Separable Differential Equations | Free Calculus 2 Lesson (4 College Credits) | UPI Study

UPI Study › Courses › Calculus 2 › Free lessons › Lesson 5.3 Free lesson

Separable Differential Equations

Calculus 2 · Chapter 5: Differential EquationsSection 5.3 · lesson 24 of 5410 min · free

← Slope Fields and Euler's MethodNext lesson: Exponential Growth and Decay →

Earn 4 college credits for Calculus 2Recommended for credit by NCCRS · Math 265 · self-paced · one payment, lifetime access
Enroll in Calculus 2 · $250 →Download the syllabus (PDF)

In this lesson

Splitting the variables to solve a rate equation.

  • Recognise when an equation is separable
  • Separate, integrate both sides, and solve
  • Apply an initial condition to fix the constant
  • Tell an implicit solution from an explicit one

Transcript

Everything said in the video, section by section (1,519 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 5.3

Separable Differential Equations

A cup of coffee cooling on a desk. A population of bacteria multiplying in a dish. A radioactive sample decaying in a lab. The balance of a savings account earning continuous interest. All four are governed by the same kind of mathematical statement. An equation that relates a quantity to its own rate of change. Such an equation is a differential equation. And solving it means finding the actual function the quantity follows over time. The most important technique for doing that by hand — and the one that unlocks every one of those four examples — is separation of variables. Section five point one introduced differential equations. Section five point two used slope fields to picture solutions without solving. This section presents the first general solution method.

The Form

Recognising a Separable Equation

First, how do you recognize when the method applies? A first-order differential equation is separable if its right-hand side can be written as a product of a function of x alone, and a function of y alone. In symbols: d y d x equals g of x, times h of y. The defining feature is that the x-parts and the y-parts can be completely pulled apart. Every x collected on one side of the equation. Every y on the other. Not every equation allows this. d y d x equals x y is separable — here g of x is x, and h of y is y. A clean product. d y d x equals e to the x over y squared is separable. And d y d x equals one plus y squared, times cosine x, is separable. But d y d x equals x plus y is not. Because a sum cannot be factored into an x-function times a y-function. There's no way to pull them apart. That's the test. Product, yes. Sum, no.

The Method

Separate, Integrate, Solve

Once an equation is recognized as separable, the solution procedure has three moves. First, separate. Rewrite the equation so that all y-terms, including d y, sit on one side. And all x-terms, including d x, sit on the other. Treating d y d x as a ratio of differentials is the informal but completely reliable way to do this. Second, integrate both sides. Each with respect to its own variable. And a single constant of integration suffices — because the constants from the two sides can be combined into one. Third, solve for the constant, and if possible, for y. Let's do the most important example there is. Solve d y d x equals y. This is separable with g of x equal to one, and h of y equal to y. Separate by dividing by y and multiplying by d x. Valid where y is not zero. You get one over y, d y, equals d x. Integrate both sides. The left gives log absolute y. The right gives x plus C. Now solve for y by exponentiating both sides. Absolute y equals e to the x plus C. Which is e to the C, times e to the x. Write e to the C as a new positive constant, absorb the sign of the absolute value, and you get y equals A e to the x. Where A is an arbitrary nonzero constant. And notice — A equals zero recovers the solution y equals zero, which was set aside when we divided. So the formula covers that case too.

Figure 5.3

One Equation, a Whole Family of Curves

Look at what that general solution actually is. y equals A e to the x is not one curve. It's an entire family. One curve for every value of A. In the picture you can see them threading through the slope field. Positive A values give curves above the axis, growing. Negative A values mirror below. And every single one of them is tangent to the slope segments at every point it passes. This is the hallmark of a differential equation. Its general solution always contains an arbitrary constant. So the equation alone tells you the shape of every possible behavior — but not which one you're actually looking at. For that, you need one more piece of information.

