Integral Calculus 201-SN3-RE • CEGEP, Montreal

Revision sheet: differential equations (201-SN3-RE)

This sheet does not re-explain what a differential equation is: you know that. It deals with what decides the mark on the last question of the 201-SN3-RE (formerly 201-NYB) final exam, namely how the constant of integration is handled, as it changes form between the logarithm and the exponential, and putting the equation in standard form before any integrating factor.

It is written for CEGEP students in the Science program (Sciences de la nature). Once the sheet is read, the corrected exercise set on the same chapter puts every reflex to the test.

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The thread of the chapter

Solving a differential equation means finding a FAMILY of functions; an initial condition selects a single one. Almost every lost mark comes from the constant: lost, added at the wrong moment, or determined before the solving is finished.

This chapter is part of 201-SN3-RE, Integral Calculus

The essentials

General solution, particular solution

• The GENERAL solution of a first-order equation contains ONE arbitrary constant: it is a family of curves.
• An INITIAL CONDITION $y(x_{0})=y_{0}$ selects a single curve of this family.
• The constant is determined AFTER writing out the complete general solution, never before.
• Verifying a solution is immediate: differentiate it and substitute it into the equation.
• A second-order equation has TWO constants, so two conditions are needed.

Four solutions of

y'=y

for

C=-1

\frac{1}{2}

1

and

2

: the constant MULTIPLIES the exponential, it is not added to it, and only one curve passes through a given point.

The curves of the family never cross when the equation has the form $y'=f(x,y)$ with $f$ regular: exactly one solution passes through each point.

Separation of variables, in five moves

• Recognize the form $\frac{dy}{dx}=g(x)h(y)$ , a PRODUCT of a function of $x$ and a function of $y$ .
• Separate: $\frac{dy}{h(y)}=g(x)\,dx$ , noting the values of $y$ that make $h$ zero.
• Integrate both sides, with ONE single constant, on the $x$ side.
• Isolate $y$ , transforming the constant when it passes through an exponential.
• Check the CONSTANT solutions lost in step 2, those where $h(y)=0$ .

The constant changes nature as it passes through the exponential: $\ln\left|y\right|=x+C$ gives $y=\pm e^{C}e^{x}$ , which is rewritten $y=Ke^{x}$ with $K$ of either sign.

Linear equation and integrating factor

• STANDARD form is compulsory: $y'+P(x)y=Q(x)$ , with the coefficient of $y'$ equal to $1$ .
• Integrating factor: $\mu(x)=e^{\int P(x)\,dx}$ , with no constant of integration here.
• Multiplying the equation by $\mu$ makes the left-hand side equal to $\left(\mu y\right)'$ .
• What remains is $\mu y=\int\mu Q\,dx+C$ , then divide by $\mu$ .
• Usual models: $y'=ky$ gives $y=Ce^{kx}$ ; $y'=k(y-A)$ gives $y=A+Ce^{kx}$ ; the logistic $y'=ky\left(1-\frac{y}{K}\right)$ gives an S-shaped curve with ceiling $K$ .

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Adding the constant instead of multiplying by it

the whole question, since the function obtained is not a solution

What not to write

“ $\ln\left|y\right|=x+C$ , so $y=e^{x}+C$ .”

What to write

“ $y=e^{x+C}=e^{C}e^{x}=Ke^{x}$ : the constant becomes a FACTOR.”

Both curves start from

(0, 2)

, but only

2e^{x}

(green) satisfies

y'=y

;

e^{x}+1

(red) has derivative

e^{x}

, not

y

Why: The exponential of a sum is a product, not a sum. Immediate check: $y=e^{x}+C$ gives $y'=e^{x}=y-C\neq y$ , whereas $y=Ke^{x}$ gives $y'=Ke^{x}=y$ ✓. One second of differentiation eliminates the wrong answer.

2. Using the initial condition too early

1 to 2 marks, and often an expression that cannot be simplified

What not to write

“ $\ln\left|y\right|=x+C$ , and $y(0)=3$ gives $C=\ln 3$ , so $\ln\left|y\right|=x+\ln 3$ .”

What to write

“First I isolate $y=Ke^{x}$ , THEN $y(0)=3$ gives $K=3$ , so $y=3e^{x}$ .”

Why: Determining the constant before isolating $y$ forces you to drag a logarithm along to the end, and the constant takes a needlessly complicated form there. The order is always the same: complete general solution, then initial condition.

3. Losing the constant solution by dividing

1 to 2 marks, and the equilibrium solution of the model disappears

What not to write

“ $\frac{dy}{dx}=y(2-y)$ , I divide by $y(2-y)$ and integrate.”

What to write

“I first note that $y=0$ and $y=2$ are CONSTANT solutions, then I divide on the intervals where $y\neq 0$ and $y\neq 2$ .”

y=0

and

y=2

(green) make

y(2-y)

zero: they are solutions, which dividing by

y(2-y)

wipes out. The other solutions approach them.

