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J. Wyss,

Justin’s lecture notes

MATH 246: Chapter 0 Section 0

Justin Wyss-Gallifent

1. What is a differential equation and what does it mean to solve one?

(a) The most straightforward definition of a differential equation (a DE) is that it’s an equa- tion involving some or all of the following: An unknown function, derivatives of that function, and other functions of the same variable(s).

Example: f ′^ ( t )+ f ( t ) = 10 in which f is our unknown function of t. Example: y ′′^ + 3 y ′^ xy = 6 in which y is our unknown function of x.

Example: t^2 f ′′( t ) =

5 f ′( t )(sin t ) in which f is our unknown function of t.

Example: 17 dy^ xd^ y

xy in which y is our unknown function of x.

2 dx −^ dx^2 = Example: ∂ xu + sin ( x )∂ yu = y^3 ∂ xyu in which u is our unknown function of both x and y.

(b) Solving a DE means finding a function which makes the DE true when you plug that function in. For the following don’t worry about where the solutions came from, just notice that they work. Example: f ( t ) = et^ is a solution to the DE f ( t ) f ′( t ) = 0.

Example: f ( t ) = sin( t ) is a solution to the DE f

( t ) + f ′′( t ) = 0.

Example: f ( t ) = t + e^2 t^ is a solution to the DE f ′′( t )+ 4 t = 4 f ( t ).

Just as regular equations can have more than one solution ( x^2 DE. In fact usually a DE will have infinitely many solutions.

— 9 = 0 has two) so can a

Example: f ( t ) = 487 et^ is another solution to the DE f ( t ) f ′( t ) = 0. You can probably

see lots more now.

2. Associated definitions

(a) A DE is called ordinary (so an ODE) if the unknown function is just a function of one variable. Otherwise it’s partial (so a PDE).

Example: f ′( t )+ 3 tf ′′( t ) = et^ is an ODE. Example: ux ( x, y )+ uyx ( x, y )+ y = 3 is a PDE.

(b) The order of a DE is the highest derivative that appears in it. We say things like first- order and second-order and so on.

Example: x^7 f ′( x )+ (cos x ) f ( x )+ x = ex^ is first-order. Example: tf ( t )+ etf ′′( t ) = 1 f ′( t ) is second-order. −

(c) A DE is linear if it can be written as a sum/difference of some or all of:

• An unknown f multiplied by a coefficient.
• Derivatives of the unknown f multiplied by coefficients.
• Coefficients.

By coefficients we mean they can be other functions of the same variables, including just constants, including 0.

Example: The DE 5 tf ( t )+ (ln t ) f ′( t ) = 5 is linear. Example: The DE (tan t ) f ( t )+ t^3 f ′( t )+ 7 f ′′( t ) = 1 is linear. Example: The DE f ( t )

t + (1 t ) f ′′′( t ) = f ′( t ) is linear.

Example: The DE f ( t )^2 + f ′( t )

= 7 is nonlinear because the f ( t )^2 is not permitted. Example: The DE sin( y ′)+ y ′^ + y = x is nonlinear because the sin( y ′) is not permitted.

Example: The DE y ′ y + y = xy is nonlinear because the y ′ y is not permitted.

MATH 246: Chapter 1 Section 1

Justin Wyss-Gallifent

1. Introductory overview of first-order ODEs.

(a) A first-order ODE is permitted to have an unknown function y (of a single variable, say t ) its derivative y ′^ and then some other functions of t.

Example: t ( y ′)^2 + y = sin t

Example: y ′^ − ty = e^2 t ′ Example: sin( y ′)+ ey^ = t

(b) In general these can be very hard! For now let’s restrict ourselves a bit to first-order ODEs that have the form y ′^ = f ( t, y ) because in first-order ODE getting to this point is usually algebra.

That f is confusing, it’s not the unknown function but rather it just represents the fact that we can have a bunch of y and t on the right hand side. In other words things like this:

Example: y ′^ = ty

Example: y ′^ = 4 t − 8 y

Example: y ′^ = y^. t

2. Explicit first-order DEs. 3. (a) Because solving even first-order ODEs is hard we’ll go down even further and look at explicit

first-order ODEs that have the form y ′^ = f ( t ).

Example: y ′^ = t^2.

Example: y ′^ = 4 t + sin t.

(b) At this point you might have an epiphany and realize that often you can solve these because solving these is as easy as integrating the right side.

Example: y ′^ = t^2. To solve this we integrate to get y = 1 t^2 + C for any constant C. 3

Example: y ′^ = 4 t + sin t. To solve this we integrate to get y = 2 t^2 − cos t + C for any constant C.

