2x^2 \ge 0$ ; here the sign is $+$ .

4. When is the best moment to use the initial condition in separation of variables?

Before integrating, to choose which antiderivative to use.
At the end, after solving for $y$ and then adding $C$ .
It is not needed for a separable equation.
As soon as $C$ appears: substitute the point, find $C$ , then solve for $y$ .

Show the answer

Substituting right after integrating finds $C$ in the simplest form and avoids sign and algebra slips.

Before integrating, to choose which antiderivative to use.: You cannot find C before it exists.
At the end, after solving for $y$ and then adding $C$ .: A $C$ added after solving gives the wrong family.
It is not needed for a separable equation.: Without it you have a family, not one solution.
Correct: As soon as $C$ appears: substitute the point, find $C$ , then solve for $y$ .: $C$ is found in the simplest equation, before rearranging.

5. Find the particular solution of $\frac{dy}{dx} = y\cos x$ with $y(0) = 2$ .

Type an expression for $y =$ .

Show the answer

$\ln|y| = \sin x + C$ ; at $(0, 2)$ , $C = \ln 2$ . So $y$ $= e^{\sin x + \ln 2}$ $= 2e^{\sin x}$ .

Answer: $y =$ $2e^{\sin(x)}$

6. The particular solution of $\frac{dy}{dx} = -\frac{x}{y}$ with $y(0) = 3$ is $y = \sqrt{9 - x^2}$ . What is its domain?

$-3 < x < 3$
$-3 \le x \le 3$
$x \ge 0$
$(-\infty, \infty)$

Show the answer

$9 - x^2 > 0$ and $y \ne 0$ require $-3 < x < 3$ , an open interval containing $x = 0$ .

Correct: $-3 < x < 3$ : At $x = \pm 3$ , $y = 0$ and $-\frac{x}{y}$ is undefined, so the endpoints are out.
$-3 \le x \le 3$ : At $x = \pm 3$ , $y = 0$ and the differential equation divides by 0.
$x \ge 0$ : The solution is defined for negative $x$ too, back to $-3$ .
$(-\infty, \infty)$ : For $|x| > 3$ the root is not real.

Part 9 · Summary

Summary

A particular solution satisfies the differential equation and an initial condition. Separate, integrate, add

C

, substitute the initial point immediately, then solve for

y

with the sign that fits. Give its domain: the open interval around the initial

x

where it is defined and differentiable.

Part 10 · Up next

What comes next

Part 11 · Connections