Assigned problems for MAA134 Differential Equations

Transcription

Assigned problems for
MAA134 Differential Equations
at the course occasion in period 3
Academic Year 2014/15
Version 2015-06-16
Block 1 – ODE of first order
1.
Introduction to differential equations
..............................
ZC 1.1–1.3
Determine the order of the differential equation and whether it is linear or not (with respect
to y).
√ 2 2
1) (5x2 − 2)y 000 − 2xy 0 + y = x4
8) 3y 00 = 7yy 0
15) y (4) e− x +y + 2y 0 = y
INL 1.1
2) y (4) y = x3 y
3) x4 y (3) + x7 y = 0
2 00
0
4) x y + 5xyy + y = sin(x)
0
x 3
5) yy = e y
6) cos(x)y 00 + x2 y 0 + 3y = 0
7) x2 y (5) − ln(x)y 000 + xy = 0
2.
9) ex y (4) + tan(x)y 00 + x3 y = 1
p
10) xy 0 + y 00 /x = x2 + 1
√
11) xy (3) = x5 y 0 + x2 y
0
12) y − 4yy
000
=0
√
13) y 00 /(1 + x2 ) = y x cot(x)
p
14) 7y 0 + 2 1 + y 2 = 0
17) 5y = y (5) y 00
√
18) x7 xy (4) + cos(x)y 00 = y
19) e9x y 00 + sin(x)y 0 + xy = 0
20) y 000 y 0 − xy 00 y = 0
First-order ordinary differential equations
INL 2.1
....................
ZC 2.1–2.5
Sketch the phase portrait of the autonomous differential equation and classify the stationary
points. Alse, state the equilibrium solutions and sketch the typical solution curves.
1)
dy
= −(y + 2)y 2 (y − 1)3
dx
8)
dy
= y 3 (y − 2)(y − 7)
dx
2)
dy
= (y + 1)y(y − 1)2
dx
9)
dy
= −(y + 3)2 (y + 1)3 y
dx
3)
dy
dy
= −(y + 3)2 (y + 1)(y − 2)3 10)
= (y + 4)y 2 (y − 2)3
dx
dx
4)
√
√
16) ln(x)y 00 + y/ x = tan( x)y 0
dy
= (y + 2)y 3 (y − 3)3
dx
dy
5)
= −y 2 (y − 2)(y − 4)
dx
11)
15)
dy
= (y + 4)y 3 (y − 1)2
dx
16)
dy
= (y + 2)2 (y + 1)y 2
dx
17)
dy
= (y + 3)3 y(y − 5)2
dx
18)
dy
= y 4 (y − 2)3 (y − 4)2
dx
19)
dy
= (y + 1)4 y(y − 2)3
dx
20)
dy
= (y + 3)2 y 3 (y − 6)3
dx
dy
= −(y + 1)y(y − 3)2
dx
dy
12)
= (y + 1)2 y 2 (y − 1)
dx
6)
dy
= (y + 1)y(y − 4)3
dx
13)
dy
= (y + 2)3 y 2 (y − 4)
dx
7)
dy
= −(y + 5)y 2 (y − 1)
dx
14)
dy
= (y + 1)2 y 4 (y − 2)3
dx
2
2.. FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS
INL 2.2
3
Solve the initial-value problem. Also, determine the interval of existence for the solution.
1) yy 0 = −x ,
6) y 0 = 5y + y 2 ,
y(2) = −3 .
2) y 0 = y 2 − 9 ,
16) y 0 = 2x3 e−2y ,
y(0) = −4 .
√
12) y 0 = y 2 cos(x) ,
17) 2xyy 0 = 1 + y 2 ,
7) y 0 =
y(0) = −1 .
2
11) yy 0 = xe−y ,
p
y(0) = ln(2) .
xy ,
y(1) = 1 .
y(0) = 2 .
y(π/6) = 2/3 .
y(5/3) = −2 .
√
2
3) y 0 = xex −2y ,
√
y( 2) = ln(2e) .
8) y 0 = 2xey ,
y(2) = − ln(2) .
9) xy 0 + 2y = y 2 ,
4) y 0 = (y − 1)2 ,
y(1) = 3 .
y(−1) = 0 .
5) y 0 = y 2 + 4 ,
√
y(0) = 2 3 .
INL 2.3
10) y 0 ln(y) = xy ,
y(0) = 1/e2 .
13) y 0 = (x + 1) y ,
18) y 0 = x3 y 4 ,
y(1) = 4 .
14) y 0 =
y−2
x+1
y(1) = 2 .
2
,
√
19) y 0 = y 2 x ,
y(4) = 3 .
y(−2) = 3 .
p
15) y 0 = x 1 − y 2 ,
√
y( π) = 1/2 .
3
1) xy 0 + 3y = 3 ,
6) xy 0 − 5y = 2x ,
y(1) = 8 .
y(1) = 5 .
2) y 0 + 3xy = x ,
7) x2 y 0 + 2xy = 1 ,
y(0) = 0 .
y(2) = 1 .
3) y 0 − 9x2 y = 27x2 ,
8) y 0 − 4xy = 2x ,
y(0) = 9 .
y(0) = 1 .
4) x2 y 0 − xy = 3 ,
9) y 0 − 8x3 y = 16x3 ,
y(1) = −1 .
5) y 0 + 6x2 y = x2 ,
y(0) = 3 .
10) x2 y 0 + 5xy = −2 ,
y(0) = 1/3 .
y(1) = 1 .
√
11) xy 0 − 2y = x x ,
√
16) xy 0 + 4y = 2 x ,
y(4) = 80 .
y(1) = −1 .
2
12) y 0 + 6xy = e−3x ,
17) y 0 − 12x3 y = x3 ,
y(0) = −2 .
y(0) = 1/12 .
13) y 0 − 7xy = 14x ,
18) x2 y 0 + 9xy = 8/x4 ,
y(1) = 3 .
y(0) = 5 .
14) y 0 + 15x4 y = −3x4 , 19)
y(0) = 4/5 .
√
xy 0 − y = 1 ,
y(4) = 1 .
√
15) x2 y 0 − 6xy = 16/x , 20) xy 0 + 8y = x2 x ,
y(−1) = 3 .
y(1) = −1 .
INL 2.4
2
20) yy 0 + x2 e2x +y = 0 ,
p
y(0) = ln(3) .
Solve the initial-value problem where the function U is defined by U (x−ξ) =
1) y 0 + 8y = 16 U (x − 2) ,
y(0) = 1 .
2) y 0 + 3y = −9 U (x − 3) ,
y(0) = 1 .
3) y 0 + 9y = 27 U (x − 4) ,
y(0) = 1 .
4) y 0 + 6y = −12 U (x − 5) ,
y(0) = 1 .
5) y 0 + 2y = 6 U (x − 6) ,
y(0) = 1 .
6) y 0 + 5y = −5 U (x − 7) ,
y(0) = 1 .
7) y 0 + 7y = 14 U (x − 8) ,
y(0) = 1 .
8) y 0 + 3y = −6 U (x − 9) ,
y(0) = 1 .
9) y 0 + 10y = 30 U (x − 6) ,
y(0) = 1 .
10) y 0 + πy = 2 U (x − 5) ,
y(0) = 1 .
11) y 0 + 2y = −4 U (x − 4) ,
y(0) = 1 .
12) y 0 + 12y = 24 U (x − 3) ,
y(0) = 1 .
0,
1,
x < ξ,
.
x ≥ ξ.
13) y 0 + 4y = −20 U (x − 7) ,
y(0) = 1 .
14) y 0 + 8y = 24 U (x − 5) ,
y(0) = 1 .
15) y 0 + 5y = −15 U (x − 8) ,
y(0) = 1 .
16) y 0 + 7y = 28 U (x − 9) ,
y(0) = 1 .
17) y 0 + 9y = −18 U (x − 5) ,
y(0) = 1 .
18) y 0 + 6y = 42 U (x − 4) ,
y(0) = 1 .
4
19) y 0 + 13y = −26 U (x − 6) ,
y(0) = 1 .
INL 2.5
20) y 0 + 4y = 12 U (x − 3) ,
y(0) = 1 .
Prove that the differential equation is exact and determine (at least on implicit form) the
general solution.
1) (10xy 3 + 35x4 y 3 ) dx + (15x2 y 2 + 21x5 y 2 ) dy = 0 11) (1 + x)ex y − y 3 dx + (xex − 3xy 2 ) dy = 0
2) (2x/y 3 + y 2 ) dx + (2xy − 3x2 /y 4 ) dy = 0
12) (6x2 y + 10xy 3 ) dx + (2x3 + 15x2 y 2 ) dy = 0
3) (y 5 + 6x2 ln(y)) dx + (5xy 4 + 2x3 /y) dy = 0
13) (6x/y − 2y 3 /x3 ) dx + (3y 2 /x2 − 3x2 /y 2 ) dy = 0
4) (16x3 y 2 − 2xy 3 ) dx + (8x4 y − 3x2 y 2 ) dy = 0
√
√
14) (6y 2 /x4 +12x y) dx+(3x2 / y −4y/x3 ) dy = 0
5) (2xe3y − 9x2 y 2 ) dx + (3x2 e3y − 6x3 y) dy = 0
15) (5x4 y 3 + 5xy 4 ) dx + (3x5 y 2 + 10x2 y 3 ) dy = 0
6) (3x2 y 2 + 6xy 5 ) dx + (2x3 y + 15x2 y 4 ) dy = 0
√
√
16) (6x2 y y−2y 3 /x3 ) dx+(3y 2 /x2 +3x3 y) dy = 0
7) (y 2 /x + 4xy) dx + (2 ln(x)y + 2x2 ) dy = 0
17) (8xy 2 − 15x4 ln(y)) dx + (8x2 y − 3x5 /y) dy = 0
8) (3y 3 − 4x3 y 2 ) dx + (9xy 2 − 2x4 y) dy = 0
18) (y 2 /x − 3y 3 /x2 ) dx + (9y 2 /x + 2y ln(x)) dy = 0
√
√
9) (6x3 + y 3 + 2x ln(y)) dx + (3xy 2 + x2 /y) dy = 0 19) (6x2 y − y 5 ) dx + (x3 / y − 5xy 4 ) dy = 0
20) (4x3 /y 2 − 2xy 3 ) dx − (2x4 /y 3 + 3x2 y 2 ) dy = 0
10) (2xy 3 − 2y 2 ) dx + (3x2 y 2 − 4xy) dy = 0
INL 2.6