Why: Dividing by an expression that can be zero makes the corresponding solutions disappear. In a logistic model, $y=K$ is the equilibrium state, that is, the answer to the question “what happens to the population in the long run”. Losing it means losing the interpretation.

4. Applying the integrating factor without the standard form

the whole question

What not to write

“ $xy'+2y=x^{3}$ : I take $\mu=e^{\int 2\,dx}=e^{2x}$ .”

What to write

“First I divide by $x$ : $y'+\frac{2}{x}y=x^{2}$ , so $P(x)=\frac{2}{x}$ and $\mu=e^{2\ln\left|x\right|}=x^{2}$ .”

Why: The integrating factor formula assumes the coefficient of $y'$ is $1$ . Without this rearrangement, $P$ is wrong and everything else collapses. The check: after multiplying by $\mu$ , the left-hand side must be exactly the derivative of $\mu y$ .

5. Adding a constant to the integrating factor

wasted time, and sometimes an algebra error

What not to write

$\mu(x)=e^{\int P\,dx+C}$

What to write

$\mu(x)=e^{\int P\,dx}$ , with no constant: any integrating factor will do, so take the simplest.

Why: A multiplicative constant in $\mu$ cancels on both sides of the equation: it serves no purpose. The only constant that matters is the one from the FINAL integration, the one that generates the family of solutions.

6. Dropping the absolute value without discussion

1 mark, and half of the solutions disappear

What not to write

“ $\ln\left|y\right|=x+C$ gives $\ln y=x+C$ .”

What to write

“ $\left|y\right|=e^{C}e^{x}$ , so $y=\pm e^{C}e^{x}$ , written $y=Ke^{x}$ with $K$ any real number, possibly negative.”

Why: Without the $\pm$ , you lose all the negative solutions, including those that match a negative initial condition. Going through a $K$ of either sign settles the matter once and for all, and it must be written, not implied.

7. Confusing exponential growth with logistic growth

the whole interpretation question

What not to write

“The population grows in proportion to its size, so it tends to infinity.”

What to write

“The LOGISTIC model $y'=ky\left(1-\frac{y}{K}\right)$ slows down as it approaches $K$ : the population tends to $K$ , not to infinity.”

The two curves start together, then the logistic bends and hugs its ceiling

K=10

while the exponential keeps climbing: only the long term tells them apart.

Why: The two models coincide at the start, when $y$ is small compared with $K$ , which makes them indistinguishable on the first data. It is the long-term behaviour that separates them, and that is precisely what the question asks you to interpret.

8. Not checking the solution found

variable, but it is the cheapest check of the whole semester

What not to write

“I have isolated $y$ , I am done.”

What to write

“I differentiate my solution and substitute it into the original equation, then I check the initial condition.”

Why: Unlike most chapters, a differential equation can be checked completely in thirty seconds: differentiate, substitute, compare. A paper that skips this hands in sign errors that the computation never reveals.

Which method to choose

Which method, depending on the form of the equation

Put the equation in a recognizable form before any computation. Two methods, separation and the integrating factor, handle most CEGEP questions.

If the equation can be written $\frac{dy}{dx}=g(x)h(y)$ , a product → SEPARATION of variables, without forgetting the constant solutions

Example: $\frac{dy}{dx}=xy$ gives $\frac{dy}{y}=x\,dx$
If the equation can be written $y'+P(x)y=Q(x)$ after dividing → INTEGRATING FACTOR $\mu=e^{\int P}$

Example: $y'+\frac{2}{x}y=x^{2}$ gives $\mu=x^{2}$
If the equation is $y'=ky$ → EXPONENTIAL growth or decay: $y=Ce^{kx}$ , directly

Example: radioactive decay, continuously compounded interest
If the equation is $y'=k(y-A)$ with $A$ constant → $y=A+Ce^{kx}$ : this is Newton's law of cooling, $A$ being the room temperature

Example: a coffee cooling in a room at $20$ degrees
If the equation is $y'=ky\left(1-\frac{y}{K}\right)$ → LOGISTIC model: separation then partial fractions, ceiling $K$

Example: a population limited by its resources

An equation that fits none of these forms after rearranging calls for a substitution: $v=\frac{y}{x}$ for a homogeneous equation, $u=y^{1-n}$ for a Bernoulli equation, or $x$ viewed as a function of $y$ (exercises 1 and 7 of the set). If none of them applies either, an algebraic step is missing.

What to do with the constant, step by step

The constant changes form three times during a solution, and each transformation is a trap.