4. General solutions, initial value problems, specific (particular) solutions

(a) We’ve started to notice that we can have many solutions to a DE. In the explicit DEs above get a constant C which can be anything.

(b) A general solution to a DE is a solution involving constants and for which different constants will give all solutions.

(c) A specific solution or a particular solution is a solution in which a specific (particular) choice of constant(s) has been made.

Example: The general solution to y ′^ = t^2 is y = 1 t^3 + C. Some specific solutions are 3 y = 1 t + 1, y = 1 t − 107 and y = 1 t + π. 3 3 3

(d) Often when we encounter a DE it comes pre-packaged with an initial value , or IV. In our simple exact case (and in many future cases) this will be an insistance that y ( tI ) = yI for specific tI and yI. The DE and the IV together form an initial value problem or IVP. It’s very common that tI = 0 but this isn’t always the case!

Example: y ′^ = 2 t with y (0) = 3 is an IVP.

Example: y ′^ = 2 t with y (0) = 5 is an IVP with the same DE but different IV.

Example: y ′^ = 2 t with y (1) = 3 is an IVP with again the same DE but different IV.

(e) When we’re given an IVP the idea will be to first solve the DE to get the general solution and then use the IV to get the specific solution.

Example: y ′^ = 2 t with y (0) = 3. First we find y = t^2 + C , the general solution, and then

y (0) = 02 + C = 3 so C = 3 and the specific solution is y = t^2 + 3.

Example: y ′^ = 2 t with y (1) = 3. First we find y = t^2 + C , the general solution, and then

y (1) = 12 + C = 3 so C = 2 and the specific solution is y = t^2 + 2.

5. Some Theory which will arise again and again.

(a) Finding a solution to y ′^ = f ( t, y ) ; we will be a little more specific. We will not just say that y is a solution but that y is a solution on an interval ( a, b ). This will mean:

i. The derivative y ′^ (derived from y ) is defined for every t in ( a, b ).

ii. The right side, f ( t, y ), is defined for every t in ( a, b ) when y is plugged in.

iii. The DE is true for every t in ( a, b ).

Example: Consider the DE y ′^ = −. We claim that y =

1 − t^2 is a solution on (− 1 , 1). t y Checking the three things above:

i. y =

1 − t^2 so y ′^ = −√^ t 2 which^ is^ defined^ for^ every^ t^ in^ (−^1 ,^ 1). 1 − t ii. The right side is − t which equals − t which is defined for every t in (− 1 , 1). y

√ 1 − t^2 iii. The DE holds true for every t in (0 , ∞). Comment: This y is not a solution on [− 1 , 1] because even though y itself is defined at

t = ± 1 we see that y ′^ is not!

(b) Intervals of Existence and Theorey for Explicit IVPs:

We now know that solving the explicit ODE given by y ′^ = f ( t ) is as easy (or hard) as inte- grating f ( t ). However the Fundamental Theorem of Calculus tells us something interesting. It states that if a function is continuous then it is integrable. This means that even if we can’t actually integrate f ( t ) using techniques that we know, we still know there is a solution. Moreover that solution will exist on an interval where f ( t ) is continuous.

The practical upshot of this is that when we’re solving an IVP with y ′^ = f ( t ) and y ( tI ) = yI , there will be a solution on the largest interval ( a, b ) which contains tI and on which f ( t ) is continuous. This interval is called the interval of existence of the solution and holds whether or not we can actually, in practice, find that solution.

Example: y ′^ = 1 with y (1) = 5. We notice the largest open interval containing t = 1 on t^2 I which 1 is defined is (0 , ∞) and so this is the IE. Notice that we don’t need to solve it, but t^2 we could, since the general solution is y = − (^) t + C and then y (1) = − 1 + C = 5 so C = 4 1 and the specific solution is y = − (^) t + 4.

Example: y ′^ = t^ with t (0) = 17. We notice the largest open interval containing ( t −3)( t +6) tI = 0 on which t^ is defined is (− 6 , 3) so this is the IE. We could possibly solve this ( t −3)( t +6) with some messy partial fractions but we won’t.

Here’s one with an IVP:

Example: Consider y ′^ − 6 y = et^ with y (0) = 2. We see that a ( t ) = − 6 so A ( t ) = − 6 t and the general solution is

∫

y = e −(− 6 t ) e t e − 6 t dt ∫

= e^6 t^ e −^5 t^ dt $ %

= e^6 t^ −

e −^5 t^ + C 5

et^ + Ce^6 t 5

At this point y (0) = − 1 e^0 + Ce^0 = − 1 + C = 2 so that C = 11 so the specific solution is 5 5 5

y = −

et^ +

e^6 t 5 5

3. At this point you can probably see tha∫t solving a first-order linear ODE is as easy (or as hard)

as first finding A ( t ) and then finding f ( t ) eA ( t )^ dt.

