Active Calculus · 7: Differential Equations

7.4: Separable Differential Equations

Motivating Questions
  • What is a separable differential equation?
  • How can we find solutions to a separable differential equation?
  • Are some of the differential equations that arise in applications separable?

In Sections 7.2 and 7.3, we have seen several ways to approximate the solution to an initial value problem. Given the frequency with which differential equations arise in the world around us, we would like to have some techniques for finding explicit algebraic solutions of certain initial value problems. In this section, we focus on a particular class of differential equations (called separable) and develop a method for finding algebraic formulas for their solutions.

A separable differential equation is a differential equation whose algebraic structure allows the variables to be separated in a particular way. For instance, consider the equation

dydt=ty. \frac{dy}{dt} = ty\text{.}

We would like to separate the variables tt and yy so that all occurrences of tt appear on the right-hand side, and all occurrences of yy appear on the left, multiplied by dy/dt.dy/dt\text{.} For this example, we divide both sides by yy so that

1ydydt=t. \frac1y \frac{dy}{dt} = t\text{.}

Note that when we attempt to separate the variables in a differential equation, we require that one side is a product in which the derivative dy/dtdy/dt is one factor and the other factor is solely an expression involving y.y\text{.}

Not every differential equation is separable. For example, if we consider the equation

dydt=t−y, \frac{dy}{dt} = t-y\text{,}

it may seem natural to separate it by writing

y+dydt=t. y + \frac{dy}{dt} = t\text{.}

As we will see, this will not be helpful, since the left-hand side is not a product of a function of yy with dydt.\frac{dy}{dt}\text{.}

Preview Activity 7.4.17.4.1

In this preview activity, we explore whether certain differential equations are separable or not, and then revisit some key ideas from earlier work in integral calculus.

  1. Which of the following differential equations are separable? If the equation is separable, write the equation in the revised form g(y)dydt=h(t).g(y) \frac{dy}{dt} = h(t)\text{.}
    1. dydt=−3y.\displaystyle \frac{dy}{dt} = -3y\text{.}
    2. dydt=ty−y.\displaystyle \frac{dy}{dt} = ty - y\text{.}
    3. dydt=t+1.\displaystyle \frac{dy}{dt} = t + 1\text{.}
    4. dydt=t2−y2.\displaystyle \frac{dy}{dt} = t^2 - y^2\text{.}
  2. Explain why any autonomous differential equation is guaranteed to be separable.
  3. Why do we include the term “+C+C” in the expression
    ∫x dx=x22+C? \int x~dx = \frac{x^2}{2} + C?
  4. Suppose we know that a certain function ff satisfies the equation
    ∫f′(x) dx=∫x dx. \int f'(x)~dx = \int x~dx\text{.}

    What can you conclude about f?f\text{?}

Solving separable differential equations

Before we discuss a general approach to solving a separable differential equation, it is instructive to consider an example.

Example 7.4.17.4.1

Find all functions yy that are solutions to the differential equation

dydt=ty2. \frac{dy}{dt}= \frac{t}{y^2}\text{.}

Answer

We begin by separating the variables and writing

y2dydt=t. y^2\frac{dy}{dt} = t\text{.}

Integrating both sides of the equation with respect to the independent variable tt shows that

∫y2dydt dt=∫t dt. \int y^2\frac{dy}{dt}~dt = \int t~dt\text{.}

Next, we notice that the left-hand side allows us to change the variable of antidifferentiation 1  from tt to y.y\text{.} In particular, dy=dydt dt,dy = \frac{dy}{dt}~dt\text{,} so we now have

∫y2 dy=∫t dt. \int y^2 ~dy = \int t~dt\text{.}
This is why we required that the left-hand side be written as a product in which dy/dtdy/dt is one of the terms.

This equation says that two families of antiderivatives are equal to each other. Therefore, when we find representative antiderivatives of both sides, we know they must differ by an arbitrary constant C.C\text{.} Antidifferentiating and including the integration constant CC on the right, we find that

y33=t22+C. \frac{y^3}{3} = \frac{t^2}{2} + C\text{.}

It is not necessary to include an arbitrary constant on both sides of the equation; we know that y3/3y^3/3 and t2/2t^2/2 are in the same family of antiderivatives and must therefore differ by a single constant.

Finally, we solve the last equation above for yy as a function of t,t\text{,} which gives

y(t)=32 t2+3C3. y(t) = \sqrt[3]{\frac 32 \, t^2 + 3C}\text{.}

Of course, the term 3C3C on the right-hand side represents 3 times an unknown constant. It is, therefore, still an unknown constant, which we will rewrite as C.C\text{.} We thus conclude that the funtion

y(t)=32 t2+C3 y(t) = \sqrt[3]{\frac 32 \, t^2 + C}

is a solution to the original differential equation for any value of C.C\text{.}

Notice that because this solution depends on the arbitrary constant C,C\text{,} we have found an infinite family of solutions. This makes sense because we expect to find a unique solution that corresponds to any given initial value.