If you have just integrated both sides → ONE single constant, on the $x$ side

Example: $\ln\left|y\right|=\frac{x^{2}}{2}+C$
If you apply the exponential → the constant becomes a FACTOR: $e^{C}$ , renamed $K$

Example: $\left|y\right|=e^{C}e^{x^{2}/2}$
If you remove the absolute value → $K$ can take EITHER SIGN, and $K=0$ gives back the zero constant solution

Example: $y=Ke^{x^{2}/2}$ for every real $K$
If you have an initial condition → substitute and solve for $K$ , only at that moment

Example: $y(0)=3$ gives $K=3$

The value $K=0$ often recovers the constant solution lost in the division: it is the check that closes the question cleanly.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

A separation of variables, written up for the marker

When to use it: Any equation of the form $\frac{dy}{dx}=g(x)h(y)$ .

1 Spot and STATE the constant solutions: “ $h(y)=0$ for $y=0$ , which is a solution.”
2 Separate the variables, writing the differentials on both sides.
3 Integrate, with a single constant, and keep the absolute value inside the logarithm.
4 Isolate $y$ , transforming the constant at each pass through the exponential.
5 Apply the initial condition, then CHECK by differentiating.

Concluding sentence

The general solution is $y=Ke^{x^{2}/2}$ for any real $K$ , and the condition $y(0)=3$ gives $y=3e^{x^{2}/2}$ .

The trap: Writing a constant on each side. One is enough: $C_{1}$ and $C_{2}$ combine at once into $C=C_{2}-C_{1}$ , and keeping two suggests you have not understood what a family of solutions is.

Marking: Typically 1 mark for the constant solutions, 1 mark for the separation, 2 marks for the integration and isolating $y$ , 1 mark for the initial condition.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Differentiate the solution and substitute it

Compute $y'$ from the solution found, then replace $y$ and $y'$ in the original equation: the two sides must agree identically.

$y=3e^{x}$ gives $y'=3e^{x}=y$ ✓, whereas $y=e^{x}+3$ gives $y'=e^{x}=y-3\neq y$ .
Check the initial condition

Substitute $x_{0}$ into the particular solution: you must get exactly $y_{0}$ .

$y(0)=3e^{0}=3$ ✓. A different value signals an error in determining the constant.
Check the long-term behaviour

Let $x$ tend to infinity and check that the result matches the physical meaning of the problem.

A coffee must tend to the room temperature: $T(t)=20+70e^{-kt}\to 20$ ✓. A limit of zero or infinity would signal a badly set-up model.
Find the constant solutions within the family

Give the constant the value $0$ and check that you land on the equilibrium solution lost in the division.

$y=Ke^{x}$ with $K=0$ gives $y=0$ , which is indeed a solution of $y'=y$ ✓.
Check the units and the order of magnitude

In an applied problem, check that the time constant and the values obtained make sense: a decay that takes a century for a coffee is absurd.

$k=0.0336$ per minute gives a half-life of about $20$ minutes ✓, consistent with a cooling coffee.

The typical problem, taken apart

Newton's law of cooling, from the model to the prediction

A coffee at $90$ degrees is placed in a room at $20$ degrees. After $10$ minutes, it is at $70$ degrees.

What will its temperature be after $30$ minutes?

The curve starts at

90

degrees, passes through the measurement

(10, 70)

and approaches the line

y=20

without crossing it: it is the GAP to the room temperature that decays exponentially.

Step 1

Model: the rate of cooling is proportional to the gap with the room temperature, so $\frac{dT}{dt}=-k\left(T-20\right)$ with $k>0$ .

Why

It is the GAP that drives the cooling, not the absolute temperature: a coffee at $21$ degrees in a room at $20$ hardly cools any more. Writing $\frac{dT}{dt}=-kT$ would model a coffee heading for 0 degrees whatever the room.

Step 2

Separation: $T=20$ is a constant solution; elsewhere, $\frac{dT}{T-20}=-k\,dt$ , so $\ln\left|T-20\right|=-kt+C$ .

Why

The constant solution $T=20$ is noted before dividing: it is the equilibrium temperature, and it will be recovered by $K=0$ at the end. The absolute value is kept; it will disappear cleanly at the next step.

Step 3

Exponential: $\left|T-20\right|=e^{C}e^{-kt}$ , so $T-20=Ke^{-kt}$ with $K$ of either sign, and $T(t)=20+Ke^{-kt}$ .

Why

The constant becomes a factor, then takes either sign: both transformations in one line. The form obtained is the one the tree announced for $y'=k(y-A)$ , which confirms the consistency.

Step 4

Conditions: $T(0)=90$ gives $20+K=90$ , that is $K=70$ . Then $T(10)=70$ gives $20+70e^{-10k}=70$ , that is $e^{-10k}=\frac{5}{7}$ and $k=\frac{\ln\frac{7}{5}}{10}\approx 0.0336$ per minute.

Why

Two conditions for two unknowns, $K$ and $k$ , applied in order: the simpler one first. Keep $\frac{5}{7}$ as an exact fraction, which will make the final computation immediate.

