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Fall 2011 offering of

AMATH 250Introduction to Differential Equations

Professor: J. WestUniversity of Waterloo

LaTeXer: M. L. Baker Updated: September 21, 2011

Contents

1 Introduction

1.1 Dimensions (units) of physical quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Principle of dimensional homogeneity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Separable first-order DEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 First-order linear DEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.4.1 General procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 More curve sketching (1.2.4), Undetermined coefficients (1.2.5) . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.5.1 General procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Undetermined coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

These notes are currently a work in progress, and as such may be incomplete or contain errors.

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Administrative

Assignments due every Tuesday, Box #6 (slot 8-12) (last name)

Office hours: MWF 9:50 - 10:10, 11:30 - 12:00; F 14:45 - 16:00.

1 Introduction

Definition 1.1. A differential equation is an equation involving an unknown function and its derivative(s).

Example 1.2.dydx = y(x).

Solution. Some solutions are y = 0, y = ex, y = 2ex. Hence the general solution is y = Cex.

Example 1.3 (Skydiver DE). A skydiver with downward velocity v(t), where we use the convention that downward is positivThe forces are gravity (Fg = mg where m is mass in kg, and g is the acceleration due to gravity, that is 9.8 metres per seconsquared).

Suppose air drag Fd is proportional to the velocity, so Fd = v. Newtons second law therefore says

d

dt(mv) = mg v.

Assuming the mass is constant, this gives the skydiver DE:

m dvdt

= mg v.

Solution. Find v(t) given m, g, . We could divide by m:

dv

dt= g

mv.

The naive approach is to integrate to get

v(t) =

g

mv

dt

but v is an unknown function. So lets look at some qualitative solutions:

1.1 Dimensions (units) of physical quantitiesBasic units: M (mass), L (length), T (time). Others include current, temperature , and so on.

Example 1.4. We use square-bracket notation to denote the dimension of a quantity:

[v] =L

T, [acceleration] =

dv

dt

=

L

T2, [force] = [ma] = M

L

T2= M LT2, [energy] = [mgh] = ML2T2.

1.2 Principle of dimensional homogeneity

Can only add or subtract quantities with the same dimensions.

3 kg + 4m

sis meaningless

but3 kg + 4 g = 3.004 kg.

Example 1.5. Find [] in the Skydiver DE. We can set

[mg] = [v]

ML

T2= []

L

T

[] =M

T.

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Example 1.6. What is [] if et appears somewhere? (t is time). Well, look at the Taylor expansion

et = 1 + (t) +(t)2

2+ . . .

hence obtain

[t] = 1 [] =1

T.

1.3 Separable first-order DEs

Standard form: dy

dx= f(x, y).

Separable DE:dy

dx= A(x)B(y).

Example 1.7. We have the following rules:

dydx = x cos y is separable.

dydx

y = yex is separable.

The Skydiver DE is separable.

dxdt = t + x is not separable.

Example 1.8. How to solve a separable DE: we divide by B(y) to obtain

1

B(y)

dy

dxdx = A(x)

Integrate w.r.t. x: 1

B(y)

dy

dxdx =

A(x) dx

Change of variables/substitution rule gives 1

B(y)dy =

A(x) dx

Then solve for y. Check where B(y) = 0 separately.

Example 1.9. For dydx = 2xy, if y = 0, the DE is satisfied. This is called an equilibrium (constant) solution. If y = 0, thendy

y=

2x dx

which givesln |y| = x2 + C

and hence|y| = ecex

2

therefore y = kex2

where k = 0. This can be checked just by substituting back into the DE.

Example 1.10. dydx = y(y 2). The equilibrium solutions are y = 0 and y = 2. Otherwise,1

y(y 2)dy =

dx

therefore 1

2

y+

12

y 2

dy = x + C

and so

1

2ln |y| +

1

2ln |y 2| = x + C

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or1

2ln

y 2y = x + C

thusy 2

y= ke2x.

Solving,

y =2

1 + ke2x.

Hence in conclusion the solutions are

y = 0 and y = 21 + ke2x

, k R.

Example 1.11 (Newtons Law of Cooling). Consider the DE

dT

dt= k(T TA)

where k is a proportionality constant, T is temperature of the object and TA is the (constant) ambient temperature. We haan initial condition

T(0) = T0.

Solve:

dT

T TA= k dt

and getln |T TA| = kt + C1

henceT TA = C2e

kt

Setting t = 0, we see thatC2 = T0 TA.

Therefore,T(t) = TA + (T0 TA)e

kt.

1.4 First-order linear DEs

General form is dy

dx+ k(x)y = f(x).

Example 1.12.dy

dx+

1

xy =

1

xcos y, x > 0.