Prove that the differential equation is homogeneous and determine (at least on implicit form)
all solutions.
1) (x + y)y 0 = x − y
8) (y + xey/x )dx = xdy
√
2) (y + 2 xy)dx = xdy
p
3) xyy 0 = y 2 + x 4x2 + y 2
9) xyy 0 = x2 + 3y 2
p
10) (y 2 + x x2 + y 2 )dx = xydy
2
15) x2 + xyy 0 = 2y 2
16) x2 dy = (2xy + y 2 )dx
17) 4x2 + 2xyy 0 = 3y 2
0
2
4) (x3 + y 3 )dx = xy 2 dy
11) 3x + 2y + 4xyy = 0
5) x(x + y)y 0 = y(x − y)
12) y(3x + y)dx + x(x + y)dy = 0
6) (x + 3y)dx = (y − 3x)dy
13) x2 = y 2 + xyy 0
19) y 2 − 3xy − 2x2 = (x2 − xy)y 0
7) (x − y)y 0 = x + y
14) 4xydx + x2 dy = y 2 dx
20) (3xy+y 2 )dx+(x2 +xy)dy = 0
INL 2.7
18) (x2 + 2y 2 )dx = xydy
1) 2xy 0 + y(1 + xy) = 0 , 6) 2xy 0 = 5(1/y − y) ,
y(1) = 1/2 .
2) y 0 = (xy − 1)y ,
y(0) = 1 .
y(−2) = −3 .
7) y 0 = y + xy 1/3 ,
y(0) = −8 .
11) xy 0 = y(y − 1) ,
y(−1) = 1/3 .
12) y 0 = (ex y − 1)y ,
y(−1) = −1 .
3) (x + x2 )y 0 = y(1 − y) , 8) 3y 0 = −4x3 y(1 + 2y 3 )13)
xy 0 = 2y(1 + 2y 2 ) ,
,
y(1) = 2 .
y(1) = 1/4 .
y(0) = 1/2 .
4) y 0 = y + 1/y ,
y(0) = −1 .
9) y 0 = 2xy(1 − y) ,
y(0) = 3 .
5) y 0 = y(2 − y) ,
10) xy 0 + y = x2 y −2 ,
y(0) = 1 .
y(1) = 1 .
16) xy 0 = y + 2xy −3 ,
y(3) = −1 .
17) xy 0 + x3 y 2 = 2y ,
y(−1) = 2 .
18) 2xy 0 + x2 y 3 + 2y = 0 ,
y(−e) = −1/(2e) .
14) y 0 + xy(2 + y 3 ) = 0 , 19) xy 0 + 4y = x4 y 2 ,
y(0) = −1 .
y(1) = 1 .
15) xy 0 = (x4 y − 1)y ,
y(−1) = 6 .
20) xy 0 = 2y + y −1 ,
y(1) = −2 .
3.. MODELING WITH FIRST-ORDER DE
INL 2.8
5
The differential equation has an integrating factor which only depends on x (odd-numbered
problems) or y (even-numbered problems). Find the equation for the solution curve which
contains the point (1, 2) .
1) (12x + y 3 /x) dx + 3y 2 dy = 0
11) (x + 3y 2 ) dx + 2xy dy = 0
2) 10x/y dx + (5x2 /y 2 − 6/y) dy = 0
12) 2xy dx + (3x2 + 2/y) dy = 0
3) (4y 3 /x2 + 20x) dx + 12y 2 /x dy = 0
13) (6xy + 14y 3 ) dx + (2x2 + 21xy 2 ) dy = 0
4) 9x2 y dx + (6x3 + 16y 2 ) dy = 0
14) (8x3 y + 3x2 y 3 ) dx + (6x4 + 5x3 y 2 ) dy = 0
5) (16y 3 /x − 6/x) dx + 24y 2 dy = 0
15) (3y 4 + 15x2 y 2 ) dx + (4xy 3 + 6x3 y) dy = 0
6) 7y 3 dx + (28xy 2 − 10) dy = 0
16) (6xy 2 − 4x3 y) dx + (9x2 y − 2x4 ) dy = 0
7) (45x3 y 2 + 16x2 ) dx + 18x4 y dy = 0
17) (28x2 y + 24xy 2 ) dx + (7x3 + 16x2 y) dy = 0
8) 18x2 y dx + (18x3 − 12y 3 ) dy = 0
18) (18x5 y − 10xy 2 ) dx + (9x6 − 20x2 y) dy = 0
9) (4y 7 − 30x4 ) dx + 14xy 6 dy = 0
19) (27y 2 + 20x2 y 3 ) dx + (18xy + 12x3 y 2 ) dy = 0
10) 10xy 3 dx + (20x2 y 2 + 16y 2 ) dy = 0
3.
20) (8xy 4 − 5y 2 ) dx + (20x2 y 3 − 15xy) dy = 0
Modeling with first-order DE
INL 3.1
..............................................
ZC 3.1–3.3
Modeling with first-order linear differential equations
1) A rapidly falled ill person with a high fever
requires immediate diagnosis. One has access to only a relatively slow thermometer
and therefore need to be able to calculate
the fever long before the thermometer reads
the temperature.
It is assumed that the
rate at which the temperature read by the
thermometer changes per unit time, at any
time is proportional to the difference between
temperature read by the thermometer and the
temperature of the medium which surrounds
the measuring point. A reasonable assumption
is also that the smallness of the thermometer
and the corresponding negligible heat content
not appreciably disturb the temperature in the
body of the ill person.
The thermometer is normally kept at room
temperature (20 degrees). Three (3) seconds
after the thermometer has been inserted into
the mouth of the patient it reads 30 degrees.
Another three seconds later, the thermocouple
meter reads 35 degrees. Using the collected
data, make a quick calculation of the person’s
temperature. What is the conclusion, i.e. how
high fever has the ill person?
2) A radioactive material decays at a rate proportional to the amount present at time t. Assume
that the half-life is 50 years, and that at a given
time there is 100 mg of the material. How long
ago was it 10 g of the material?
3) In a simplified model (No. 1) by how much a
student learns from a particular course material
it is assumed that the learning per unit time
at a given time is proportional (α) to the
difference of how much it is possible to learn
(the total amount of course material M ) and
how much has been learned up to that time.
In a somewhat more realistic model (No. 2),
it is also taken into account that a student has
time to forget parts of the previously learned,
and then at a rate which is proportional (β) to
the student’s knowledge at the time. Formulate
and solve the differential equations for the two
models. In so doing, assume that the student’s
knowledge is zero at the time 0. Compare the
solutions for very large t. Furthermore, describe
the meaning of the special cases α β (α much
greater than β) and α β.
4) Temperature changes in a thermometer are
assumed to be described by Newton’s cooling/warming law. A feverish person has with
the help of such a thermometer measured its
own body temperature to be 38o C. What does
the thermometer read three minutes after the
fever measurement if it shows 22o C two minutes
after the fever measurement and if the room
temperature is 20o C? It is considered reasonable
to suppose that the thermometer’s small extent
not substantially disturb the temperature of the
surrounding medium (the person’s body and
6
the air respectively).
9) In an isolated population where nobody dies,
individuals fall ill with a mild disease at a rate
which in each moment is proportional to the
number of healthy individuals. Let the constant
of proportionality be equal to 10−8 s−1 , the
number of individuals in the entire population