Step 5

$T(30)=20+70e^{-30k}=20+70\left(e^{-10k}\right)^{3}=20+70\left(\frac{5}{7}\right)^{3}=20+70\cdot\frac{125}{343}=20+\frac{8750}{343}\approx 45.5$ degrees.

Why

$30$ minutes are three times $10$ minutes, so the exponential is the CUBE of the one already known. This shortcut avoids handling the decimal value of $k$ and gives an exact answer as a fraction.

Step 6

Checks: $T$ decreases ✓, stays above $20$ ✓, and tends to $20$ as $t\to+\infty$ ✓. The sequence $90$ , $70$ , $55.7$ , $45.5$ at times $0$ , $10$ , $20$ , $30$ shows gaps $70$ , $50$ , $35.7$ , $25.5$ whose ratio is constant and equals $\frac{5}{7}$ ✓.

Why

A constant ratio of the GAPS over equal intervals is the signature of exponential decay: it is the most telling check of the chapter, and it needs no extra computation.

The conclusion, written out

The temperature follows $T(t)=20+70e^{-kt}$ with $k=\frac{\ln\frac{7}{5}}{10}\approx 0.0336$ per minute; after $30$ minutes, the coffee is at $20+70\left(\frac{5}{7}\right)^{3}\approx 45.5$ degrees.

The classic mistake on this problem: Setting $\frac{dT}{dt}=-kT$ and forgetting the room temperature. The solution would be $T=90e^{-kt}$ , which tends to $0$ degrees: the coffee would freeze, which no room at $20$ degrees allows.

Learn by heart

• The general solution contains ONE constant per order of the equation: it is a family.
• Initial condition AFTER isolating $y$ , never before.
• $\ln\left|y\right|=x+C$ gives $y=Ke^{x}$ : the constant becomes a FACTOR of either sign.
• Separation: note the constant solutions BEFORE dividing, and recover them with $K=0$ .
• Integrating factor: first put the equation in the form $y'+Py=Q$ , then $\mu=e^{\int P}$ , with no constant.
• $y'=ky$ gives $Ce^{kx}$ ; $y'=k(y-A)$ gives $A+Ce^{kx}$ ; the logistic levels off at $K$ .
• Cooling: it is the GAP to the room temperature that decays exponentially.
• Thirty-second check: differentiate, substitute, check the initial condition.

Frequently asked questions

Why does the solution of a differential equation contain a constant?

Because the equation only constrains the derivative, and infinitely many functions share the same derivative up to a constant. The general solution is therefore a family of curves. An initial condition, which sets the value at one point, selects exactly one of them.

Why does the constant become a factor after taking the exponential?

Because the exponential of a sum is a product of exponentials. So if the logarithm of the absolute value of y equals x plus C, then the absolute value of y equals e to the power C times e to the power x. This factor is renamed K, of either sign once the absolute value is removed.

When should I use separation of variables and when the integrating factor?

Separation applies when the equation can be written as a product of a function of x and a function of y. The integrating factor applies to first-order linear equations, of the form y prime plus P of x times y equals Q of x, after dividing so that the coefficient of y prime is one.

Why should I note the constant solutions before separating?

Because separating requires dividing by an expression in y, and this division wipes out the values that make it zero. These values are exactly the constant solutions, often the equilibrium states of the model, and therefore the answer to the question about long-term behaviour.

What is the difference between exponential and logistic growth?

The exponential grows without limit, while the logistic curve slows down as it approaches a ceiling called the carrying capacity. The two curves are almost identical at first, when the population is small compared with that ceiling, and only differ in the long run. That distinction is what interpretation questions look for.

Practise it

Corrected exercises: Differential equations, 201-SN3-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

17 corrected exercises
170 points
255 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-calc2-differential-equations. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: differential equations (201-SN3-RE)

The essentials

General solution, particular solution

Separation of variables, in five moves

Linear equation and integrating factor

The mistakes that cost marks

1. Adding the constant instead of multiplying by it

2. Using the initial condition too early

3. Losing the constant solution by dividing

4. Applying the integrating factor without the standard form

5. Adding a constant to the integrating factor

6. Dropping the absolute value without discussion

7. Confusing exponential growth with logistic growth

8. Not checking the solution found

Which method to choose

Which method, depending on the form of the equation

What to do with the constant, step by step

How the answer is expected to be written

A separation of variables, written up for the marker

Check before you hand in

The typical problem, taken apart

Newton's law of cooling, from the model to the prediction

Learn by heart

Frequently asked questions

Why does the solution of a differential equation contain a constant?

Why does the constant become a factor after taking the exponential?

When should I use separation of variables and when the integrating factor?

Why should I note the constant solutions before separating?

What is the difference between exponential and logistic growth?

Corrected exercises: Differential equations, 201-SN3-RE

See also

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