CEGEP Calculus II Separable Differential Equations Free · no sign-up

CEGEP Calculus II — Separable Differential Equations Worksheet

By Marius — private math tutor in Montreal. Updated 30 August 2026

This is where the integral gets pointed backwards. A differential equation describes how something changes; a solution is a function that satisfies that description at every point of an interval. The set covers the separable kind and only the separable kind — the ones in which every y can be moved to one side and every x to the other — from checking that a proposed function really is a solution, through general solutions and initial value problems, to a stirred tank and a decaying tracer. Read the questions on this page, then print the free PDF when you want to write on it.

Practice worksheet — free PDF

4 pages 5 questions Letter size, print-ready

Download the worksheet Read the questions here

No email, no account, no watermark. Teachers: photocopy it for your classes freely. Worked solutions and 4 harder problems come with the CEGEP Calculus II bundle.

All 5 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

Q1Verifying a Solution of a Differential Equation

Consider the differential equation $\frac{d y}{d x} = 2 x (1 + y^{2}) .$
1. Show that $y = \tan (x^{2} + C)$ satisfies this equation for every constant $C$ , and state the restriction on $x$ that the verification needs.
2. Decide whether $y = x^{3}$ is a solution of the same equation. Justify your answer by substitution.
Q2Solving a Separable Equation

Find the general solution of each equation. Give $y$ explicitly in terms of $x$ .
1. $\frac{d y}{d x} = \frac{\cos x}{y^{2}}$
2. $\frac{d y}{d x} = \frac{x y}{1 + x^{2}}$
Q3Initial Value Problems

Solve the initial value problem $\frac{d y}{d x} = 3 x^{2} e^{- y}, y (0) = \ln 2 .$ Give $y$ explicitly and state the largest interval on which your solution is valid.
Q4Growth, Decay and Mixing Applications

A tank holds $200 L$ of pure water. Brine containing $3 g / L$ of salt runs in at $4 L / min$ ; the tank is kept well stirred and drains at the same $4 L / min$ , so its volume never changes. Let $S (t)$ be the mass of salt in the tank, in grams, after $t$ minutes.
1. Explain why $\frac{d S}{d t} = 12 - \frac{S}{50}$ , then solve for $S (t)$ .
2. How much salt is in the tank after $50$ minutes? Give your answer to three decimals.
3. Find ${lim}_{t \to \infty} S (t)$ and say, with units, what it means about the liquid leaving the tank.
Q5Synthesis — drawing on several topics in this set

A short-lived isotope is used to trace sediment movement in a reservoir. Its mass $A (t)$ , in milligrams, decays at a rate proportional to the mass present: $\frac{d A}{d t} = - k A, k > 0, t in years .$
1. Verify by substitution that $A (t) = A_{0} e^{- k t}$ satisfies both the equation and the condition $A (0) = A_{0}$ , then derive the same formula by separating the variables.
2. The half-life of the isotope is $30$ years. Find $k$ exactly.
3. A $12 mg$ sample is released. How long until $2 mg$ remain? Give the answer in years to three decimals, and say what it means.

The 4 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

Which stream is this for? This is CEGEP Calculus II — 201-NYB-05 under the legacy numbering, 201-SN3-RE under the current one — and the competency is 0M03. The ministerial devis names separable equations and stops there, so that is the whole of this set: no integrating factors, no linear first-order machinery, no exact or second-order equations, no Euler's method and no slope fields. If you have been sent looking for those, they belong to a later course and are not on this sheet. What the devis does assume is that you can already integrate — the x-side of a separation is an ordinary integral, substitution and all, and the differential equation adds nothing to it.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

What it means to be a solution, and how you check (Q1)

Q1 sets dy/dx = 2x(1 + y²) against a proposed family, y = tan(x² + C), and asks you to show it satisfies the equation for every constant C. There is nothing to solve. A verification is always the same two-column job: differentiate the candidate to obtain the left side, substitute the candidate into the right side, and compare the two expressions as functions. Part (b) puts a second candidate against the same equation and asks you to decide, again by substitution.

Proving and disproving are not symmetric. To establish that a function is a solution, the two sides must agree at every point of an interval — you need an identity, and the write-up has to show both sides in full. To establish that a function is not a solution, one value of x at which the sides differ is a complete argument. So the efficient strategy is to test a convenient input first: if it fails, you are finished in two lines; if it holds, you still owe the general argument.