Pinning It Down

Applying an Initial Condition

The general solution describes every possible behavior the system could have. To single out the one that matches a specific situation, an initial condition is supplied. A known value of y at some value of x. Substituting that point into the general solution produces an equation for the constant, which is then solved. The problem of solving a differential equation together with an initial condition is called an initial value problem. Let's do one. Solve d y d x equals six x squared y, with y at zero equal to five. Separate, by dividing by y. You get one over y d y equals six x squared d x. Integrate. Log absolute y equals two x cubed, plus C. Exponentiate to get the general solution: y equals A, e to the two x cubed. Now apply the initial condition. Substitute x equals zero and y equals five. Five equals A times e to the zero. Which is A. So A is five. The particular solution is y equals five, e to the two x cubed. Notice the order of operations there, because it matters. Apply the initial condition after writing the general solution. Not before integrating. That's the standard order, and it avoids carrying an unknown constant through unnecessary algebra. One more thing. Equations of the form d y d x equals k y, whose solutions are y equals A e to the k x, are worth recognizing on sight. Because they model so many real processes. A positive k gives exponential growth — a bacterial culture, or a continuously compounded account. A negative k gives exponential decay — radioactive material, or a cooling object's temperature difference from its surroundings. In each case the initial condition supplies the starting amount A, and the rate constant k is measured from data. The same three-step separation procedure handles all of them. Only the constants change.

The Form of the Answer

Implicit and Explicit Solutions

One more distinction, about the form of the answer. After integrating, the result is often an equation relating x and y in which y is not isolated. That's an implicit solution. When the equation can be rearranged to express y explicitly as a function of x, the result is an explicit solution. Both are legitimate. Sometimes solving for y is easy. Sometimes it's messy. And occasionally it's impossible in elementary terms — in which case the implicit form is the final answer. An implicit solution still fully determines the relationship between the variables. It simply leaves y entangled with x rather than standing alone. Let's see both. Solve d y d x equals x over y, with y at zero equal to three. Separate by multiplying both sides by y and by d x. You get y d y equals x d x. Integrate: y squared over two equals x squared over two, plus C. That relation is an implicit solution. Apply y at zero equals three to find C. At x equals zero, y is three. So nine halves equals zero plus C. C is nine halves. Multiply through by two: y squared equals x squared plus nine. Now solve for y to get the explicit form. And here's the subtlety. Taking a square root gives plus or minus. But because y at zero equals three is positive, the positive square root is the correct branch. So y equals root x squared plus nine. The initial condition did double duty there. It fixed the constant, and it selected the correct branch. The implicit form describes a hyperbola. The explicit form keeps only the upper branch — the one passing through zero comma three.

Key Takeaways

What to Carry Forward

Here's something to think about. In the first worked example, dividing both sides by y was the step that separated the variables. But that division is only valid when y is not zero. So check directly: does the constant function y equals zero also satisfy d y d x equals y? Its derivative is zero. And y is zero. So yes, it does. Why does the general solution y equals A e to the x still capture that case? And what general lesson does that carry about dividing by a variable while separating? Five takeaways. An equation is separable when it can be written as d y d x equals g of x, times h of y. The x-stuff and the y-stuff factor apart. The method is three steps. Separate the variables. Integrate both sides, with one constant sufficing. Then solve for the constant and for y. The general solution contains an arbitrary constant and describes a whole family of curves. An initial condition selects one particular solution. Apply the initial condition after writing the general solution, substituting the known point to solve for the constant. And answers may be implicit — a relation in x and y — or explicit, with y isolated. The initial condition also fixes which branch, and which sign, to keep. Next up is section five point four — Exponential Growth and Decay. The most important special case of everything we've just done.

About Calculus 2 at UPI Study

Calculus 2 (Math 265) is a self-paced online course of 54 video lessons in 10 chapters, from integration techniques to series, parametric curves and vectors. It carries a NCCRS credit recommendation for 4 semester credits.

Every lesson video and transcript is free on pages like this one. Enrolling adds a practice quiz after each lesson, the graded work and final exam that earn the credit, lifetime access, and an official transcript on completion. UPI Study has helped over 130,000 students with college credits. ACE and NCCRS have over 2,100 cooperating colleges that will consider accepting credits; the final decision is with the universities.

Enroll now → Course details Download the syllabus Every course with PRO →

Why take it here

Why students take Calculus 2 with UPI Study

Open the door to higher mathCalculus 2 is the gateway to Calculus 3, differential equations, statistics and most engineering, physics, economics and computer-science degrees. Completing it now lets you qualify for those courses at your university.
Credit at a fraction of the priceThe same 4 credits can cost $750 or more at a US college. Here it is one payment, self-paced, with lifetime access.
Finish your degree soonerTransfer credits reduce the courses you still owe, so you can graduate earlier or lighten a heavy semester.
Strengthen an applicationA completed college-level course on an official transcript shows admissions and scholarship committees you can do the work.
Study on your own scheduleNo start dates, no deadlines. Watch the lessons, practice, and take the final when you are ready.
130,000+students helped with college creditsACE and NCCRS have over 2,100 cooperating colleges that will consider accepting credits; the final decision is with the universities.

