# Separable Differential Equations

A **separable differential equation** is a common kind of differential equation that is especially straightforward to solve. Separable equations have the form $\frac{dy}{dx}=f(x)g(y)$, and are called separable because the variables $x$ and $y$ can be brought to opposite sides of the equation. Then, integrating both sides gives $y$ as a function of $x$, solving the differential equation. Separable differential equations are useful because they can be used to understand the rates of chemical reactions, the growth of populations, the movement of projectiles, and many other physical systems.

#### Contents

## Solving Separable Equations

The idea behind solving a separable equation is to move the variables to opposite sides, and then integrate. In doing this, we treat the derivative $\frac{dy}{dx}$ like a fraction, which is a valid manipulation in this case (even though it isn't in general).

Solve the differential equation $\frac{dy}{dx}=xy$, with the initial condition $y(0)=3$.

We can rearrange this equation as $\frac{1}{y}\, dy= x\, dx,$ treating $dx$ and $dy$ as variables. Then, integrating both sides, we have $\int \frac{1}{y}\, dy=\int x\, dx\implies \ln |y|=\frac{x^2}{2}+C\implies y=Ce^{x^2/2}.$ Plugging in the initial condition gives $C=3$, so the solution to our equation is $y=3e^{x^2/2}$.

In general, the trick is to rearrange $\frac{dy}{dx}=f(x)g(y)$ into $\frac{1}{g(y)} \, dy =f(x)\, dx,$ and then integrate. Sometimes however, it can be difficult to identify the functions $f$ and $g$.

Solve the differential equation $\frac{dy}{dx}=xy+x$.

We have to factor the right-hand side; this gives $\frac{dy}{dx}=x(y+1)\implies \frac{1}{y+1}=x\, dx \implies \ln|y+1|=\frac{x^2}{2}+C\, dx\implies y=Ce^{x^2/2}-1,$ and this is the solution to our equation.

Occasionally, it's not even necessary to separate the variables.

Find $y$ in terms of $x$ where $\dfrac{dy}{dx} = \frac {x}{\sqrt{x^2 + 9}}$ and $y(4)=2.$

We have $\begin{aligned} \frac{dy}{dx}&=\frac{x}{\sqrt{x^2+9}}\\ dy&=\frac{x}{\sqrt{x^2+9}}dx\\ \int dy&=\int\frac{x}{\sqrt{x^2+9}}dx\\ y&=\sqrt{x^2+9}+C. \end{aligned}$

Substituting $y(4)=2$ gives

$2=\sqrt{4^2+9}+C=5+C\Rightarrow C=-3,$

and therefore our answer is

$y=\sqrt{x^2+9}-3.\ _\square$

If $\frac{dy}{dx}=y^2+2$, express $y$ in terms of $x$.

Multiplying both sides by ${dx}$ and dividing both sides by ${y^2+2}$ we get: $\frac{dy}{y^2+2}={dx}.$ Taking the integral of both sides gives $\int\frac{dy}{y^2+2}=\int{dx},$ which now becomes the following if we integrate: $\dfrac{\sqrt{2}}{2}\arctan{ \left ( \dfrac{\sqrt{2}}{2}y \right ) }=x+c.$ We can now rearrange to get: $y=\sqrt2\tan(\sqrt2(x+c)).$

Simplify the differential equation $x^2 \dfrac{dy}{dx}=\dfrac{x^2 +1}{3y^2 + 1}.$

We have $\begin{aligned} x^2 \frac{dy}{dx}&=\frac{x^2 +1}{3y^2 + 1}\\ (3y^2+1)dy&=\left(1+\frac{1}{x^2}\right)dx\\ \int(3y^2+1)dy&=\int\left(1+\frac{1}{x^2}\right)dx\\ \Rightarrow y^3+y&=x-\frac{1}{x}+C, \end{aligned}$ where $C$ is the constant of integration. $_\square$

## Word Problems

Many problems involving separable differential equations are word problems. These problems require the additional step of translating a statement into a differential equation.

When reading a sentence that relates a function to one of its derivatives, it's important to extract the correct meaning to give rise to a differential equation. The key is to search for phrases like "rate of change" because they indicate that there is a derivative involved. In fact, the term "derivative" and phrase "rate of change" are synonymous, so these should be kept in mind when constructing a differential equation that models a word problem.

How can we write the following sentences in the form of a differential equation?

The rate of growth of a bacteria population is directly proportional to the current bacteria population.

The rate of change of an object's temperature is directly proportional to the difference of the object's current temperature and the temperature of the surrounding environment.

The force felt by an object in free-fall is the difference of its weight and drag force, where the drag force is proportional to the object's current velocity.