4. Theory!

By the FTOC if both f ( t ) a∫nd a ( t ) are continous on on an interval then not only will A ( t )

exists but so will e − A ( t )^ and f ( t ) eA ( t ). This means that if we have an initial value y ( tI ) = yI then the interval of existence of the solution will be the largest open interval containing tI on which both f ( t ) and a ( t ) are continouous. As before this lets us find the IE even when we can’t solve the IVP.

Example: Consider y ′^ + 1 y = 1 with y (2) = 17. Here a ( t ) = 1 and f ( t ) = 1. The t t − 5 t t − 5 largest open interval containing tI = 2 on which both are continuous is (0 , 5) so this is the IE of the solution. Finding the solution is a different matter entirely, but it exists on (0 , 5)!

g ( y )

MATH 246: Chapter 1 Section 3

Justin Wyss-Gallifent

1. Separable ODEs.

A DE is separable if it can be written in the form y ′^ = f ( t ) g ( y ). The word separable comes from the fact that the right side is separated into a product of a function of t and a function of y.

Example: y ′^ = ty is separable - it is already separated!

Example: ty ′^ + y ′^ = y^2 is separable because it can be separated, first by factoring y ′( t +1) = 2!^ " y^2 and then dividing y ′^ = y and thinking of it as y ′^ = 1 y^2. t +1 t +

2. The method of solving a separable DE for the non-constant solutions is really slick:

y ′^ = f ( t ) g ( y )

dy = f ( t ) g ( y ) dt 1

g ( y )

dy = f ( t ) dt ∫ 1

g ( y )

dy = f ( t ) dt

Where the integral on the left is with respect to y and the integral on the right is with respect to t. Since both indefinite integrals should get their own constant, instead we just put a single

• C on the right.

The 1 looks really icky to integrate but in our examples it generally works out pretty nicely.

Example: Solve y ′^ = t

. We work as follows: y +

dy t

dt

y +

( y + 1) dy = tdt ∫ ∫

y +1 dy = t dt

y^2 + y =

t^2 + C 2

3. Constant Solutions:

Notice that we divided by g ( y ). What if g ( y ) = 0? A really important thing to notice with separable ODEs is that if there are some y with g ( y ) = 0 then those values of y , taken as functions, are solutions. This is because for those functions the derivative will be zero and the ODE will be satisfied. They are called the constant solutions to the ODE.

Example: Consider y ′^ = ( y^2 4) t^2. Here g ( y ) = y^2 4 will equal 0 when y = ±2.

Thus y = 2 and y = 2 are con

stant solutions (actual fu

nctions, which are constants and

solutions) to the ODE

6. Two Small Notes:

(a) The effect of initial values:

When we solve a separable ODE and get an implicit solutions for which there seems to be more than one explicit solution, an initial value usually tells us which one it is:

Example: Solve y ′^ = t with y (1) = 3. First we solve the DE: r y −

dy t

dt

y

y dy = t dt ∫ ∫

y dy = t dt

1 y^2 =

t^2 + C 2 2 uy^2 = t^2 + 2 C √ y = ± t^2 + 2 C

Then y (1) = ±

12 + 2 C = − 3 so we are forced to use the negative in front of the square

root. Thus −

12 + 2 C = −3 so 1 + 2 C = 9 and C = 4. Then the explicit solution is

y = −

t^2 + 8. Note: There are no constant solutions here since g ( y ) = 1 is never 0. y

(b) Uniqueness (?) of solutions:

The existence of constant solutions often leads to non-unique solutions to IVPs. This can happen when a constant solution satisfies the DE but also the procedural method gives another solution. The way to manage this is to not forget to find your constant solutions and check if they satisfy the IV.

Example: Solve y ′^ = y^2 /^3 with y (0) = 0. Notice that y = 0 is a constant solution which also satisfies the DE. However the DE is separable:

dy = y 2 / 3 dt y −^2 /^3 dy = 1 dt ∫ ∫

y −^2 /^3 dy = 1 dt

3 y^1 / 3 = t + C % & 3

y =

t +

C

Then y (0) = C = 0 so C = 0. This gives the additional solution y = t = 1 t^3. 3 3 27

7. Overlap

At this juncture it might be helpful to notice that an ODE doesn’t need to be just one of the categories we’ve looked at - explicit, first-order linear, and separable - it could fall into more than one category.