equal to 50 000, and the number of diseased at
the time t0 equal to 30. Find the rate at which
people become ill as a function of time t. Also,
determine the number of diseased individuals
for large times.
5) In an isolated population where nobody dies,
individuals fall ill with a mild disease at a rate
which in each moment is proportional to the
number of healthy individuals. Let the constant
of proportionality be equal to 10−7 s−1 , the
number of individuals in the entire population
equal to 100 000, and the number of diseased at
the time t0 equal to 20. Find the rate at which
people become ill as a function of time t. Also,
determine the number of diseased individuals
for large times.
10) In a simplified model (No. 1) by how much
satiation a rabbit gets by a certain amount of
6) A rapidly falled ill person with a high fever
food, it is assumed that the change of the satiety
requires immediate diagnosis. One has acper unit time at a given time is proportional
cess to only a relatively slow thermometer
(α) to the difference of how great satiation
and therefore need to be able to calculate
that is possible to get (maximum value S)
the fever long before the thermometer reads
and the amount of satiation that has been
the temperature.
It is assumed that the
achieved up to that time. In a somewhat more
rate at which the temperature read by the
realistic model (No. 2), it is also taken into
thermometer changes per unit time, at any
account that the rabbit’s hunger all the time
time is proportional to the difference between
”depletes” the current satiety, and then at a
temperature read by the thermometer and the
rate that is proportional (β) to the rabbit’s
temperature of the medium which surrounds
satiety at the time. Formulate and solve the
the measuring point. A reasonable assumption
differential equations for the two models. In so
is also that the smallness of the thermometer
doing, assume the rabbit’s satiety is zero at the
time 0. Compare the solutions for very large
not appreciably disturb the temperature in the
t. Furthermore, describe the meaning of the
body of the ill person.
special cases α β (α much greater than β)
and α β.
The thermometer is normally kept in a heated
state of 30 degrees. Five (5) seconds after the
thermometer has been inserted into the mouth 11) The water in a small lake is constant in volume
(Vlake ), but is translated by two streams with a
of the patient it reads 35 degrees. Another
flow on each u0 /2 (volume per unit time) into
five seconds later, the thermocouple meter reads
the lake, and by a third stream with a flow u0
37,6 degrees. Using the collected data, make a
bringing water from the lake. One of the two
quick calculation of the person’s temperature.
streams which represent the in-flow is from time
What is the conclusion, i.e. how high fever has
0 polluted with the impurity content η (amount
the ill person?
of impurity per unit volume). Assume that the
7) A radioactive material decays at a rate proporimpurity diffusion speed in the lake is much
tional to the amount present at time t. Assume
greater than the average rate at which water
that the half-life is 15 years, and that at a given
flows in (through) the lake. How much pollution
time there is 10 g of the material. How long will
is there in the lake at the time t, and how large
it be before there is only 0.7 g remaining of the
is the amount pollution after a long period of
material?
time?
8) Temperature changes in a thermometer are 12) The number of bacteria in a bacterial culture is
assumed to be described by Newton’s coolincreasing at a rate that is proportional to the
ing/warming law. A feverish person has with
number at the time t. At a certain moment,
the help of such a thermometer measured its
there are 5 000 bacteria while an hour before
own body temperature to be 39o C. What does
there was 2 000. How many will there be two
the thermometer read four minutes after the
hours after the first time?
fever measurement if it shows 25o C one minute
after the fever measurement and if the room 13) Temperature changes in a thermometer are
temperature is 19o C? It is considered reasonable
assumed to be described by Newton’s coolto suppose that the thermometer’s small extent
ing/warming law. A feverish person has with
the help of such a thermometer measured
not substantially disturb the temperature of the
its own body temperature to be 40o C. What
surrounding medium (the person’s body and
does the thermometer read five minutes after
the air respectively).
the fever measurement if it shows 25o C three
7
minutes after the fever measurement and if the
person’s body and the air respectively).
room temperature is 21o C? It is considered
reasonable to suppose that the thermometer’s 14) A radioactive material decays at a rate proportional to the amount present at time t. Assume
small extent and the corresponding negligible
that
the half-life is 30 years, and that at a given
heat content not substantially disturb the
time
there is 0.3 mg of the material. How long
temperature of the surrounding medium (the
ago was it 5 mg of the material?
INL 3.2
Modeling with first-order non-linear differential equations
1) At time 0, there are 50 grams of chemical A
and 80 grams of chemical B. The chemicals
are combined whereby they react together in
the proportions 1 : 2 forming the chemical C,
i.e. for each 3 grams of the final product C,
it will be used 1 gram of substance A and 2
grams of substance B. The rate of the reaction
is assumed to be proportional to the product of
the remaining amounts of the chemicals A and