Expect the two sides to arrive in different clothes. Differentiating a tangent produces a secant squared, while the equation is written in terms of y², so nothing cancels — the step that closes the verification is a Pythagorean identity, not algebra. Recognising that early saves a page of fruitless rearranging, and it generalises: whenever a candidate comes from a trigonometric, exponential or logarithmic family, look for the identity that converts the derivative into the form the right-hand side is written in.

Where the verification is claimed matters. Q1(a) asks for the restriction on x, and that is part of the answer rather than a footnote. A candidate built from a function with vertical asymptotes is only defined and differentiable where its argument stays away from them, and the verification asserts nothing outside that region. The same instinct is what the interval-of-validity questions later on the sheet are training.

Separating the variables, in the order that keeps it clean (Q2)

Q2 asks for the general solution of two equations and, in both cases, for y given explicitly in terms of x. Separation applies when the right-hand side factors into a function of x times a function of y — a quotient counts, since dividing by y² is multiplying by y⁻². It does not apply when x and y are tangled in a sum, and nothing on this sheet is of that kind.

dy/dx = f(x) · g(y) → ∫ dy / g(y) = ∫ f(x) dx

Five steps, and the order is the point

Worked on dy/dx = xy², which is not on the sheet — so you can run the same loop on Q2 yourself.

1
Confirm it separates
Write the right-hand side as a product: xy² is x times y², so it does. Do this before touching anything else. The minute spent here is what stops you separating an equation that cannot be separated.
2
Move the y terms left and the x terms right, and name the divisor
y⁻² dy = x dx. You divided by y², so write "assuming y ≠ 0" in the margin. That clause is not politeness — it is the assumption the next callout is about.
3
Integrate both sides, with one constant
∫y⁻² dy = ∫x dx gives −1/y = x²/2 + C. Put + C on one side only; two constants collapse into one anyway, and writing both is how a sign goes missing. Whatever technique the x-side needs — the substitution u = 1 + x² in Q2(b), for instance — is ordinary integration, and the differential equation neither helps nor hinders it.
4
Apply an initial condition now, while the form is still implicit
Q2 has none, so C stays a letter. Where there is one, as in Q3, substituting it into the implicit equation gives a clean number in a short equation. Doing it after you have unwrapped an exponential or a logarithm leaves the constant buried inside one, which is where the algebra usually goes wrong.
5
Solve for y, then differentiate your answer back
−1/y = x²/2 + C gives y = −2/(x² + 2C). Q2 says "give y explicitly", so an implicit relation is not yet an answer to the question asked. Then check: differentiate what you wrote and see the original equation come back. It costs one line and catches almost every slip.

Dividing to separate can throw a solution away — and it is often the interesting one. Step 2 divided by g(y), which is illegal wherever g(y) = 0. And any constant y = c with g(c) = 0 is a solution: a constant function has derivative 0, and the right-hand side is 0 there too. These constant, or equilibrium, solutions are discarded before a single integral is taken. y = 0 is the classic — it vanishes the moment you divide by y in Q2(b), and it satisfies that equation perfectly well. What you do next depends on the shape of the family you end up with. A family of the form y = A·(something) contains it at A = 0, so allowing the constant to be zero quietly closes the gap. A family of the form y = 1/(something) is never zero, and there the lost solution has to be stated separately or it is simply missing from your answer. Ask the same two questions every time: what did I divide by, and does setting it to zero solve the equation? Notice, too, how often the discarded solution is the physically meaningful case — the tank at its steady concentration, the sample that has fully decayed, the population that never started.

Initial value problems, and the interval you are allowed to claim (Q3)

Q3 is a full initial value problem: dy/dx = 3x²e⁻ⁿ with y(0) = ln 2, y wanted explicitly, and the largest interval on which the solution is valid. The general solution is a whole family of curves; the initial condition names the single one passing through the given point. Notice how the negative exponent behaves under separation — e⁻ⁿ on the right puts eⁿ dy on the left, and that integral is as easy as integrals get, so the real work in this question is in the last two lines rather than the first two.