Example:

y ′^ = ty is both separable and first-order linear.

Example: y ′^ = t^2 is all of explicit, separable and first-order linear.

MATH 246: Chapter 1 Section 5

Justin Wyss-Gallifent

The overarching goal of this section is to find things out about solutions to DEs without actually

solving them explicitly. Instead we attack them graphically.

1. Phase Line Portraits for Autonomous Differential Equations.

(a) As we’ve seen, autonomous differential equations look like y ′^ = g ( y ). Usually they have constant solutions when g ( y ) = 0. But what about when g ( y ) ̸= 0?

• If g ( y ) = 0 then y is constant.
• If g ( y ) > 0 then y ′^ > 0 and y increases.
• If g ( y ) < 0 then y ′^ < 0 and y decreases.

Moreover as solutions move toward the constant solutions we see g ( y ) tends toward 0 and so the graphs flatten out and become asymptotic.

What this means is that we can analyze the behavior of the solutions by looking at g ( y ), specifically at the sign chart.

(b) Stability. Moreover we’ll see something happen near the constant solutions. Specifically one of three things can happen.

i. If nearby solutions move away from the constant solution on both sides then the constant solution is unstable.

ii. If nearby solutions move toward the constant solution on both sides then the constant solution is stable.

iii. If there is different behavior on each side then the constant solution is semistable.

Example: Consider y ′^ = y ( y − 3)( y + 2)^2. This has the following sign chart:

When y is − 2 0 3 y ′^ = g ( y ) is + 0 − 0 + 0 + so y ( t ) is inc cons dec const inc const inc

And hence the following families of solutions (not drawn in Matlab because Matlab

couldn’t do a good job):

y=

y=

y=−

From these families of solutions we can draw all sorts of conclusions:

• The constant solution y = − 2 is stable, the constant solution y = 0 is unstable, and the constant solution y = 3 is semistable.
• The particular solution y ( t ) satisfying y (0) = α has lim t →∞

y ( t ) = 2 when −∞ < α < 0.

• The particular solution y ( t ) satisfying y (0) = α is decreasing when − 2 < α < 0.
• y = −2 is stable.
• y = 0 is unstable.
• y = 3 is semistable.

2. Contour Plots of Implicit Solutions

(a) When we solve a separable DE we often get an implicit solution with a C in it. This implicit solution is an equation. If we pick various values of C and plot the resulting equations we get a contour plot.

What’s useful about these contour plots is that the parts of the curves that form functions are explicit solutions to the DE because they’re functions ( y in terms of t ) that satisfy the implicit solution. This means that we can pick a point on a curve and follow it as far left and right as possible and the result is the graph of an explicit solution to the DE. Example: Consider dy^ = 1. This is separable with general solution y^2 − 6 y = x + C. dx 2 x − 6 This is not as bad as it looks:

y^2 − 6 y = x + C

y^2 − 6 y +9 = x + C + 9

( y − 3)^2 = x + C + 9

These are all parabolas opening right with their vertices at y = 3. If we sketch a few of these (note that they extend out forever, this is just a subset):

10

8

6

4

2

0

• 2
• 4
• 6
• 8
• 10
• 10 - 5 0 5 10

From this contour plot we can draw all sorts of conclusions:

• Solutions extend infinitely far to the right but not the left.
• The specific solution y ( t ) satisfying y (0) = 0 is a decreasing function with lim t →∞

y ( t ) =

−∞.

• Solutions are either always increasing or always decreasing.

For class handout: Direction field for (^) dt dy (^) = t − y (^2) :

MATH 246: Chapter 1 Section 6

Justin Wyss-Gallifent

The overarching goal of this section is to finally see some applications of DEs. We’ll focus on three:

Population dynamics, tanks, and motion. All three of these result in either first-order linear or

separable differential equations which we can solve.

1. Population dynamics:

(a) Introduction: In precalculus you probably learned that if a population grows at rate 5% then it obeys the formula P = Ae 0_._ 05 t

But why? The answer is that to say “a population grows at rate 5%” means that the change in population at any time equals 5% of the actual population, meaning:

p ′^ = 0_._ 05 p

This is a first order linear differential equation. If we rewrite it as p ′^ − 0_._ 05 p = 0 then a ( t ) = − 0_._ 05 so A ( t ) = − 0_._ 05 t and the solution is:

∫

p ( t ) = e −(− 0_._ 05 t ) 0 dt = Ce (^0). (^05) t

That’s why!