B not converted to C. It is also observed that 60
grams of chemical C is formed in 1/30 minutes.
How much is formed in 1/15 minutes?
2) At time 0, there are 100 grams of substance A.
The substance disintegrates into two substances
B and C, formed in a ratio of 5:3, i.e. for
each decayed gram of original substance A it
is created 5/8 gram of subject B and 3/8 gram
of substance C. The rate of the disintegration is
assumed to be proportional to the square of the
remaining amount of substance A, and the time
until the amount of original substance has been
halved equals 10 minutes. How many grams
are there of substance B at time t (in minutes
counted) if the grams at time 0 equals zero.
it will be used 3 grams of substance A and 1
gram of substance B. The rate of the reaction
4) At time 0, there are 50 grams of substance A.
each decayed grams of original substance A it
is created 2/7 gram of subject B and 5/7 gram
are there of substance C at time t (in minutes
6) At time 0, there are 90 gram of substance A.
each decayed grams of original substance A it is
created 4/7 grams of subject B and 3/7 grams
gram of substance B. The rate of the reaction
8
it will be used 1 gram of substance A and 3
i.e. for each 7 grams of the final product C, 14) At time 0, there are 40 gram of substance A.
created 2/3 grams of subject B and 1/3 grams 15) At time 0, there are 70 grams of chemical A
of substance C. The rate of the disintegration
is assumed to be proportional to the square of
the remaining amount of substance A, and the
time until the amount of original substance has
been halved equals 8 minutes. How many grams
the proportions 5 : 2 forming the chemical C, 16) At time 0, there are 55 gram of substance A.
each decayed grams of original substance A it is 17) At time 0, there are 65 grams of chemical A
the proportions 1 : 3 forming the chemical C, 18) At time 0, there are 60 gram of substance A.
9
until the amount of original substance has been 20) At time 0, there are 45 gram of substance A.
INL 3.3
Modeling with systems of first-order differential equations
Write a system of differential equations which at the time t describes how the amount
of salt in a liquid varies in three containers A, B and C respectively. Also, state the
corresponding initial values. It is assumed that the rates by which salt and liquid become
well-mixed within the different containers are much greater than the flow-rates by which
liquids are streaming through the system of containers.
1) Three different tanks A, B and C are with tubes
interconnected so that liquid in the system flows
from A to B, from B to C and from C to A.
It also flows liquid from the environment into
A, and from B to the environment. At the
time 0 there are 200 liters of liquid in each
of the containers, of which that in B is evenly
mixed with 2 grams of salt. The liquid flowing
into A has the flow rate φin = 4 liter/min and
salinity ρin = 1 gram/liter. The other flow rates
are φA→B = 2 liter/min, φB→C = 1 liter/min,
φC→A = 3 liter/min and φut = 2 liter/min.
from A to B, from B to C and from C to B.
A, and from C to the environment. At the
of the containers, of which that in A is evenly
φC→B = 1 liter/min and φut = 4 liter/min.
from A to B, from B to C and from B to A.
B, and from C to the environment. At the
into B has the flow rate φin = 4 liter/min and
φB→A = 2 liter/min and φut = 3 liter/min.
10
7)
8)
9)
10)
time 0 there are 200 liters of liquid in each 11) Three different tanks A, B and C are with tubes
from B to A, from B to C and from A to C.
Three different tanks A, B and C are with tubes
are φB→A = 3 liter/min, φB→C = 1 liter/min,
φA→C = 2 liter/min and φut = 4 liter/min.
of the containers, of which that in C is evenly
from A to B, from B to A, from B to C and from
C to B. It also flows liquid from the environment
into C, and from A to the environment. At
the time 0 there are 200 liters of liquid in each 13) Three different tanks A, B and C are with tubes
from C to B, from B to A and from A to C.
into C has the flow rate φin = 3 liter/min and
C, and from A to the environment. At the
are φA→B = 2 liter/min, φB→A = 1 liter/min,
φB→C = 3 liter/min, φC→B = 1 liter/min and
φut = 1 liter/min.
are φC→B = 2 liter/min, φB→A = 3 liter/min,
A, and from C to the environment. At the 14) Three different tanks A, B and C are with tubes
from A to B, from B to C, from B to A and from
C to A. It also flows liquid from the environment
into C, and from A to the environment. At
the time 0 there are 200 liters of liquid in each
φB→A = 3 liter/min, φC→A = 1 liter/min and
φut = 3 liter/min.
from A to B, from B to C and from A to C.
11
C to B. It also flows liquid from the environment
into A, and from C to the environment. At
are φA→B = 4 liter/min, φB→A = 2 liter/min,
φB→C = 3 liter/min, φC→B = 2 liter/min and
φut = 1 liter/min.
from A to C, from C to B, from B to C and from
B to A. It also flows liquid from the environment
into A, and from C to the environment. At 19) Three different tanks A, B and C are with tubes
from A to C, from C to B and from B to A.
C, and from B to the environment. At the
are φA→C = 3 liter/min, φC→B = 2 liter/min,
φB→C = 2 liter/min, φB→A = 1 liter/min and
φut = 1 liter/min.
salinity
ρin = 2 gram/liter. The other flow rates
are
φ
= 5 liter/min, φC→B = 2 liter/min,
A→C
φ
=
3
liter/min
and φut = 7 liter/min.
B→A
A, and from B to the environment. At the 20) Three different tanks A, B and C are with tubes
from A to B, from B to A, from B to C and from
Block 2 – ODE of higher orders,
Difference equations
4.
Linear ODE of higher orders
1)


