General Solutions by Separation of…

Part 1 · Hook

Why this matters

Checking that

y = Ce^{x^2/2}

solves

\frac{dy}{dx} = xy

takes one line. Finding it in the first place takes a trick: put all the

y

's on one side with

dy

, all the

x

's on the other with

dx

, and integrate. When an equation lets you do that, it is separable, and you can solve it with the antiderivatives you already know.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. $\int \frac{1}{x}\,dx =$

$\ln|x| + C$
$-\frac{1}{x^2} + C$
$\ln x$
$\frac{x^0}{0} + C$

Show the answer

The derivative of $\ln|x|$ is $\frac{1}{x}$ for $x \ne 0$ ; the absolute value keeps negative $x$ allowed.

Correct: $\ln|x| + C$ :
$-\frac{1}{x^2} + C$ :
$\ln x$ :
$\frac{x^0}{0} + C$ :

2. $\int 2xe^{x^2}\,dx =$

$e^{x^2} + C$
$x^2e^{x^2} + C$
$2e^{x^2} + C$
$\frac{e^{x^3}}{3} + C$

Show the answer

With $u = x^2$ , $du = 2x\,dx$ : $\int e^u\,du = e^u + C$ .

Correct: $e^{x^2} + C$ :
$x^2e^{x^2} + C$ :
$2e^{x^2} + C$ :
$\frac{e^{x^3}}{3} + C$ :

Part 4 · See it

See it first

The slope field of dy/dx = xy on −2.5 ≤ x ≤ 2.5 and −4 ≤ y ≤ 4, with the curves y = Ce^(x²/2) for C = 0.25, 0.5, 1, 2 in crimson above the x-axis, for C = −0.25, −0.5, −1, −2 in blue below it, and C = 0, the x-axis itself, in green. Each curve is flattest at x = 0 and follows the segments. — Separating $\frac{dy}{dx} = xy$ gives $y = Ce^{x^2/2}$ . Each curve is one value of $C$ ; $C = 0$ is the $x$ -axis. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

The right side factors as $g(x)h(y)$ the equation is separable: $\frac{1}{h(y)}\,dy = g(x)\,dx$
Integrate both sideseach side becomes an antiderivative; the two constants combine into one C
An antiderivative of $\frac{1}{y}$ is $\ln|y|$ keep the absolute value; exponentiating gives $|y| = e^{C}e^{G(x)}$ , so $y = Ae^{G(x)}$ with $A$ any constant
Solve for $y$ when you canthe general solution: a family of curves, one for each constant

Part 6 · Key ideas

Key ideas

Separable means the right side is a product (or quotient) of a function of $x$ and a function of $y$ . $\frac{dy}{dx} = x + y$ is not separable.
Separate completely: only $y$ and $dy$ on one side, only $x$ and $dx$ on the other. $dy$ and $dx$ stay in the numerators.
Integrate both sides and write $+ C$ once, right away. Adding it after solving for $y$ gives a different, wrong family.
$\int \frac{dy}{y} = \ln|y|$ . Then $|y| = e^{C}e^{\dots}$ , and $\pm e^{C}$ becomes a constant $A$ that may be any sign.

Part 7 · Misconception

A common mistake

The wrong idea: Solve $\frac{dy}{dx} = 2xy^2$ : $-\frac{1}{y} = x^2$ , so $y = -\frac{1}{x^2} + C$ .

What actually happens: The constant belongs in the integration step: $-\frac{1}{y} = x^2 + C$ , so $y = -\frac{1}{x^2 + C}$ . Adding $C$ at the end gives functions that do not satisfy the equation.

Part 8 · Check yourself

Check yourself

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

Table

Four differential equations

Four differential equations, labeled I to IV
Label	I	II	III	IV
Equation	$\frac{dy}{dx} = 2xy^2$	$\frac{dy}{dx} = x + y$	$\frac{dy}{dx} = e^{x - y}$	$\frac{dy}{dx} = x^2 - y^2$

1. Which equations in the table are separable? Select all that apply.

I
II
III
IV

Show the answer

Separable means the right side is $g(x)h(y)$ . I and III factor that way; II and IV do not.

Correct: I: $2xy^2 = (2x)(y^2)$ , a product of a function of $x$ and a function of $y$ .
II: $x + y$ is a sum; it cannot be written as $g(x)h(y)$ .
Correct: III: $e^{x - y} = e^{x} \cdot e^{-y}$ , a product.
IV: $x^2 - y^2$ is a difference of a function of $x$ and a function of $y$ , not a product of one with the other.

2. Which is a correct separation of equation III?

$e^{-y}\,dy = e^{x}\,dx$
$dy = e^{x - y}\,dx$
$e^{y}\,dy = e^{-x}\,dx$
$e^{y}\,dy = e^{x}\,dx$

Show the answer

$e^{x - y} = \frac{e^{x}}{e^{y}}$ , so $e^{y}\,dy = e^{x}\,dx$ .

