A differential equation is called linear if it can be written in the form
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The order of a differential equation is the order of the highest derivative appearing in the equation. A differential equation is called linear if it can be written in the form The coefficients a j (x) and b(x) can be arbitrary functions of x, but a linear equation cannot have terms such as y 3 , yy , or siny. 1 1 1 0 ' n n n n a xy a xy a xy a xy bx '
description
The order of a differential equation is the order of the highest derivative appearing in the equation. A differential equation is called linear if it can be written in the form. - PowerPoint PPT Presentation
Transcript of A differential equation is called linear if it can be written in the form
The order of a differential equation is the order of the highest derivative appearing in the equation.
A differential equation is called linear if it can be written in the form
11 1 0'n n
n na x y a x y a x y a x y b x
The coefficients aj(x) and b(x) can be arbitrary functions of x, but
a linear equation cannot have terms such as y3, yy , or siny. '
43. Solve 20 0.ydyt e
dt
Unit 10 Review
5
4
4
5
20
20
4 0
ln 4
y
y
y
e dy t dt
e dy t d
y
e t
C
C
t
t
2
1/22
2
2
2
12 4
1 4' 12 4 8
2 12 44
' 4 0 12 4 4 012 4
y x x
xy x x
xx
yy x x xx
24. ' 4 0, 12 4yy x y x
Verify that the given function is a solution of the differential equation.
5. " 2 ' 5 0, sin 2
' sin 2 2 cos 2
" sin 2 2 cos 2 2 cos 2 2 sin 2
3 sin 2 4 cos 2
x
x x
x x x x
x x
y y y y e x
y e x e x
y e x e x e x e x
e x e x
Verify that the given function is a solution of the differential equation.
" 2 ' 5
3 sin 2 4 cos 2 2 sin 2 2 cos 2 5 sin 2
3 2 5 sin 2 4 4 cos 2 0
x x x x x
x x
y y y
e x e x e x e x e x
e x e x
V
Use separation of variables to find the general solution.
26. ' 0y x y
3 3
2 2
33 3ln
3
xC
x
dy dyx y x dx
dx y
y Cex
y C y e
Use separation of variables to find the general solution.
3 27. ' 4 0t y y
3 2
2 2
2 3
1
2 /
4 0
4
1 4
2
dyt ydt
y dy t dt
Cy t
yt C
Use separation of variables to find the general solution.
8. 8dy
ydt
1/2
1/2 1
2
/2
8
4
2 8 4
y dy dt
y t
y t
t
C
C y C
Solve the initial value problem.
2 39. , 1 0dy
y x ydx
2 3
33
2
32
2
32
3
1 3
3 2 2
3
2
3 30
3
2 22
3
2
y dy x dx
yC y C
x x
y Cx
C C yx
10. Below is the slope field for sin sin . Sketch the graphs of the
solutions with initial conditions 0 1. Show that 0 is
a solution and add its graph to the plot.
y y t
y y t
0 ' 0y t y
sin sin sin 0sin 0
0 is a solution of sin sin .
y y t t
y t y y t
11. Consider the differential equation .y t y
(a) Sketch the slope field in 1 3, 1 3.t y
we
Th
h
e is
ave
ocline o
slope
f slope is
on the l ne
i s .C y
C t
t C
y C
1,3
3, 10C
1C
2C
3C 1C
2C
3C
t
y
11. Consider the differential equation .y t y
(b) Show that 1 is a solution for all . Since the lim 0,
these solutions approach the particular solution 1 as . Explain
how this behavior is reflected in your sl
t t
ty t Ce C e
y t t
ope field.
1 1 and
1 1
t t
t t
y t Ce y Ce
y t y t t Ce Ce QED
0 0 0 0
0 0 0 0
0 0
If , is above 1, the particular solution through , 1
as .
If , is below 1, the particular solution through , 1
as .
If , is a point on 1, the par
P t y y t P t y y t
t
P t y y t P t y y t
t
P t y y t
0 0ticular solution through ,
is the line 1.
P t y
y t
`
12. Let be the solution to satisfying 0 0.
(a) Use Euler's Method with 0.1 to approximate 0.5 .
yy t y te y
h y
n nt
0
1
2
3
4
5
0
0.1
0.2
0.3
0.4
0.5
0
00 0 0, 0 0.1 0 0y hF t y e
01 1 1, 0 0.1 0.1 0.01y hF t y e
0.012 2 2, 0.01 0.1 0.2 0.0298y hF t y e
0.02983 3 3, 0.0298 0.1 0.3 0.0589y hF t y e
0.05894 4 4, 0.0589 0.1 0.09660.4y hF t y e
(b) Use separation of variables to find 0.5 exactly.y
2 2 2
ln , 0 0 1 ln 12 2 2
y y
y
dyte e dy tdt
dt
t t te C y C y C y
0.5 0.118y
1 1 1,n n n ny hF t y y
13. Use Euler's Method to approximate 2.95 ,
given , 2.7 5, and 0.05.
y
y t y y h
n nt
0
1
2
3
4
5
1 1 1,n n n ny hF t y y
2.7
2.75
2.8
2.85
2.9
2.95
5
0 0 0, 5 0.05 2.7 5 5.1387y hF t y
1 1 1, 5.1387 0.05 2.75 5.1387 5.2791y hF t y
2 2 2, 5.2791 0.05 2.8 5.2791 5.4212y hF t y
3 3 3, 5.4212 0.05 2.85 5.4212 5.5650y hF t y
4 4 4, 5.565 0.05 2.9 5.56 5.7105y hF t y
6
6
2 3 6 1 6 & 33
3The general solution is
1 /10 10
10 3 7
3
1 7 /10t
t
yy y y A
ye
y
k
C
Ce
0
0
1 & 1 /kt
ydy y Aky y C
dt A e C y A
14. Find the solution of 2 3 , 0 10, if grows according
to the logistics equation.
y y y y y t
0.3
0.3 0.3 0.
0.3
30.3
500' 0
500
1 9
5.973 day
.3 1500 1 /
50 1
50 500 9500 5 3 1
200 1 9 91 9 2 2 6
0.3 l 1/ 6 sn
t
t
t t tt
P tP t P t P t
e C
C
e e ee
P te
tt
1
0
15. A population grows according to the logistics equation with growth
constant 0.3 day and carrying capacity 500.
(a) Find assuming 50.
(b) After how many days will
P t
k A
P t P
P t
reach 200?
0
0
1 & 1 /kt
ydy y Aky y C
dt A e C y A
