DE-BOOK Verifying Separable

Section 4.2 Verifying Separable

Many differential equations will not be handed to you in separable form, so various algebra techniques will be required to put the equations into this form. The following set of examples illustrate this idea.

Example 7.

\frac{d y}{d x} = x + x y^{2}, \frac{d P}{d t} = \frac{P t}{P^{2} + 1}, \frac{d y}{d x} = 11 e^{4 x + 3 y} .

Solution 1. Factor

Since

x

is common in both terms of the sum, we can factor it out as

\frac{d y}{d x} = \underset{n o t s e p a r a t e d}{\underset{⏟}{x + x y^{2}}} \overset{a l g e b r a}{⟶} \frac{d y}{d x} = \underset{s e p a r a t e d}{\underset{⏟}{x (1 + y^{2})}}

showing that the differential equation is separable.

Solution 2. Split the Fraction

This example uses the following rule for multiplying fractions:

\frac{A \cdot B}{C \cdot D} = \frac{A}{C} \cdot \frac{B}{D} .

First, we split

P

and

t

in the numerator and denominator, then use the rule above to separate the fraction, like so

\frac{d P}{d t} = \underset{n o t s e p a r a t e d}{\underset{⏟}{\frac{P t}{P^{2} + 1}}} = \frac{P \cdot t}{(P^{2} + 1) \cdot (1)} = \underset{s e p a r a t e d}{\underset{⏟}{\frac{P}{P^{2} + 1} (\frac{t}{1})}}

and the equation is separable.

Solution 3. Separate the Exponents

Applying the rule

e^{A + B} = e^{A} \cdot e^{B}

to this differential equation allows us to show it is separable as follows:

\frac{d y}{d x} = \underset{n o t s e p a r a t e d}{\underset{⏟}{11 e^{4 x + 3 y}}} = 11 (e^{4 x} \cdot e^{3 y}) = \underset{s e p a r a t e d}{\underset{⏟}{11 e^{4 x} \cdot e^{3 y}}} .

Example 8.

\frac{d^{2} y}{d x^{2}} \sin x = y, t - \frac{d s}{d t} = - s, x \frac{d y}{d x} + 10 x^{2} y = 6 x^{2} .

Solution 1. Attention to Detail

This is sort of a trick question to get you in the habit of checking the order of a differential equation. Since this DE is not first-order, it is not separable.

Solution 2. Isolate the Derivative First

We see that this DE is first-order, so now we need to see if it can be rearranged in separable form. Isolating

\frac{d s}{d t}

we get

\frac{d s}{d t} = s + t

and it should be clear that there is no way to separate

t

and

s

by multiplication, and therefore this DE is not separable.

Solution 3. Isolate, Simplify, & Factor

As before, let’s isolate

d y / d x

on the left side of the equation, like so

x \frac{d y}{d x} = 6 x^{2} - 10 x^{2} y

\frac{d y}{d x} = \frac{1}{x} (6 x^{2} - 10 x^{2} y)

\frac{d y}{d x} = 6 x - 10 x y .

Now that the derivative is isolated on the left, we attempt to separate the variables on the right. Factoring

2 x

out of both terms gives

\frac{d y}{d x} = 2 x \cdot (3 - 5 y)

and we see that the differential equation is separable.

In general, there is more than one way to separate a differential equation. This example could have been separated in either of the following ways:

\frac{d y}{d x} = x \cdot (6 - 10 y), \frac{d y}{d x} = (- 2 x) \cdot (5 y - 3)

The only thing that matters is that

x

and

y

are separated.

Reading Questions Check-Point Questions

1. The differential equation, $\frac{d y}{d x} - x (y^{2} + y) = 0,$ is in separable form.

The differential equation, $\frac{d y}{d x} - x (y^{2} + y) = 0,$ is in separable form.

True.
This equation is not currently in separable form. However, it can be put in separable form by moving the $x (y^{2} + y)$ term to the right-side of the equation.
False.
This equation is not currently in separable form. However, it can be put in separable form by moving the $x (y^{2} + y)$ term to the right-side of the equation.

2. Match each equation to the algebra technique needed to show the equation is separable.

Drag the algebra technique to the corresponding differential equation.

$e^{A + B} = e^{A} \cdot e^{B}$
$w^{'} = w e^{t + w}$
$\frac{A \cdot B}{C \cdot D} = \frac{A}{C} \cdot \frac{B}{D}$
$\frac{d y}{d θ} = \frac{3 y}{(y + 9) θ}$
Isolate the derivative
$x r^{'} + 2 r = 0$
Factor
$\frac{d P}{d t} = e^{t} P^{2} - e^{t}$
No technique is needed
$w^{'} = e^{Q} e^{x}$

3. Select the Separable Differential Equations.

Click on each of the separable differential equations below.

A differential equation is separable if you can rearrange it to express it as a product of functions involving only

x

and

y

separately. Look for equations that can be manipulated into this form, even if they do not initially appear separable.


$\frac{d y}{d x} = \frac{x + y}{x y}$	$\frac{d y}{d x} = \frac{1}{x^{2}} + \frac{y}{x^{2}}$	$\frac{d y}{d x} = x^{2} + y^{2}$

$\frac{d y}{d x} = x (\frac{y}{x} + 1)$	$\frac{d y}{d x} = \frac{\sin (x + y)}{x}$	$\frac{d y}{d x} = y \cdot e^{- x}$

$x^{2} \frac{d y}{d x} - y = x^{2} y$	$\frac{d y}{d x} = y^{2} - x y$	$\frac{d y}{d x} = \cos (x y)$

$\frac{d y}{d x} = \frac{y}{\sqrt{x^{2} + 1}}$	$\frac{d y}{d x} = x + \frac{y^{2}}{x}$	$\frac{d y}{d x} = \frac{1}{x} + \frac{1}{y}$

4. Write the differential equation, $\frac{d y}{d x} - y \cos (x) = y,$ in separable form.

Write the differential equation,

\frac{d y}{d x} - y \cos (x) = y,

in separable form by identifying

f (x)

and

g (y) .

$\frac{d y}{d x} = ($		$) \cdot ($		$)$
	$f (x)$		$g (y)$

5. Write the differential equation, $t P \frac{d P}{d t} = e^{t} P^{2} - e^{t} y,$ in separable form.

Write the differential equation,

t P \frac{d P}{d t} = e^{t} P^{2} - e^{t},

in separable form by identifying

f (t)

and

g (P) .

$\frac{d P}{d t} = ($		$) \cdot ($		$)$
	$f (t)$		$g (P)$

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