Single-Variable-Calculus · Unit 13 · Video 1 · Interactive Practice

Separating the Variables: Splitting dy/dx to Solve Differential Equations

IKey Formulas

Formula	Step	For $\dfrac{dy}{dx} = -xy$
$\dfrac{dy}{dx} = f(x)\,g(y)$	Separable form	$f(x) = -x$ , $g(y) = y$
$h(y)\,dy = f(x)\,dx$ , with $h = \dfrac{1}{g}$	Separate	$\dfrac{dy}{y} = -x\,dx$
$H(y) = F(x) + c$	Antidifferentiate (implicit)	$\ln y = -\dfrac{x^2}{2} + c$
$y = H^{-1}\big(F(x) + c\big)$	Invert (explicit)	$y = a\,e^{-x^2/2}$

Key Insight: along a solution, the chain rule gives $\dfrac{d}{dx}H(y) = h(y)\dfrac{dy}{dx} = f(x) = F'(x)$ , so $H(y)$ and $F(x)$ have the same derivative and differ by a constant $c$ .

IIVisualization 1 — One Solution Through Every Point

Every constant $a$ gives a solution $y = a\,e^{-x^2/2}$ of $\dfrac{dy}{dx} = -xy$ , and exactly one passes through each point.

💡 Dividing by $y$ needs $y \neq 0$ , yet the check by differentiating never uses the sign of $a$ : $a = 0$ gives the solution $y = 0$ , and negative $a$ gives the mirror-image bells below the axis.

IIIVisualization 2 — The Recipe, General and Specific

Form, separate, antidifferentiate, invert: each general step has a concrete counterpart in every separable equation.

Equation Step 1

General

1 · Form: $\dfrac{dy}{dx} = f(x)\,g(y)$

This equation

$\dfrac{dy}{dx} = -xy$ : $\;f(x) = -x$ , $\;g(y) = y$

IVVisualization 3 — Why the Two Sides Differ by a Constant

On a solution, $H(y) - F(x) = \ln y + \tfrac{x^2}{2}$ stays fixed at $c$ ; on a non-solution it drifts.

Curve a 1.00 x −2.50

VQuiz Questions

Problem 1 · Separate the Variables

Given: $\dfrac{dy}{dx} = x^2 y$ — find the separated form.

✅ Correct! Here

f(x) = x^2

and

g(y) = y

; dividing by

g(y)

and multiplying by

dx

leaves only

y

with

dy

and only

x

with

dx

❌ Close, but… moving

y

to the left requires dividing by it, not multiplying:

\dfrac{dy}{dx} = x^2 y

gives

\dfrac{1}{y}\dfrac{dy}{dx} = x^2

❌ Not separated yet. This is the differential

dy = y'\,dx

, but

y

still sits on the right with

dx

, so the right side cannot be integrated in

x

alone.

❌ Not quite. The variables are still mixed:

y

appears beside

dx

and

x

beside

dy

. Each variable must sit with its own differential.

Show solution

The equation has the form $\dfrac{dy}{dx} = f(x)\,g(y)$ with $f(x) = x^2$ and $g(y) = y$ . Divide both sides by $y$ and multiply both sides by $dx$ :

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

Every $y$ is on the left with $dy$ , every $x$ on the right with $dx$ . (Integrating next gives $\ln y = \tfrac{x^3}{3} + c$ , so $y = a\,e^{x^3/3}$ .)

Problem 2 · Where the Constant Goes

Given: $\dfrac{dy}{dx} = 2xy$ — find the general solution.

✅ Correct! Check:

\dfrac{d}{dx}\big(a\,e^{x^2}\big) = 2x \cdot a\,e^{x^2} = 2xy

❌ Close, but… the constant

c

belongs to

\ln y = x^2 + c

, so it goes up into the exponent:

e^{x^2 + c} = e^{c}e^{x^2}

, a constant factor. Check:

\dfrac{d}{dx}\big(e^{x^2} + c\big) = 2x\,e^{x^2}

, which equals

2xy

only when

c = 0

❌ Not quite. An antiderivative of

2x

x^2

, not

\tfrac{x^2}{2}

— the

\tfrac12

belongs to

\int x\,dx

in the video's example.

❌ Not quite. That integrates

2x

alone, as if the equation were

\dfrac{dy}{dx} = f(x)

. The right side also contains

y

, so divide by

y

first.