(b) General Approach: Our general formula will involve a population with a certain growth rate R but in addition some new amount may arrive or depart every time period, maybe by being introduced, removed, etc. So in general we have

p ′^ = Rp + a ( t )

Our rate will always be constant but the amount that are introduced or subtracted may vary.

(c) Examples:

Example: A population of monkeys starts with 100. It has a growth rate of 4% per year but an additional 8 monkeys join each year from a neighboring troop. Find the number of monkeys after t years. Solution: Here we have p ′^ = 0_._ 04 p + 8 with p (0) = 100. The solution (work ommitted) is p = 300 e (^0). (^04) t − 200.

Example: In a certain neighborhood there is a mosquito problem. The population starts at 10M and has a growth rate of 20% monthly. Traps are put out and these traps kill 3M monthly. Find the number of mosquitos after t months and determine when the mosquitos will be wiped out.

Solution: Here we have p ′^ = 0_._ 2 p − 3 with p (0) = 10. The solution (work ommited) is p = − 5 e^0_.^2 t^ + 15 and if we solve − 5 e^0.^2 t^ + 15 = 0 we get t = 5 ln(3) ≈ 5._ 49 months.

dt

E −

3. Motion:

(a) Introduction: In calculus you probably learned that a falling object with no air resistance

has a ( t ) = − 9_._ 8

But why? The answer comes from equating two forces. if the object has acceleration a ( t ) and mass m then the force on it is ma ( t ). The force from gravity is 9_._ 8 m. When we equate these we get

and then we cancel the m.

ma ( t ) = − 9_._ 8 m

(b) Adding some air: When we add air resistance there are now two forces. First there is gravity pulling down and then drag (air resistance for example) pushing up. These two forces combine to form the total force. We know the total force is ma ( t ) = mv ′( t ).

Therefore ma ( t ) = force of gravity + drag force

The force of gravity is 9_._ 8 m. The drag force is harder, it’s mkv 2 where k is the drag coefficient. Thus we have 2 ma ( t ) = − 9_._ 8 m + mkv

or, cancelling the m again: 2 a ( t ) = − 9_._ 8+ kv

Finally we replace a ( t ) by v ′( t ) to get:

dv = 9_._ 8+ kv 2

dt

(c) General Approach: We’ll generally just use the IVP

dv = − 9_._ 8+ kv 2 with v (0) = 0 (usually)

to find v at time t and answer questions from that. Things to note:

• v (0) might not be 0 if there is some initial velocity.
• In the Metric system k will be in m − 1 , mass will be in kg and 9.8 stays as-is.
• In the English system k will be in ft −^1 , mass will be in slugs and we use 32.2 (instead of 9.8).
• Terminal velocity occurs when v ′^ = 0 so this is when v =

9_._ 8 k

• If we know v then we can also find out distance travelled since v = h ′^ and so h ( t ) ∫ (^) t E h ( tI ) = (^) t I v ( t ) dt
• We can certainly change from air to some other substance or from earth gravity to some other standard. Information would have to be given.
• It’s insanely useful to know that

1 1 −^1 x e +1 −^1

1+ z

2 z x^2 − a^2 dx^ =^ −^ a tanh^ a +^ C^ where^ tanh( z )^ =^ e^2 z^ − 1 and^ tanh^ ( z )^ =^2 ln^1 − z

(d) Examples:

Example: A skydiver leaps out of a plane at 3000m. The drag coefficient is 0_._ 002 m − 1 . What is the IVP here? What is her terminal velocity? Find her velocity at time t. Solution: We have dt dv = − 9_._ 8+ 0_._ 002 v^2 with v (0) = 0. " Her terminal velocity is v = 9_._^8 =

4900 = 70 m/s. 0_._ 002 The solution to the IVP is shown here:

dv (^) 2

dt

= − 9_._ 8+ 0_._ 002 v

dv (^) 2

dt

= 0_._ 002(− 4900 + v ) 1 dv = 0_._ 002 dt

∫

v^2 − 4900 ∫ 1 dv = 0_._ 002 dt v^2 − 4900 (^1) − 1

v

tanh 70

= 0_._ 002 t + C $ v

tanh− 1 = − 0_._ 14 t + C 70 v = tanh(− 0_._ 14 t + C ) 70 v = tanh(− 0_._ 14 t + C ) 70 v = 70 tanh(− 0_._ 014 t + C )

− 1 Then v (0) = 70 tanh( C ) = 0 so C = tanh 0 = 0 and our final answer is

v = − 70 tanh (0_._ 14 t )