Where the credit goes

Universities that work with NCCRS

Calculus 2 carries a NCCRS credit recommendation. These universities work with NCCRS; some are UPI Study's signed partners. ACE and NCCRS have over 2,100 cooperating colleges that will consider accepting credits; the final decision is with the universities.

Thomas Edison State UniversityThomas Edison State UniversitySigned partnerUniversity of the PeopleUniversity of the PeopleSigned partnerCharter Oak State CollegeCharter Oak State CollegeSigned partnerStrayer UniversityStrayer UniversitySigned partnerColumbia Southern UniversityColumbia Southern UniversitySigned partnerIndiana TechIndiana TechSigned partnerPenn StatePenn StateWorks with NCCRSSouthern New Hampshire UniversitySouthern New Hampshire UniversityWorks with NCCRSSUNYSUNYWorks with NCCRS +many more cooperating universitiesFind my college →

Find my college → Not listed? Ask Athena in the corner of this page, or book a 15-minute call.

FAQ

Questions students ask us

About Separable Differential Equations

How do you know if a differential equation is separable?
A first-order differential equation is separable if its right-hand side can be written as a product g(x)\,h(y), a function of x alone times a function of y alone. For example, \frac{dy}{dx} = xy and \frac{dy}{dx} = (1+y^2)\cos x are separable. However, \frac{dy}{dx} = x + y is not, because a sum cannot be factored into an x-part times a y-part.
What are the steps to solve a separable differential equation?
Separate the variables, integrate both sides, then solve. First move all y-terms, including dy, to one side and all x-terms, including dx, to the other, giving \frac{1}{h(y)}\,dy = g(x)\,dx. Next integrate each side with respect to its own variable. One constant of integration is enough because the two constants combine. Finally, solve for y if possible, and use any initial condition to find the constant.
How do you solve dy/dx = 6x^2 y with the initial condition y(0) = 5?
The solution is y = 5e^{2x^3}. Dividing by y gives \frac{1}{y}\,dy = 6x^2\,dx, and integrating gives \ln|y| = 2x^3 + C. Exponentiating yields the general solution y = Ae^{2x^3}. Substituting x = 0 and y = 5 gives 5 = Ae^0 = A, so A = 5. Apply the initial condition after writing the general solution, not before integrating.
What is the difference between an implicit and an explicit solution of a differential equation?
An implicit solution is a relation between x and y in which y is not isolated, while an explicit solution expresses y directly as a function of x. For \frac{dy}{dx} = \frac{x}{y} with y(0) = 3, integration gives the implicit solution y^2 = x^2 + 9. Taking the square root introduces \pm, but since y(0) = 3 is positive, the explicit solution is y = \sqrt{x^2 + 9}.

About the Calculus 2 course

How many college credits is Calculus 2?
Calculus 2 (Math 265) carries a NCCRS credit recommendation for 4 semester credits. ACE and NCCRS have over 2,100 cooperating colleges that will consider accepting credits; the final decision is with the universities.
How much does Calculus 2 cost and what is included?
$250, paid once, with lifetime access: all 54 video lessons, a practice quiz after each, the graded work and final exam, and an official transcript when you finish. Students who want several courses choose UPI Study PRO instead.
Which universities accept UPI Study credits?
UPI Study has signed partner agreements with universities such as Thomas Edison State University, University of the People and Charter Oak State College, and works with universities such as Penn State, SNHU and SUNY. ACE and NCCRS have over 2,100 cooperating colleges that will consider accepting credits; the final decision is with the universities. Use "Find my college" on this page to check yours.
How long does Calculus 2 take?
It is self-paced with no deadlines. The 54 lessons total about 9 hours of video; most students add reading and practice time and finish in a few weeks to a few months, on their own schedule.
How are the exams done?
Quizzes and assignments are online in the course. The final exam is taken online with proctoring; a person on the UPI Study team checks assignments, usually within two days.
How do my credits reach my college?
When the course shows 100% complete, request the official transcript from your dashboard. It is sent to the college you name, and you can send it to more than one. Check your college first with "Find my college" on this page, or ask Athena.

Ready for the full course? Earn 4 credits in Calculus 2.

54 video lessons, a quiz after each one, and the graded work that earns the credit. Recommended for credit by NCCRS.

Enroll in Calculus 2 →Download the syllabusEvery course with PRO →
Calculus 2
4 credits · NCCRSEnroll →