Show solution

Separate: $\dfrac{dy}{y} = 2x\,dx$ .

Antidifferentiate (taking $y > 0$ for now):

\int \frac{dy}{y} = \int 2x\,dx \quad\Longrightarrow\quad \ln y = x^2 + c.

Exponentiate:

y = e^{x^2 + c} = e^{c}\,e^{x^2} = a\,e^{x^2}.

Differentiating, $y' = a \cdot 2x\,e^{x^2} = 2x\,y$ , which never uses the sign of $a$ , so every constant $a$ (positive, negative or $0$ ) works.

Problem 3 · An Implicit Answer with a Starting Value

Given: $\dfrac{dy}{dx} = \dfrac{x}{y}$ with $y(0) = 2$ — find the implicit solution and then $y(2)$ .

Which equation does the solution satisfy?

What is $y(2)$ ?

✅ Correct!

y = \sqrt{x^2 + 4}

(positive root, since

y(0) = 2 > 0

), so

y(2) = \sqrt{8} = 2\sqrt{2}

❌ Close, but… from

\dfrac{y^2}{2} = \dfrac{x^2}{2} + c

the starting value gives

c = 2

; multiplying by

2

doubles the constant too:

y^2 = x^2 + 4

❌ Not quite. At

x = 0

this gives

y^2 = 8

, not

4

. Find

c

by substituting:

\dfrac{2^2}{2} = 0 + c

, so

c = 2

❌ Not quite. Separate first:

y\,dy = x\,dx

, then antidifferentiate each side in its own variable.

❌ Check the value. Substitute

x = 2

into

y^2 = x^2 + 4

and take the positive root.

Show solution

Separate: multiply both sides by $y\,dx$ : $\;y\,dy = x\,dx$ .

Antidifferentiate:

\int y\,dy = \int x\,dx \quad\Longrightarrow\quad \frac{y^2}{2} = \frac{x^2}{2} + c.

Starting value: $x = 0$ , $y = 2$ gives $2 = c$ . Multiply by $2$ :

y^2 = x^2 + 4.

Invert: $y = \pm\sqrt{x^2 + 4}$ , and $y(0) = 2 > 0$ picks the plus sign. Check: $y' = \dfrac{x}{\sqrt{x^2+4}} = \dfrac{x}{y}$ . Then

y(2) = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}.

Problem 4 · Separating an Exponential

Given: $\dfrac{dy}{dx} = e^{x - y}$ with $y(0) = \ln 2$ — find the explicit solution.

✅ Correct! At

x = 0

\ln(1 + 1) = \ln 2

✓, and

y' = \dfrac{e^x}{e^x + 1} = \dfrac{e^x}{e^{y}} = e^{x - y}

✓.

❌ Close, but… from

e^{y} = e^{x} + c

the inverse is

y = \ln(e^{x} + c)

, and the logarithm of a sum does not split:

\ln(e^x + c) \neq x + \ln c

. Check:

y' = 1

, but

e^{x - y} = \tfrac12

❌ Not quite. This does solve the equation, but at

x = 0

it gives

\ln 3

. Fit

c

from

e^{\ln 2} = e^{0} + c

\;2 = 1 + c

, so

c = 1

❌ Not quite.

e^{x-y} = \dfrac{e^{x}}{e^{y}}

, so separating means multiplying by

e^{y}

\;e^{y}\,dy = e^{x}\,dx

— not

e^{-y}\,dy

Show solution

Form: $e^{x - y} = e^{x}\,e^{-y}$ , so $f(x) = e^{x}$ and $g(y) = e^{-y}$ .

Separate: $h(y) = \dfrac{1}{g(y)} = e^{y}$ , giving $\;e^{y}\,dy = e^{x}\,dx$ .

Antidifferentiate:

e^{y} = e^{x} + c.

Starting value: $e^{\ln 2} = e^{0} + c$ , so $2 = 1 + c$ and $c = 1$ .

Invert ( $H(y) = e^y$ , so $H^{-1}$ is $\ln$ ):

y = \ln\!\big(e^{x} + 1\big).

Check: $y' = \dfrac{e^{x}}{e^{x} + 1} = \dfrac{e^{x}}{e^{y}} = e^{x - y}$ ✓.

Solved: 0 / 4