2)





3)





4)





5)





6)





7)





ZC 4.1–4.4, 4.7, 4.10
Determine whether the functions f1 , f2 , f3 are linear independent on R (on R+ in versions
10 and 13) or not.


3
f1 (x) = 3 + x


 f1 (x) = 7x − x
 f1 (x) = |x| + 5
2
f2 (x) = 2x + 5x3
f2 (x) = x + 5
f2 (x) = x − 2x
15)
8)




f3 (x) = 9x + 4x3
f3 (x) = |x| − 5
f3 (x) = 3x2 − 1

f1 (x) = e2x


 f1 (x) = x
−x
x

−2x
 f1 (x) = 2e + 3e
f2 (x) = 2x + 3x2
f2 (x) = e
9)


f2 (x) = e−x − 2ex
16)
f3 (x) = 4x − x3
f3 (x) = cosh(2x)


f3 (x) = e−x + ex

f1 (x) = x

 f1 (x) = ln(x)
2

f2 (x) = ln(x2 )
f2 (x) = 3x − x
10)


 f1 (x) = x + |x|

f3 (x) = x ln(x)
f3 (x) = 2x + x2
f2 (x) = 2|x| − 3x
17)



2
f3 (x) = x − |x|
f1 (x) = |x| + 5

 f1 (x) = x − x
2
f2 (x) = 2 + 5x
f2 (x) = x + 5
11)



2
f3 (x) = x − 3
f3 (x) = x

 f1 (x) = cos(x) − 2 sin(x)

f2 (x) = sin(x)
18)
2x
f1 (x) = 1 + sin2 (x)


 f1 (x) = e − 2

f3 (x) = sin(2x)
f2 (x) = e2x cosh(2x)
f2 (x) = cos2 (x) − 3
12)


f3 (x) = 3 + e2x
f3 (x) = 5

2
4


 f1 (x) = 2x + 3x
2
f1 (x) = x

 f1 (x) = x ln(x)
f2 (x) = 7x2 − 8x4
19)

2
3

f2 (x) = x + ln(x)
f2 (x) = 4x + x
13)
f3 (x) = x2


f3 (x) = x
f3 (x) = 2x3 − 6x2


3x
x
2
f1 (x) = sinh(3x)


 f1 (x) = e + e
 f1 (x) = |x| − 2x
f2 (x) = |x|
f2 (x) = e2x
f2 (x) = e3x
20)
14)




2
2x
f3 (x) = ex − 3e2x
f3 (x) = |x| + 3x
f3 (x) = e
INL 4.1



..................................
12
4.. LINEAR ODE OF HIGHER ORDERS
INL 4.2
13
Find the general solution of the differential equation. Hint: Begin by testing whether
the DE has a solution on any of the forms eβx , x + β, xβ or (ln(x))β .
1) x(x + 1)y 00 + (2 − x2 )y 0 − (x + 2)y = 0
11) (2x + 1)y 00 + 4xy 0 − 4y = 0
2) x2 (1 − x)y 00 + x(x − 4)y 0 + 6y = 0
12) (x + 1)y 00 + xy 0 − y = 0
3) x2 (x + 1)y 00 + (2x + 1)(y − xy 0 ) = 0
13) x2 y 00 − (2x + 3)xy 0 + (2x + 3)y = 0
4) xy 00 − (2x + 1)y 0 − 3(x − 1)y = 0
14) xy 00 − 2(x + 1)y 0 + (x + 2)y = 0
5) x(x + 1)y 00 + (x + 2)y 0 − y = 0
15) x2 y 00 − x(x + 2)y 0 + (x + 2)y = 0
6) x2 y 00 − (4x + 3)(xy 0 − y) = 0
7) xy 00 − (2x + 1)y 0 + 2y = 0
00
0
16) xy 00 − (3x + 1)y 0 + 3y = 0
17) x2 y 00 − 2xy 0 + (x2 + 2)y = 0
Hint: The DE has a solution on the form x sin(x)
8) xy − (x + 1)y + y = 0
18) x2 y 00 + (2x2 − x)y 0 − 2xy = 0
9) x3 y 00 + xy 0 − y = 0
19) xy 00 + (x − 1)y 0 + 3(1 − 4x)y = 0
10) (x ln x)2 y 00 + x(ln x)2 y 0 − 6y = 0
INL 4.3
20) xy 00 − (x + 2)y 0 + 2y = 0
Solve the initial-value problem.
1) y 00 + y 0 − 6y = 0, y(0) = 3, y 0 (0) = −4
11) y 00 − 2y 0 − 8y = 0, y(0) = 1, y 0 (0) = −14
2) y 00 − 4y 0 + 4y = 0, y(0) = 3, y 0 (0) = 2
12) y 00 + 6y 0 + 9y = 0, y(0) = 2, y 0 (0) = −3
3) y 00 − 2y 0 − 3y = 0, y(0) = −1, y 0 (0) = 2
13) y 00 − 4y 0 − 12y = 0, y(0) = 2, y 0 (0) = 28
4) y 00 + 4y 0 + 4y = 0, y(0) = 2, y 0 (0) = 4
14) y 00 − 12y 0 + 36y = 0, y(0) = −3, y 0 (0) = −11
5) y 00 + 3y 0 − 10y = 0, y(0) = 8, y 0 (0) = −5
15) y 00 − 5y 0 + 6y = 0, y(0) = 3, y 0 (0) = 1
6) y 00 − 10y 0 + 25y = 0, y(0) = −2, y 0 (0) = −7
16) y 00 + 10y 0 + 25y = 0, y(0) = −4, y 0 (0) = 19
7) y 00 − y 0 − 12y = 0, y(0) = 1, y 0 (0) = 11
17) y 00 − 3y 0 − 28y = 0, y(0) = 7, y 0 (0) = −6
8) y 00 + 8y 0 + 16y = 0, y(0) = 3, y 0 (0) = 10
18) y 00 − 8y 0 + 16y = 0, y(0) = −1, y 0 (0) = 0
9) y 00 + y 0 − 2y = 0, y(0) = −2, y 0 (0) = −5
19) y 00 + 2y 0 − 15y = 0, y(0) = 10, y 0 (0) = −2
10) y 00 − 6y 0 + 9y = 0, y(0) = 2, y 0 (0) = 1
INL 4.4
20) y 00 + 12y 0 + 36y = 0, y(0) = 2, y 0 (0) = −10
Find the general solution of the differential equation.
1) y 00 − 10y 0 + 25y = xe−x + 3e5x
9) y 00 − 4y 0 + 4y = 3e2x − 3xe−x
2) y 00 − y 0 − 12y = e2x − 4xe4x
10) y 00 − 2y 0 − 3y = 2xe−x + 5e2x
3) y 00 + 8y 0 + 16y = xe3x − 2e−4x
11) y 00 + 4y 0 + 4y = xe3x − 4e−2x
4) y 00 + y 0 − 2y = 3xex + 5e−3x
12) y 00 + 3y 0 − 10y = 4xe−5x + 2e−x
5) y 00 − 6y 0 + 9y = 6e3x + xex
13) y 00 + 12y 0 + 36y = 2e−6x − 3xex
6) y 00 − 2y 0 − 8y = 2xe−2x − 5e3x
14) y 00 + 2y 0 − 15y = 2xe−5x + 6e−3x
7) y 00 + 6y 0 + 9y = 2xe2x + 5e−3x
15) y 00 − 8y 0 + 16y = −3e4x + 4xe2x
8) y 00 + y 0 − 6y = 7ex + xe2x
16) y 00 − 3y 0 − 28y = 5e4x − 2xe7x
14
17) y 00 + 10y 0 + 25y = e−5x + 2xe−3x
19) y 00 − 12y 0 + 36y = −4e6x + xe3x
18) y 00 − 5y 0 + 6y = −3xe2x − e−3x
20) y 00 − 4y 0 − 12y = 3e2x − 2xe6x
INL 4.5
Find the general solution of the differential equation (either for x > 0 or x < 0).
1) x2 y 00 + 9xy 0 + 16y = 0
8) x2 y 00 − xy 0 − 3y = 0
15) x2 y 00 − 11xy 0 + 36y = 0
2) x2 y 00 + 2xy 0 − 2y = 0
9) x2 y 00 + 5xy 0 + 4y = 0
16) x2 y 00 − 3xy 0 − 12y = 0
3) x2 y 00 − 5xy 0 + 9y = 0
10) x2 y 00 + 4xy 0 − 10y = 0
2 00
0
2 00
17) x2 y 00 + 13xy 0 + 36y = 0
0
4) x y − xy − 8y = 0
11) x y − 9xy + 25y = 0
5) x2 y 00 + 7xy 0 + 9y = 0
12) x2 y 00 − 2xy 0 − 18y = 0
6) x2 y 00 + 2xy 0 − 6y = 0
13) x2 y 00 + 11xy 0 + 25y = 0
19) x2 y 00 − 7xy 0 + 16y = 0
7) x2 y 00 − 3xy 0 + 4y = 0
14) x2 y 00 − 4xy 0 + 6y = 0
20) x2 y 00 − 2xy 0 − 28y = 0
4.
Non-linear ODE of higher orders
INL 4.6
18) x2 y 00 + 3xy 0 − 15y = 0
...........................................
ZC 4.10
Find all solutions of the differential equation.
1) (y − 2)y 00 = (y 0 )2
6) (y + 1)y 00 = 2(y 0 )2
11) y 2 y 00 = y 0
2) yy 00 = 3(y 0 )2
7) yy 00 = (y − 1)(y 0 )2
12) (2y − 1)(y 0 )2 = y 2 y 00 17) y 00 = 12y 2 (y 0 )3
3) yy 00 + (y 0 )2 = yy 0
8) yy 00 + (y 0 )2 = 0
13) y 0 + (y 0 )2 = yy 00
18) 2yy 00 = 1 + (y 0 )2
4) y 00 = (y 0 )2
9) y 2 y 00 = (y 0 )3
14) yy 00 + (y 0 )3 = 0
19) yy 00 +2(y 0 )3 = 2(y 0 )2
5) y 00 = 2y(y 0 )3
10) (y 2 + 1)y 00 = 2y(y 0 )2 15) (y 2 + 1)y 0 = y 2 y 00
16) y(y−1)y 00 +(y 0 )2 = 0
20) y 00 = y 0 + (y 0 )3
7.. LAPLACE TRANSFORM
7.
Laplace transform
2)
3)
4)
5)
6)
7)
8)
9)
10)
.......................................................................
ZC 7.1–7.6
Solve the differential equation problem for t ≥ 0 and with y(0) = 2.