We get

xdy

dx+ y = cos x

and henced

dx(xy) = cos x = xy = sin x + C = y =

1

xsin x +

C

x.

1.4.1 General procedure

First multiply the general form equation by I(x) (integrating factor). We get

Idy

dx+ I k(x)y = I f(x)

We would like this to becomed

dx(I y) = I f(x).

i.e.

Idy

dx+

dI

dxy = If(x)

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The first and third equations are equivalent ifdI

dx= I k(x).

But this is a separable DE. HencedI

I=

k(x) dx = ln |I| =

k(x) dx = I(x) = Ce

Rk(x) dx.

We can set C = 1 for simplicity. Hence the formula for the integrating factor is

I(x) = e

Rk(x) dx

.Then the DE becomes

d

dx(I(x)y) = I(x)f(x).

Then integrate and solve for y.

Example 1.13. Considerdy

dx3k(x)

y = 2exf(x)

.

The integrating factor is eR3 dx = e3x. The DE becomes

e

3x dy

dx 3e

3x

y = 2e

x

e

3x

.

This becomesd

dx

e3xy

= 2e2x.

Finally, integrate w.r.t. x to get

e3xy =

2e2x dx = ex + C.

Isolating for y, we gety = ex + Ce3x.

Example 1.14. Consider

xdy

dx

= 3y + x5ex, x > 0.

Dividing through by x and rearranging givesdy

dx

3

xy = x4ex

which is in standard form. Integrating factor is

I(x) = eR

3

xdx = e3 lnx = x3.

Multiply the DE by x3:

x3dy

dx 3x4y = xex.

This becomesd

dx x3y = xex.

(Check). Integrate and get

x3y =

xex dx = xex

(1)(ex) dx = xex ex + C.

Hencey = x4ex x3ex + Cx3.

Check that LHS=RHS in the DE.

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Example 1.15 (Skydiver DE). Consider the (separable) DE

mdv

dt= mg v.

It is also linear:dv

dt+

mv = g.

Integrating factor is

exp

mdt .

We get

e(/m)tdv

dt+

me(/m)tv = ge(/m)t.

Becomesd

dt

e(/m)tv

= ge(/m)t.

This gives

e(/m)tv =mg

e(/m)t + C

hencev(t) =

mg

+ Ce(/m)t.

We perform the dimensional check: mg

=

M LT2MT

=L

T= velocity = [v].

Also m

t

=MT

MT = 1.

1.5 More curve sketching (1.2.4), Undetermined coefficients (1.2.5)

1.5.1 General procedure

1. Equilibrium solutions

2. Exceptional solutions* (C = 0 usually)

3. Slope

4. Asymptotics*

5. Concavity

*: if solution known.

Example 1.16. dydx = 3y ex. Solve and sketch. Well, we get

dy

dx 3y = ex

which is linear. Solve (exercise) to get y = Ce3x

+

1

2ex

. To sketch:

Equilibrium solutions: set dydx = 0. Then 3y ex = 0 hence y = 13e

x, which is not constant. So there are no equilibriusolutions.

Exceptional solutions: C = 0 yields y = 12ex. Notice

u =1

2ex + Ce3x

> 1

2ex if C > 0

< 12ex if C < 0

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Slope:

dy

dx= 3(y

1

3ex)

> 0 if y > 13ex

= 0 if y = 13ex

< 0 if y < 13ex

Asymptotics:

1

2ex + Ce3x

Ce3x as x +

12

ex as x

Figure 1: See notebook 2011.09.21.1

Remark 1.17. Theoretical remark. First order DE: dydx = f(x, y). If f is C1 then unique solution through each point.

Example 1.18. Pathological examples:

dydx = y/x. This is solved by y = Cx (x = 0). [Diagram 2011.09.21.2]

dydx = 3y2/3 implies y = (x + c)3 or y = 0. [Diagram 2011.09.21.3]

1.6 Undetermined coefficients

Consider constant-coefficient linear DE:dy

dx+ ky = f(x).

Example 1.19. Find a particular solution tody

dx+ 3y = x2 + 1.

Trial function: y = Ax2 + Bx + C. Plug into DE:

(2Ax + B) + 3(Ax2 + Bx + C) = x2 + 1.

We get3Ax2 + (2A + 3B)x + (B + 3C) = x2 + 1.

Equate coefficients to get

3A = 1

2A + 3B = 0

B + 3C = 1

A = 13 , B =29 , C =

1127 (exercise). So a particular solution

yp =1

3x2

2

9x +

11

27.

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To get the general solution, consider the homogeneous DE: (RHS = 0)

dy

dx+ 3y = 0 = yh = Ce

3x.

The above is called the homogeneous solution. Turns out that the general solution is obtained by

y(x) = yh(x) + yp(x).

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