$e^{-y}\,dy = e^{x}\,dx$ : $e^{-y}$ is on the right side of the equation; moving it to the left divides by it, giving $e^{y}$ .
$dy = e^{x - y}\,dx$ : $y$ is still on the $dx$ side: not separated.
$e^{y}\,dy = e^{-x}\,dx$ : The $x$ factor $e^{x}$ stays as it is; only the $y$ factor moves.
Correct: $e^{y}\,dy = e^{x}\,dx$ : $\frac{dy}{dx} = \frac{e^{x}}{e^{y}}$ ; multiply both sides by $e^{y}\,dx$ .

3. What is the general solution of equation I?

$y = -\frac{1}{x^2} + C$
$y = -\frac{1}{x^2 + C}$
$y = \frac{1}{x^2 + C}$
$y = -(x^2 + C)$

Show the answer

$\int y^{-2}\,dy = \int 2x\,dx$ gives $-\frac{1}{y} = x^2 + C$ , so $y = -\frac{1}{x^2 + C}$ .

$y = -\frac{1}{x^2} + C$ : The constant was added after solving for $y$ ; it belongs right after integrating.
Correct: $y = -\frac{1}{x^2 + C}$ : $-\frac{1}{y} = x^2 + C$ , then take reciprocals and change the sign.
$y = \frac{1}{x^2 + C}$ : $\int y^{-2}\,dy = -\frac{1}{y}$ ; the minus sign was lost.
$y = -(x^2 + C)$ : From $-\frac{1}{y} = x^2 + C$ , $y$ is the reciprocal, not the negative.

4. Which is a correct first step in separating $\frac{dy}{dx} = \frac{x^2}{y}$ ?

$\frac{dy}{y} = x^2\,dx$
$y\,dy = \frac{dx}{x^2}$
$dy = \frac{x^2}{y}\,dx$
$y\,dy = x^2\,dx$

Show the answer

$y\,dy = x^2\,dx$ has only $y$ with $dy$ and only $x$ with $dx$ .

$\frac{dy}{y} = x^2\,dx$ : $y$ is in the denominator on the right; moving it multiplies, giving $y\,dy$ .
$y\,dy = \frac{dx}{x^2}$ : $x^2$ is in the numerator; it stays there.
$dy = \frac{x^2}{y}\,dx$ : $y$ is still on the $dx$ side: not separated.
Correct: $y\,dy = x^2\,dx$ : Multiply both sides by $y\,dx$ .

5. Separate and integrate $\frac{dy}{dx} = y\cos x$ .

$y = \sin x + C$
$\ln|y| = \sin x + C$
$\ln|y| = -\sin x + C$
$\ln|y| = \sin x$

Show the answer

$\frac{dy}{y} = \cos x\,dx$ , so $\ln|y| = \sin x + C$ , and $y = Ae^{\sin x}$ .

$y = \sin x + C$ : $y$ multiplies $\cos x$ ; it must be divided across before integrating.
Correct: $\ln|y| = \sin x + C$ : $\int \frac{dy}{y} = \int \cos x\,dx$ .
$\ln|y| = -\sin x + C$ : An antiderivative of $\cos x$ is $+\sin x$ .
$\ln|y| = \sin x$ : The constant of integration is missing.

6. A student writes: " $\frac{dy}{dx} = x + y$ , so $\frac{dy}{y} = x\,dx$ ." What is wrong?

$x + y$ is a sum, so dividing by $y$ leaves $\frac{x}{y} + 1$ : the equation is not separable.
Nothing: dividing by $y$ is allowed when $y \ne 0$ .
The student should have written $y\,dy = x\,dx$ .
The student forgot $+ C$ .

Show the answer

Separation needs $g(x)h(y)$ . A sum such as $x + y$ cannot be split that way.

Correct: $x + y$ is a sum, so dividing by $y$ leaves $\frac{x}{y} + 1$ : the equation is not separable.: Dividing $x + y$ by $y$ does not leave $x$ .
Nothing: dividing by $y$ is allowed when $y \ne 0$ .: $\frac{x + y}{y} = \frac{x}{y} + 1$ , not $x$ .
The student should have written $y\,dy = x\,dx$ .: That would also be wrong: the right side is not a product.
The student forgot $+ C$ .: No integration has happened yet; the separation itself is wrong.

Part 9 · Summary

Summary

For a separable equation, put

y

with

dy

and

x

with

dx

, integrate both sides, and add one constant immediately. Keep

\ln|y|

; when you exponentiate, the constant absorbs the sign. The general solution is a family of curves, one for each constant.

Part 10 · Up next

What comes next

Part 11 · Connections