 0, 0 ≤ t < 4
 0, 0 ≤ t < 2
0
0
6 − 2t , 4 ≤ t < 7
5, 2 ≤ t < 5
11) y (t) + 7y(t) =
y (t) + 7y(t) =




5, t ≥ 7
2t + 1 , t ≥ 5




 2t + 9 , 0 ≤ t < 2
 5 − 3t , 0 ≤ t < 5
0
0
7, 2 ≤ t < 4
0, 5 ≤ t < 7
12) y (t) + 9y(t) =
y (t) + 4y(t) =




0, t ≥ 4
9, t ≥ 7




 7 − t, 0 ≤ t < 6
 6, 0 ≤ t < 3
0, 6 ≤ t < 8
7t − 4 , 3 ≤ t < 6 .
13) y 0 (t) + 5y(t) =
y 0 (t) + 8y(t) =




3, t ≥ 8
0, t ≥ 6




 6, 0 ≤ t < 4
 0, 0 ≤ t < 6
4t + 3 , 4 ≤ t < 9
3t + 2 , 6 ≤ t < 9
14) y 0 (t) + 2y(t) =
y 0 (t) + 3y(t) =




0, t ≥ 9
4, t ≥ 9




 0, 0 ≤ t < 3
 8 − 5t , 0 ≤ t < 4
0
0
8 − 6t , 3 ≤ t < 5
7
,
4
≤
t
<
8
15) y (t) + 3y(t) =
y (t) + 2y(t) =




7, t ≥ 5
0, t ≥ 8




 9t − 2 , 0 ≤ t < 2
 9, 0 ≤ t < 3
0
0
4, 2 ≤ t < 8
0, 3 ≤ t < 7
16) y (t) + 8y(t) =
y (t) + 5y(t) =




0, t ≥ 8
6t − 1 , t ≥ 7




 0, 0 ≤ t < 5
 4t + 7 , 0 ≤ t < 2
0
0
2 − 3t , 5 ≤ t < 9
5, 2 ≤ t < 9
17) y (t) + 5y(t) =
y (t) + 9y(t) =




8, t ≥ 9
0, t ≥ 9




 3, 0 ≤ t < 2
 2, 0 ≤ t < 5
0
0
0, 2 ≤ t < 7
0, 5 ≤ t < 8
18) y (t) + 7y(t) =
y (t) + 6y(t) =




2t + 6 , t ≥ 7
9t − 7 , t ≥ 8




 4 − 5t , 0 ≤ t < 3
 0, 0 ≤ t < 2
0
0
2, 3 ≤ t < 8
1 − 4t , 2 ≤ t < 6
19) y (t) + 3y(t) =
y (t) + 8y(t) =




0, t ≥ 8
3, t ≥ 6




 5, 0 ≤ t < 4
 8, 0 ≤ t < 3
0
0
0, 4 ≤ t < 6
0, 3 ≤ t < 9
20) y (t) + 6y(t) =
y (t) + 4y(t) =




7t − 5 , t ≥ 6
3t + 8 , t ≥ 9
INL 7.1
1)
15
16
Solve the integral equation for t ≥ 0.
(
Z t
0, 0 ≤ t < 2
y(t − w) cos(w) dw =
1) y(t) + 2
e2−t , t ≥ 2
0
INL 7.2
t
Z
e−2τ cos(2τ ) y(t − τ ) dτ
2) y(t) = cos(3t) + 4
0

