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Class notes MATH 2080 (MATH2080)

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Provides a general overview of solving methods for differential equations with examples. Examples include all work and the answers.

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  • December 28, 2024
  • 21
  • 2024/2025
  • Class notes
  • Leo rebolz
  • All classes
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williams04mackenzie
. 4-motion
3 of a falling body

ef e')
+




Gmm
& u = =


- bu




= C

with-
v(0)

endup
+
M =
vo



mg
W




distance traveled :

x(t) X(0) = 0
,




set origin as release height
x( +) = fv()d +




3 Ns/m
example. 3
kg 500m b =
m= Xo =




Vo 0 m/s 9 814/s2
g
= = -
.




3 =
39 -
3v u = e = et x( +) = fg() -

e
-



)d+
+




g(1 et)
"




5
+ v =
g v(t) =
-


x( +) =
g + +
ge +C


v(0) = 0 0 =
g(0 +
e) +

C= g -




C = -
9 81.




500 =
g
+ +
ge -




g
++ -

500
+ ~
from
Hab
+ = 51 .
97

, Chapter 4 -
2nd order ODEs


a y" + by +
cy = 0
·


a, b
, c are constants
O right hand side is
"homogenous and order" system
·
will consider flt) on right as well

linearly independence
definition -
two functions y , (t) + yalt) are linedly
independent if one function is not a


constant multiple of the other



y , ( + ) E Cyz( +) for all +




e . Sinc + cos(t) are linearly independent
ex - eat-eat are linearly independent


to verify-calculate Wranghian


WClyvel] = Y, Ye =
Y , Ye'- yey : O
Y : 'Ye'

* if = O , linearly dep.

ex .
ritra , yalth = et
, yaltlagat are


linearly independant

W2y .. y2] :
e
+
(rie2 )-ea (re")
+ +



(2) +
N2e(
(2) +
e(
+ +
= -
r,

editrel (re
+
= -
1) = 0
u
20

, (a = 0)
Solving ay" + by +
cy = 0


suppose y(t) : e't for some constant
plug y' wert ree'
+
, y"
:
in y =
,




=> 2e + bre
+
"
a - + + ce = 0


Il factor eut
O
neuer :




-e +
(ara + br + c) = 0


2
a + br + c = 0 -
characteristic equation


2) solve quadratic


3 cases of intrest
·
2 real roots (ritra)
I repeated roo+ (vi)
2 complex roots (r Fra) .




Superposition principle
for and order ODE, 2 linearly independent solutions
Y , (t) + ya(t) yield a
general solutions


( +) C,
y
=
y ,
+
(2ye(t) for all ( , ki


·

we can verify this as a solutio
a [(yi + (2y2]" + b(xy +
+
ceyz] + c(( ,
+ (eye]

* a 2(y, + Ce ya)" =
a2Gy ,
"
+ Ceye"]

7 C(ay ,
"
+ by , +
2y , ) + G(ayz" = by + cye)
thus is a solution for in
y any 1111

* some can be done for 3rd order ODE's

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