A formula is not a solution everywhere it happens to be defined. A solution is a differentiable function satisfying the equation at every point of one interval containing the initial point — not a union of separated pieces, because nothing in the equation ties values on one piece to values on another. Two things can shrink that interval. The formula may stop being defined: a logarithm needs its argument positive, a denominator must not vanish, a root needs a non-negative radicand, so you solve that condition and take the largest interval around the initial input on which it holds. Or the formula stays perfectly well defined and quietly stops satisfying the equation — squaring during the algebra can lose a sign condition, and past the point where that condition fails the expression still evaluates while no longer solving anything. When in doubt, substitute and compare the signs of the two sides. Give the answer in standard interval brackets, (−∞, 4] or (3, ∞), and say which input it contains.

Rate in minus rate out: setting up the mixing tank (Q4)

Q4 holds 200 L of pure water, runs brine in at a fixed concentration and drains the well-stirred mixture at the same flow rate. Part (a) asks you to explain why dS/dt = 12 − S/50 before solving anything, and that derivation is the mark-bearing part of the question — the integration that follows is Q2 again.

Building the equation out of the words

The order below is the one that keeps the units honest.

1
Track the mass, never the concentration
Let S(t) be the grams of salt in the tank. Concentration is a derived quantity — mass over volume — and writing the differential equation for it directly is the single most common way this question falls apart.
2
Rate in = concentration in × flow in
Grams per litre times litres per minute is grams per minute. Multiply the units out on the page; if they do not come out as the units of dS/dt, the term is wrong before any calculus starts.
3
Rate out = (S / V) × flow out
This is where "well stirred" earns its place in the wording: it is the assumption that the draining liquid has the same concentration as the tank as a whole. Without it, S/V would be a guess rather than a model.
4
Check whether the volume moves
Here the two flows are equal, so V stays at 200 L, the coefficient of S is a constant and the equation separates. Had the flows differed, V would be a function of t and the problem would leave this sheet entirely.
5
Assemble, and read the initial condition off the words
dS/dt = (rate in) − (rate out). "The tank holds pure water" is not scene-setting; it is S(0), and you need it to pin the constant down.

The long-run value is available without solving anything. Set dS/dt = 0 and solve for S: that constant is the equilibrium solution of this equation — the same kind of solution the division in a separation discards — and it is the value part (c)'s limit approaches. Two habits make that part worth full marks. Report the limiting mass with its units, then divide by the tank volume to turn it into a concentration and compare that with the incoming brine; the comparison is the interpretation being asked for. And say what it means physically: in the long run salt leaves as fast as it arrives, so the tank cannot end up saltier than what is being poured into it.

Growth, decay and the meaning of k (Q5)

Q5 is the synthesis question — a tracer decaying at a rate proportional to the mass present, dA/dt = −kA. Part (a) deliberately asks for both routes: verify the exponential formula by substitution, then derive the same formula by separating the variables. Do both, in that order, and keep them apart on the page. They are different skills, they are examined separately, and a student who can only quote the formula is stuck the moment the constant of proportionality turns up somewhere unfamiliar.

k is a relative rate, not an amount per unit time. Rearranged, k is (dA/dt)/A — a rate of change of mass divided by a mass — so its units are per year, never milligrams per year. It measures the fraction of what is present that goes each year; the actual milligrams lost per year is kA, and that number falls as A does. The consequence worth carrying into part (b) is this: when you impose a half-life, the initial mass appears on both sides and cancels, which is exactly why a half-life is a property of the isotope and not of the sample you happen to have in front of you.

Exact means exact, and rounding early moves the third decimal. Q5(b) asks for k exactly, so the answer is a logarithm; replacing it with a decimal answers a different question. Part (c) then asks for a time to three decimals — carry the logarithms symbolically all the way through and evaluate once, at the end. Rounding k first and substituting the rounded value is the standard way to be wrong in precisely the digit the question is marking. Finish with the sentence part (c) asks for: a time in years, and what fraction of the tracer has gone by then.

Getting the most out of it

Say what you divided by, every single time

Write "assuming y ≠ 0" — or whatever the divisor was — in the margin as you separate, and come back to it before you write the final answer. A large share of the marks lost on a topic like this one are constant solutions that were discarded at the first step and never recovered. The check is quick: set the divisor to zero and see whether that constant function satisfies the equation.

Substitute the initial condition while the equation is still implicit

In Q3 the condition is given at x = 0 and the implicit form is short. Use it there, get a clean number for the constant, and only then solve for y. Unwrapping the exponential first leaves the constant tangled inside it, and that is where the algebra goes wrong far more often than in the integration.