As we read the sentence, we are looking for words or phrases that have mathematical meaning. For instance, the phrase "The rate of growth of a bacteria population" is a chunk of words that can simply be expressed as $\frac{dP}{dt}$. As soon as we see "directly proportional", this tells us to introduce a constant of proportionality, usually $k$. Furthermore, this rate is proportional to "the current bacteria population", meaning we need to introduce $P$ here. To tie it all together, these two pieces are linked by an "=" because of the word "is". This gives $\underbrace{ \text{ The rate of growth of a bacteria population}}_{\frac{ dP}{dt} } \underbrace{\text{ is } } _{=} \underbrace{ \text{ directly proportional }}_{k} \underbrace{ \text{to the current bacterial population}}_{P}.$

This one is $\underbrace{ \stackrel{\text{The rate of change of}}{\text{an object's temperature}}}_{\dfrac{dT}{dt} } \underbrace{ \text{ is }}_{ \Large{=} } \underbrace{ \text{ directly proportional }}_{\Large{k}} \underbrace{ \stackrel{ \text{ to the difference of the object's current temperature}}{\text{ and the temperature of the surrounding environment. }}} _{\Large{(T - T_{a} )} }$

This becomes $\underbrace{ \text{ The force felt by an object in free-fall }} _{ F } \underbrace{ \text { is } }_{=} \underbrace{ \text{ is the difference of its weight and drag force. }} _{mg - bv }$

Suppose a pipe containing water with $0.2\text{ kg}$ of sugar per liter (l) runs into a tank filled with 400 l of water currently containing $10\text{ kg}$ of sugar. Furthermore, the flow from the pipe enters the tank at 10 l/min while at the same time a pipe drains the tank at the same rate. Set up the differential equation that models the amount of sugar in the tank.

Once a word problem has been written as a differential equation, it can be solved using the techniques of the previous section.

The rate at which a substance cools in moving air is proportional to the difference between the temperature of the substance and that of the air. If the temperature of the air is $290\text{ K}$ and the substance cools from $370\text{ K}$ to $330\text{ K}$ in $10\text{ min},$ when will the temperature of the substance become $295\text{ K}?$

Let $T$ denote the temperature, $T_{0}$ the initial temperature, and $t$ the time, so we have

$\underbrace { \text{The rate at which substance cools} } _{ \huge{ - \frac{ dT } { dt } } } \underbrace{\text{ is }} _{ \huge{ = } } \underbrace{ \text{ proportional to } } _{\huge{ k } } \underbrace{\text { the difference between the temps of substance and air.}} _{\huge{ (T- T_{0} ) } }.$

This is $- \frac{ dT }{ T - T_{ 0 } } = k \ dt \implies \int - \frac{ dT }{ T - T_{ 0 } } = \int k \ dt \implies - \ln ( T - T_{ 0 } ) = kt + C.$

To find the initial conditions, we use the fact that we are given $T_{ 0 } = 290 \text{ K}$. Since $T = 370 \text{ K}$ initially at $t = 0 \text{ s}$ and $T = 330 \text{ K}$ at $t = 600 \text{ s}$, from $(1)$ we have

$\begin{cases} -\ln (370 - 290 ) & = k \times 0 + C \\ - \ln (330 - 290) & = k \times 600 + C. \end{cases}$

From the first equation, we get $C = - \ln 80$. Substituting this in the second equation gives $k = \dfrac{ \ln 2 }{ 600 }$. Then from $(1)$ we obtain an equation purely in $T$ and $t :$

$- \ln ( T - 290 ) = \frac{ \ln 2 }{ 600 } \times t - \ln 80.$

Solving for $T$ in order to make calculations easier and see the relation more clearly, we have

$T = 80 \times 2^{ - t / 600 } + 290,$

where we can clearly see that $T$ is a decreasing function in $t$ and approaches $290$ as $t \to \infty$.

Finally, we get the time $t$ at which $T = 295 \text{ K}$ as follows:

$295 = 80 \times 2^{-t/600} + 290\implies t = 600 \log_{ 2 } ( 16 ) \text{ s}= 2400 \text{ s} = 40 \text{ min}.$

I just took my boiling water (100$^\circ$C) off of the stove because it was boiling over. In the 3 minutes I had it off the burner, the temperature decreased to 85$^\circ$C. Given that the room temperature is 25$^\circ$C and Newton's law of cooling is followed, find the temperature of the water in degrees Celsius when I leave it on the counter for 3 more minutes.

The exact same approach can be used on other problems as well.

The rate of growth of a beanstalk is proportional to the square root of its current height. If the height is 100 feet initially and it grows to 400 feet after 5 days, how tall will it be after 20 more days?

Let the height of the beanstalk (in feet) be denoted by $h$, and the time (in days) by $t$. Then $\underbrace { \text{The rate of growth of a beanstalk} } _{ \huge{ \frac{ dh }{ dt } } } \underbrace{ \text{ is } } _{ \huge{ = } } \underbrace{ \text{ proportional to } } _{ \huge{ k } } \underbrace{ \text { the square root of its current height.} } _{\huge{ \sqrt{h} } }$

Now, $\frac{ dh }{ \sqrt{ h } } = k \ dt \implies \int \frac{ dh }{ \sqrt{ h } } = \int k \ dt \implies 2 \sqrt{ h } = kt + C.$

Since $h = 100$ at $t = 0$ and $h = 400$ at $t = 5$, we have

$\begin{cases} 2 \sqrt{ 100 } & = k \times 0 + C \\ 2 \sqrt{ 400 } & = k \times 5 + C. \end{cases}$

From the first equation we have $C = 20$. Substituting this into the second equation gives $40 = 5k + 20$, which implies $k = 4$.

Thus, we now have the particular solution for the differential equation: $2 \sqrt{ h } = 4t + 20 \implies h = (2t + 10) ^{ 2 } .$

Now finding the height of the beanstalk after 20 more days is very easy. We only need to find the height of the beanstalk at $t = 5 + 20 = 25 \text { days}$. Simply substitute $t = 25$ in the equation and obtain: $h = ( 2 \times 25 + 10 ) ^{ 2 } = 60 ^{ 2 } = 3600 \text { (feet)}.$

**Cite as:**Separable Differential Equations.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/separable-differential-equations/