 0, 0 ≤ t < 3
1, 3 ≤ t < 5
y(ξ) dξ = t +

0, t ≥ 5
t
Z
3) y(t) + 2
0
4) 2e2t = 2y(t) + 5
t
Z
y(ω) sin 2(t − ω) dω
0
(
t
Z
y(ω) 5(t − ω) + 23 (t − ω)3 dω =
5) y(t) +
0
6) y(t) + e−t
(
t
Z
eξ y(ξ) dξ =
0
0, 0 ≤ t < 5
e10−2t , t ≥ 5
(
t
Z
0, 0 ≤ t < 2
4t − 8 , t ≥ 2
3
(t − ω) y(ω) dω +
7) 2y(t) = 27
0
Z
(
t
8) y(t) + 2
e
−3ξ
y(t − ξ) dξ =
0
0, 0 ≤ t < 6
e2t−12 , t ≥ 6
t
Z
0
0, 0 ≤ t < 3
t − 3, t ≥ 3
y(t − τ ) cos(3τ ) dτ
9) y (t) = 4 + 5
with y(0) = 0
0
10) e−t = y(t) + 2
t
Z
y(ξ) cos(t − ξ) dξ
0
t
Z
11)
Z
y(λ) dλ + 3
0
(
t
0, 0 ≤ t < 5
y(t − ν) e−4ν dν =
(60 − 12t)e20−4t , t ≥ 5
0
t
Z
(τ − t) y(τ ) dτ = y(t)
12) 3 sin(2t) +
0
Z
13) 3
(
t
y(ξ) dξ + y(t) =
0
Z
0, 0 ≤ t < π
13 sin(2t) , t ≥ π
t
y(t − λ) cos(5λ) dλ
14) 1 = y(t) + 8
0
(
t
Z
2
y(τ ) (t − τ ) 25 + 24(t − τ )
15) 2y(t) + 2
0
16) y 00 (t) + 5
Z
0
(
t
y 0 (t − ξ) cos(2ξ) dξ =
dτ =
0, 0 ≤ t < 5
7(t − 5)2 , t ≥ 5
0, 0 ≤ t < 4
9 sin(2(t − 4)) , t ≥ 4
(
with
y(0) = 0
y 0 (0) = 9
Z.. LINEAR DIFFERENCE EQUATIONS
INL 7.3
17
Solve the differential equation problem for t ≥ 0.
1) y 00 + 8y 0 + 16y = 2δ(t − 3) − 5U (t − 8),
y(0) = 3, y 0 (0) = 10
11) y 00 − 8y 0 + 16y = U (t − 3) − 5δ(t − 7),
y(0) = −1, y 0 (0) = 0
2) y 00 + y 0 − 2y = 4U (t − 3) + δ(t − 5),
y(0) = −2, y 0 (0) = −5
12) y 00 + 2y 0 − 15y = δ(t − 2) + U (t − 4),
y(0) = 10, y 0 (0) = −2
3) y 00 − 6y 0 + 9y = U (t − 2) − 2δ(t − 7),
y(0) = 2, y 0 (0) = 1
13) y 00 + 12y 0 + 36y = 3δ(t − 2) + 2U (t − 3),
y(0) = 2, y 0 (0) = −10
4) y 00 − 2y 0 − 8y = δ(t − 6) + 5U (t − 9),
y(0) = 1, y 0 (0) = −14
14) y 00 + y 0 − 6y = U (t − 5) − 3δ(t − 9),
y(0) = 3, y 0 (0) = −4
5) y 00 + 6y 0 + 9y = 5U (t − 6) + δ(t − 4),
y(0) = 2, y 0 (0) = −3
15) y 00 − 4y 0 + 4y = 4δ(t − 3) + 7U (t − 4),
y(0) = 3, y 0 (0) = 2
6) y 00 − 4y 0 − 12y = 5δ(t − 3) − 2U (t − 4),
y(0) = 2, y 0 (0) = 28
16) y 00 − 2y 0 − 3y = 6U (t − 2) + 2δ(t − 5),
y(0) = −1, y 0 (0) = 2
7) y 00 − 12y 0 + 36y = δ(t − 2) + 3U (t − 9),
y(0) = −3, y 0 (0) = −11
17) y 00 + 4y 0 + 4y = δ(t − 3) − U (t − 4),
y(0) = 2, y 0 (0) = 4
8) y 00 − 5y 0 + 6y = U (t − 4) + 13δ(t − 5),
y(0) = 3, y 0 (0) = 1
18) y 00 + 3y 0 − 10y = U (t − 3) + δ(t − 4),
y(0) = 8, y 0 (0) = −5
9) y 00 + 10y 0 + 25y = δ(t − 3) − 5U (t − 6),
y(0) = −4, y 0 (0) = 19
19) y 00 − 10y 0 + 25y = δ(t − 2) + U (t − 4),
y(0) = −2, y 0 (0) = −7
10) y 00 − 3y 0 − 28y = 4δ(t − 2) + 9U (t − 8),
y(0) = 7, y 0 (0) = −6
20) y 00 − y 0 − 12y = δ(t − 2) − 5U (t − 5),
y(0) = 1, y 0 (0) = 11
Z.
Linear difference equations
INL z.1
..........................................................
Notes
∞
Find the number sequence {yn }n=0 which satisfies the recurrence relation and the initial
conditions.
1) 6yn + yn−1 − 2yn−2 = 0 , y0 = 1 , y1 = 4 .
11) 12yn − 5yn−1 − 3yn−2 = 0 , y0 = 10 , y1 = 1 .
2) 4yn + 4yn−1 + yn−2 = 0 , y0 = 3 , y1 = −2 .
12) 25yn + 20yn−1 + 4yn−2 = 0 , y0 = −3 , y1 = −2 .
3) 12yn + yn−1 − yn−2 = 0 , y0 = 11 , y1 = 1 .
13) 6yn + yn−1 − yn−2 = 0 , y0 = 7 , y1 = 4 .
4) 10yn + 3yn−1 − yn−2 = 0 , y0 = 21 , y1 = 0 .
14) 12yn − yn−1 − yn−2 = 0 , y0 = 10 , y1 = 1 .
5) 9yn − 12yn−1 + 4yn−2 = 0 , y0 = −1 , y1 = 2 .
15) 25yn − 30yn−1 + 9yn−2 = 0 , y0 = 2 , y1 = −3 .
6) 20yn + yn−1 − yn−2 = 0 , y0 = 14 , y1 = 1 .
16) 8yn − 2yn−1 − 3yn−2 = 0 , y0 = 12 , y1 = 0 .
7) 9yn + 6yn−1 + yn−2 = 0 , y0 = −2 , y1 = 2 .
17) 16yn + 24yn−1 + 9yn−2 = 0 , y0 = −1 , y1 = 3 .
8) 16yn − 8yn−1 − 3yn−2 = 0 , y0 = 16 , y1 = −2 .
18) 6yn − yn−1 − yn−2 = 0 , y0 = −1 , y1 = 2 .
9) 10yn − yn−1 − 2yn−2 = 0 , y0 = 1 , y1 = 5 .
19) 9yn − 3yn−1 − 2yn−2 = 0 , y0 = 3 , y1 = −4 .
10) 16yn − 8yn−1 + yn−2 = 0 , y0 = 4 , y1 = 0 .
20) 25yn − 10yn−1 + yn−2 = 0 , y0 = 2 , y1 = 1 .
18
∞
Find the number sequence {yn }n=0 which satisfies the difference equation problem.
0 , n 6= 5 ,
0 , n 6= 6 ,
yn−1 − 3yn =
n ≥ 1,
11) yn−1 − 2yn =
n ≥ 1,
3, n = 5,
3, n = 6,
y0 = 2 .
y0 = −1 .
0 , n 6= 3 ,
0 , n 6= 4 ,
yn−1 + 5yn =
n ≥ 1,
12) yn−1 + 7yn =
n ≥ 1,
−5 , n = 3 ,
−5 , n = 4 ,
y0 = 4 .
y0 = 4 .
0 , n 6= 4 ,
0 , n 6= 7 ,
yn−1 − 4yn =