Differentiate your own answer back

Every explicit solution on this sheet can be checked in one line: differentiate what you wrote and see whether the original equation reappears, then confirm the initial condition. It catches sign errors, a mis-integrated x-side and a constant fixed in the wrong place, and it takes less time than re-reading the working.

State the interval, and put units on the interpretation

An answer to an initial value problem is a formula and the interval it holds on, written in standard brackets. And where a question is set in litres, grams, minutes or years — Q4 and Q5 both are — the closing sentence has to carry the units and say what the number means about the tank or the sample. The competency's performance criteria attach marks to that sentence, and it is the one students skip.

Want the solutions, or something more challenging?

The worksheet, the questions above and every explanation on this page stay free permanently. Three more PDFs exist for this topic — the worked answer key, a harder problem set, and the answer key to that. They come with the CEGEP Calculus II Solutions Bundle, which is what keeps the rest of the series free.

What else exists for Separable Differential Equations

Three PDFs · 9 pages · all three are in the bundle below.

Answer key — 2 pages. All 5 questions worked step by step, including the restrictions and the justifications. Not a list of final answers.
Challenge problems — 4 pages, 4 problems. A separate sheet at exam-plus difficulty covering the same 4 concepts. Harder than anything on the free sheet.
Challenge answer key — 3 pages. Every challenge problem worked to the same standard, with the checks shown.
PDF, letter size, print-ready.

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Common questions

Is this worksheet really free?

Yes — the questions are on this page to read, and the PDF downloads directly, no email and no account. The one paid item is optional: the complete CEGEP Calculus II Solutions Bundle, which covers every set at this level.

Which course is this? My timetable says 201-SN3-RE, not 201-NYB-05.

Both are the same course. 201-NYB-05 is the legacy number and 201-SN3-RE the current one, and the content is fixed by the ministerial competency 0M03 in the Sciences de la nature programme, so the material is the same across colleges whichever number is printed on your schedule.

Does this cover integrating factors, linear first-order equations, Euler's method or slope fields?

No, and that is deliberate rather than an omission. The devis for 0M03 names separable differential equations, and separable equations are the whole scope of this set. Integrating factors, linear and exact first-order equations, second-order equations, Euler's method and slope fields all belong to a later differential equations course, so if you have been sent looking for one of those, this is not the sheet.

Why does separating the variables lose the solution y = 0?

Because separating means dividing by a function of y, and that division is only legal where the function is not zero. Any constant y = c that makes it zero is itself a solution — a constant has derivative 0, and the right-hand side is 0 there too — so it is discarded before you integrate anything. Afterwards, check whether your general family recovers it. A family of the form y = A times something contains it at A = 0; a family of the form y = 1 over something never does, and there the lost solution has to be written down separately.

What is the interval of validity, and how do I find it?

It is the single interval, containing the initial input, on which your formula is differentiable and actually satisfies the equation. Find it by asking what the formula needs — a positive argument for a logarithm, a non-zero denominator, a non-negative radicand — then solving that condition and taking the largest interval around the initial input on which it holds. Two separated pieces are two different solutions rather than one, and a formula that stops satisfying the equation past some point is not a solution there even though it still evaluates.

When should I substitute the initial condition?

As soon as you have integrated both sides, while the equation is still in its implicit form. The constant then comes out as a clean number in a short equation. If you solve for y first, the constant ends up buried inside an exponential or a logarithm and has to be dug back out, which is where most of the algebra errors in an initial value problem happen.

How do I turn a mixing tank into a differential equation?

Track the mass of dissolved substance, not the concentration. The rate in is the incoming concentration times the incoming flow; the rate out is the concentration in the tank, mass over volume, times the outgoing flow — which is what the phrase "well stirred" is there to justify. The equation is rate in minus rate out, and the description of the starting contents is your initial condition. Check the volume too: equal flows in and out keep it constant, which is what makes the equation separable.

What does the constant k mean, and why doesn't the half-life depend on how much I start with?

Rearranging gives k as a rate of change divided by an amount, so its units are per unit time — a fraction of what is present, not a fixed quantity per year. When you impose a half-life, the initial amount appears on both sides of the equation and cancels, leaving k determined by the half-life alone. That is why a half-life is a property of the substance rather than of your particular sample.

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