For example, if we want to solve the initial value problem

dydt=ty2, y(0)=2, \frac{dy}{dt} = \frac{t}{y^2}, \ y(0) = 2\text{,}

we know that the solution has the form y(t)=32 t2+C3y(t) = \sqrt[3]{\frac32\, t^2 + C} for some constant C.C\text{.} We therefore must find the appropriate value for CC that gives the initial value y(0)=2.y(0)=2\text{.} Hence,

2=y(0)=32 02+C3=C3, 2 = y(0) = \sqrt[3]{\frac 32 \, 0^2 + C} = \sqrt[3]{C}\text{,}

which shows that C=23=8.C = 2^3 = 8\text{.} The solution to the initial value problem is then

y(t)=32 t2+83. y(t) = \sqrt[3]{\frac32\, t^2+8}\text{.}

The strategy of Example 7.4.17.4.1 may be applied to any differential equation of the form dydt=g(y)⋅h(t),\frac{dy}{dt} = g(y) \cdot h(t)\text{,} and any differential equation of this form is said to be separable. We work to solve a separable differential equation by writing

1g(y)dydt=h(t), \frac{1}{g(y)} \frac{dy}{dt} = h(t)\text{,}

and then integrating both sides with respect to t.t\text{.} After integrating, we try to solve algebraically for yy in order to write yy as a function of t.t\text{.}

Example 7.4.27.4.2

Solve the differential equation

dydt=3y. \frac{dy}{dt} =3y\text{.}

Answer

Following the same strategy as in Example 7.4.17.4.1, we have

1ydydt=3. \frac 1y \frac{dy}{dt} = 3\text{.}

Integrating both sides with respect to t,t\text{,}

∫1ydydt dt=∫3 dt, \int \frac 1y\frac{dy}{dt}~dt = \int 3~dt\text{,}

and thus

∫1y dy=∫3 dt. \int \frac 1y~dy = \int 3~dt\text{.}

Antidifferentiating and including the integration constant, we find that

ln⁡∣y∣=3t+C. \ln|y| = 3t + C\text{.}

Finally, we need to solve for y.y\text{.} Here, one point deserves careful attention. By the definition of the natural logarithm function, it follows that

∣y∣=e3t+C=e3teC. |y| = e^{3t+C} = e^{3t}e^C\text{.}

Since CC is an unknown constant, eCe^C is as well, though we do know that it is positive (because exe^x is positive for any xx). When we remove the absolute value in order to solve for y,y\text{,} however, this constant may be either positive or negative. To account for a possible ++ or −,-\text{,} we denote this updated constant by CC to obtain

y(t)=Ce3t. y(t) = Ce^{3t}\text{.}

There is one more technical point to make. Notice that y=0y=0 is an equilibrium solution to this differential equation. In solving the equation above, we begin by dividing both sides by y,y\text{,} which is not allowed if y=0.y=0\text{.} To be perfectly careful, therefore, we should consider the equilibrium solutions separately. In this case, notice that the final form of our solution captures the equilibrium solution by allowing C=0.C=0\text{.}

Activity 7.4.27.4.2

Suppose that the population of a town is growing continuously at an annual rate of 3% per year.

  1. Let P(t)P(t) be the population of the town in year t.t\text{.} Write a differential equation that describes the annual growth rate.
  2. Find the solutions of this differential equation.
  3. If you know that the town's population in year 0 is 10,000, find the population P(t).P(t)\text{.}
  4. How long does it take for the population to double? This time is called the doubling time.
  5. Working more generally, find the doubling time if the annual growth rate is kk times the population.
Activity 7.4.37.4.3

Suppose that a cup of coffee is initially at a temperature of 105∘105^\circ F and is placed in a 75∘75^\circ F room. Newton's law of cooling says that

dTdt=−k(T−75), \frac{dT}{dt} = -k(T-75)\text{,}

where kk is a constant of proportionality.

  1. Suppose you measure that the coffee is cooling at one degree per minute at the time the coffee is brought into the room. Use the differential equation to determine the value of the constant k.k\text{.}
  2. Find all the solutions of this differential equation.
  3. What happens to all the solutions as t→∞?t\to\infty\text{?} Explain how this agrees with your intuition.
  4. What is the temperature of the cup of coffee after 20 minutes?
  5. How long does it take for the coffee to cool to 80∘?80^\circ\text{?}
Activity 7.4.47.4.4

Solve each of the following differential equations or initial value problems.

  1. dydt−(2−t)y=2−t\displaystyle \frac{dy}{dt} - (2-t) y = 2-t
  2. 1tdydt=et2−2y\displaystyle \frac{1}{t}\frac{dy}{dt} = e^{t^2-2y}
  3. y′=2y+2,y' = 2y+2\text{,} y(0)=2y(0)=2
  4. y′=2y2,y' = 2y^2\text{,} y(−1)=2y(-1) = 2
  5. dydt=−2tyt2+1,\frac{dy}{dt} = \frac{-2ty}{t^2 + 1}\text{,} y(0)=4y(0) = 4

Summary

  • A separable differential equation is one that may be rewritten with all occurrences of the dependent variable multiplying the derivative and all occurrences of the independent variable on the other side of the equation.
  • We may find the solutions to certain separable differential equations by separating variables, integrating with respect to t,t\text{,} and ultimately solving the resulting algebraic equation for y.y\text{.}
  • This technique allows us to solve many important differential equations that arise in the world around us. For instance, questions of growth and decay and Newton's Law of Cooling give rise to separable differential equations. Later, we will learn in Section 7.6 that the important logistic differential equation is also separable.

Credit and licence

This page titled 7.4: Separable Differential Equations was last modified on Wed, 09 Nov 2022 17:00:15 GMT and is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Matthew Boelkins, David Austin, and Steven Schlicker ( ScholarWorks @Grand Valley State University ) via source content that was edited to the style and standards of the LibreTexts platform.

Source: https://math.libretexts.org/Bookshelves/Calculus/Active_Calculus_(Boelkins_et_al.)/07%3A_Differential_Equations/7.04%3A_Separable_Differential_Equations

Licence: CC BY-SA 4.0 (creativecommons.org/licenses/by-sa/4.0/). You may share and adapt it under the same licence, with this credit. WorldClass adapted the formatting only: the text is the authors', with interactive parts and links to other sites left out.

Build status · Flash Infinity AI™ is a trademark of AI-Flash Inc.. WorldClass is not affiliated with, approved or accredited by any exam board, curriculum provider or university unless a page says so.