Example 2: Consider the IVPdydx

= y x

y 0 = 0 a) Check that y x = x 1 C ex gives a family of solutions to the DE C=const). Notice that we haven't yet discussed a method to derive these solutions, but we can certainly check whether they work or not. (It's a first order linear DE, and you may have learned how to solve these in Calculus. We'll learn/review soon.)b) Solve the IVP by choosing appropriate C .c) Sketch the solution by hand, for the rectangle 3 x 3, 3 y 3 . Also sketch typical solutions for several different C-values. Notice that this gives you an idea of what the slope field looks like. How would you attempt to sketch the slope field by hand, if you didn't know the general solutions to the DE? What are the isoclines in this case?d) Compare your work in (c) with the picture created by dfield on the next page.

not separable

Wechecked a onWed

b If ylxixtltcexy.coO Itc G z Yk Ith e

y exy y Xt Y'do O

r y O X so

y 107 0I

x togetthey wordsof 9 x for the graphof a

ssgraphs

y ex

y y x solutionsy xtltCeCIO y xt1

Eo

You canuse slopefieldto draw silk 1graphs

OR I ccO

t TIfyonrepresentativesalt graphsyou canseethestonefield

Example 3a) Use separation of variables to solve the IVP

dydx

= y23

y 0 = 03b) But there are actually a lot more solutions to this IVP! (Solutions which don't arise from the separation of variables algorithm are called singular solutions.) Once we find these solutions, we can figure out why separation of variables missed them.3c) Sketch some of these singular solutions onto the slope field below.

Came back after existence uniqueness thenflag y 5 is cont soetns to Ivp existffy zj's f fry

is het along 5 o Iii p Exercise34 y.CN x fromsepvous

sqmhgigffhlfsa.sn yah O alsosolvesIVP

forseparableDE'sdd fix y3b It's evenworse If yo is any rootof0 017 0Sep of Ms Soltis were the guy yo solvestheDE

y Xtc LHS y 1 7 0134 O 2E x E3 2,3 mychoice RHS flag yo o

LH RHS

241 1273 Separationof variablesXE 2 N'lymany missesthese becauseI Lx 373 X 3 So11ns to IVP

with gio ody fladx OG

Tix f3

tryin o

ly many salts to this Np

Here's what's going on (stated in 1.3 page 22 of text as Theorem 1; partly proven in Appendix A.)Existence - uniqueness theorem for the initial value problemConsider the IVP

dydx

= f x, y

y a = b Let the point a, b be interior to a coordinate rectangle ℛ : a1 x a2, b1 y b2 in the x y

plane. Existence: If f x, y is continuous in ℛ (i.e. if two points in ℛ are close enough, then the values of f

at those two points are as close as we want). Then there exists a solution to the IVP, defined on some subinterval J a1, a2 .

Uniqueness: If the partial derivative function y

f x, y is also continuous in ℛ, then for any

subinterval a J0 J of x values for which the graph y = y x lies in the rectangle, the solution is unique!

See figure below. The intuition for existence is that if the slope field f x, y is continuous, one can follow it from the initial point to reconstruct the graph. The condition on the y-partial derivative of f x, y turns out to prevent multiple graphs from being able to peel off.

f is cont at x y means that if x yis closeenoughto ex y then fCxy is asclose

as wewantto fix y

a b

solution is unique in this interval

Example 4: Discuss how the existence-uniqueness theorem is consistent with our work in Exercises 1-3 in today's notes, and the exercise we finished from Tuesday's notes, at the start of class today.

Existence - uniqueness theorem for the initial value problemConsider the IVP

dydx

= f x, y

y a = b Let the point a, b be interior to a coordinate rectangle ℛ : a1 x a2, b1 y b2 in the x y

plane. Existence: If f x, y is continuous in ℛ (i.e. if two points in ℛ are close enough, then the values of f

at those two points are as close as we want). Then there exists a solution to the IVP, defined on some subinterval J a1, a2 .

Uniqueness: If the partial derivative function y

f x, y is also continuous in ℛ, then for any

subinterval a J0 J of x values for which the graph y = y x lies in the rectangle, the solution is unique!

Tuesday IVP:

IVP y x =

11 y2

y 0 = 0

E1)

IVP y x = x 3

y 1 = 2

E2)

IVP y x = y x

y 0 = 0

E3)

IVP y x = y23

y 0 = 0

y faux497 is continon

oh1122

T Efg CDuty522gSo exactly is alsooneSoltionq cont onRZ

fine Tk Cxsmh

f xyl y x is wet on1122OfFy I also wut

y xxl ex is onlyNP silk

f xg 855 is contmiRso solutions exist

qfy gb 3 iseIietpt

anyrectangle Ralongthe x axis

containing 10,07 containspart of the x axisin its interior so uniqueness cannotbeguaranteed

Math 2280-002Fri Jan 101.3-1.4 Slope fields and existence and uniqueness for solutions to IVPs; using separable differential equations for examples, continued.

Announcements: We'll be completing Wednesday's notes, with a few more examples in today's notes. Since I ask you to use the mathematical software "dfield" in one of your homework problems, I plan to demo it today. You can download the Java app "dfield.jar"to your own laptop from thewebsite

https://math.rice.edu/~dfield/dfpp.html

The computers in our Math Center lab have cloned versions of this app, which you can run from a command window by typing

dfieldafter the prompt in a command window, and then hitting "return".

Warm-up Exercise:

quiz score is on back side ofquiz

start lecture 1 00Find a solution to I knowyou can doSep of variables but

upY'txt y we'll study this DE

for a largeportion ofyl07 0 today's lecture

y 43 Strangethings happen

checkyet 3

2

42 LHS Talk to DE

jb g35 14 12 4 2 RHSthen soltny 01 0

Derivation of solths i 924 0

yi 1 7 0 LHSord2 y ya o R'its

J qµdxY40 this couldbe a problemfor our Ivp

3g xtc

y3

y txt CP y61 0 C o Igf243

Name________________________Student I.D.___________________Math 2280-002 Quiz 1 January 8, 2020

1) Consider the initial value problem for a function y x :y x = x y2

y 0 = 1 .

a) The differential equation in this problem is separable. Solve the initial value problem.(7 points)

b) Sketch the graph y = y x of the solution to the initial value problem above, onto the slopefield . What is the largest interval containing x0 = 0 on which the solution is defined as a differentiable function?

(3 points)

Solutions

dyt xdxly.toJdy7 fxdxty ItC

Tco I l Otc

Ly Iz i

y 22 2

y2 1 I2 l I

i tii l

graphhasvertical I

asymptotes cif

Il Iiii

iI iii 10,1iii ia Hh

rzL r2 because yCx7as Vzand as x D iz