n ≥ 1,
13) yn−1 + 8yn =
n ≥ 1,
2, n = 4,
4, n = 7,
y0 = 5 .
y0 = −3 .
0 , n 6= 7 ,
0 , n 6= 9 ,
yn−1 + 6yn =
n ≥ 1,
14) yn−1 + 5yn =
n ≥ 1,
4, n = 7,
10 , n = 9 ,
y0 = −2 .
y0 = 4 .
0 , n 6= 6 ,
0 , n 6= 8 ,
yn−1 − 7yn =
n ≥ 1,
15) yn−1 + 9yn =
n ≥ 1,
−3 , n = 6 ,
−3 , n = 8 ,
y0 = 3 .
y0 = 5 .
0 , n 6= 9 ,
0 , n 6= 5 ,
yn−1 + 4yn =
n ≥ 1,
16) yn−1 − 4yn =
n ≥ 1,
−3 , n = 9 ,
−6 , n = 5 ,
y0 = −5 .
y0 = −3 .
0 , n 6= 6 ,
0 , n 6= 4 ,
yn−1 + 2yn =
n ≥ 1,
17) yn−1 + 8yn =
n ≥ 1,
−5 , n = 6 ,
2, n = 4,
y0 = −3 .
y0 = −1 .
0 , n 6= 5 ,
0 , n 6= 9 ,
yn−1 − 5yn =
18) yn−1 − 7yn =
n ≥ 1,
n ≥ 1,
−4 , n = 9 ,
2, n = 5,
y0 = 3 .
y0 = 4 .
0 , n 6= 8 ,
0 , n 6= 7 ,
19) yn−1 − 9yn =
n ≥ 1,
yn−1 + 3yn =
n ≥ 1,
5, n = 7,
4, n = 8,
y0 = 2 .
y0 = 2 .
0 , n 6= 3 ,
0 , n 6= 8 ,
20) yn−1 − 6yn =
n ≥ 1,
yn−1 − 6yn =
n ≥ 1,
−3 , n = 8 ,
−3 , n = 3 ,
y0 = −5 .
y0 = −5 .
INL z.2
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
INL z.3
Solve the difference equation with the initial value y0 = 0 .
1) 4yn−1 + yn = (−4)n
8) 3yn−1 − yn = 3n
15) 8yn−1 + yn = (−8)n
2) 2yn−1 + yn = (−2)n
9) 5yn−1 + yn = (−5)n
16) 5yn−1 + yn = −(−5)n
3) 5yn−1 − yn = 5n
10) 4yn−1 − yn = 4n
4) 3yn−1 + yn = (−3)n
11) 6yn−1 + yn = (−6)n
5) 6yn−1 − yn = 6n
12) 7yn−1 − yn = 7n
6) 2yn−1 − yn = 2n
13) 9yn−1 − yn = 9n
19) 8yn−1 − yn = 8n
7) 7yn−1 + yn = (−7)n
14) 4yn−1 − yn = −4n
20) 3yn−1 + yn = −(−3)n
17) 9yn−1 + yn = (−9)n
18) 6yn−1 − yn = −6n
Block 3 – Systems of ODE
8.
System of first order linear DE
...........................................
ZC 8.1–8.3
Find the general solution of the linear system and sketch a representative part of the phase
portrait.
dx/dt
−2x + y
dx/dt
x + 4y
dx/dt
2x − 2y
8)
=
1)
=
15)
=
dy/dt
x − 2y
dy/dt
2x − y
dy/dt
−2x − y
!
!
− 4 x + 53 y
dx/dt
4x − 2y
dx/dt
− 38 x − 13 y
2)
=
9)
= 103
dx/dt
1
dy/dt
3x − y
dy/dt
16)
=
3 x + 3y
dy/dt
− 23 x − 73 y
!
1
dx/dt
−x
dx/dt
2 x + 9y
3)
=
10)
=
dy/dt
x − 2y
1
dx/dt
−3x + 2y
dy/dt
x
+
2y
17)
=
2
dy/dt
−2x + 2y
dx/dt
2x − y
dx/dt
7x + 6y
4)
=
11)
=
!
dy/dt
3x − 2y
12
6
dy/dt
2x + 6y
x
−
y
dx/dt
5
5
18)
=
dy/dt
− 15 x + 13
dx/dt
−3x − 2y
dx/dt
x + 2y
5 y
5)
=
12)
=
dy/dt
−x − 4y
dy/dt
3x + 2y
!
dx/dt
−4x + 6y
11
9
19)
=
x + 5y
dx/dt
2y
dx/dt
dy/dt
−3x + 5y
6)
=
13)
= 65
14
dy/dt
−x + 3y
dy/dt
x
+
y
5
5
!
2
− 10
5
dx/dt
3 x − 3y
dx/dt
dx/dt
x + 3y
−4x − 2 y
20)
=
7)
=
14)
=
dy/dt
− 43 x − 83 y
dy/dt
2x + 2y
dy/dt
3x + y
INL 8.1
Sketch for t ≥ 0 the solution curve which satisfies the initial-value problem.
!
dx/dt
3x + y
x(0)
0
− 65 x + 32 y
dx/dt
x(0)
−3
8)
=
,
=
1)
=
,
=
dy/dt
−x
+
y
y(0)
1
2
4
dy/dt
y(0)
0
−3x + 5y
dx/dt
4x + 5y
x(0)
−1
dx/dt
2x + y
x(0)
−3
9)
=
,
=
2)
=
,
=
dy/dt
−5x
−
6y
y(0)
3
dy/dt
−x
y(0)
10
dx/dt
3x − 4y
x(0)
0
dx/dt
x + 4y
x(0)
1
10)
=
,
=
,
=
3)
=
dy/dt
x−y
y(0)
1
dy/dt
−x + 5y
y(0)
0
INL 8.2
dx/dt
2x + 3y
x(0)
−2
=
,
=
dy/dt
−3x − 4y
y(0)
−2
dx/dt
4y
x(0)
0
5)
=
,
=
dy/dt
−x + 4y
y(0)
1
!
!
dx/dt
− 32 x + 14 y
x(0)
3
6)
=
,
=
y(0)
2
dy/dt
−x − 12 y
4)
dx/dt
−4x − y
7)
=
,
dy/dt
x − 2y
x(0)
2
=
y(0)
1
dx/dt
5x + 4y
=
,
dy/dt
−x + y
dx/dt
x + 3y
=
,
dy/dt
−3x − 5y
dx/dt
7x − 4y
=
,
dy/dt
9x − 5y
dx/dt
3x − y
=
,
dy/dt
25x − 7y
11)
12)
13)
14)
19
x(0)
−1
=
y(0)
1
x(0)
−2
=
y(0)
−2
x(0)
1
=
y(0)
2
x(0)
−1
=
y(0)
−6
20
15)
16)
17)
dx/dt
dy/dt
dx/dt
dy/dt
=
=
dx/dt
=
dy/dt
!
2x − 5y
,
5
4 x − 3y
x(0)
−2
=
y(0)
−3
dx/dt
5x − 3y
18)
=
,
dy/dt
3x − y
5x − 3y
,
12x − 7y
!
−x − 13 y
,
4
1
3x + 3y
dx/dt
19)
=
dy/dt
x(0)
1
=
y(0)
1
x(0)
1
=
y(0)
0
20)
x(0)
0
=
y(0)
1
!
−2x + 21 y
,
− 12 x − y
x(0)
−4
=
y(0)
0
dx/dt
−8x + 5y
=
,
dy/dt
−5x + 2y
x(0)
−1
=
y(0)
1
INL 8.3 Sketch for t ≥ 0 the solution curve which satisfies the initial-value problem.
dx/dt
3x + y
x(0)
1
dx/dt
x+y
x(0)
1
1)
=
,
=
11)
=
,
=
dy/dt
−8x − y
y(0)
1
dy/dt
−2x − y
y(0)
0
dx/dt
−x − y
x(0)
1
dx/dt
4x − 5y
x(0)
−3
12)
=
,
=
2)
=
,
=
dy/dt
9x − y
y(0)
−3
dy/dt
5x − 4y
y(0)
−3
dx/dt
−12x + 13y
x(0)
0
dx/dt
−x + 2y
x(0)
1
13)
=
,
=
3)
=
,
=
dy/dt
−13x + 12y
y(0)
5
dy/dt
−x − 3y
y(0)
0
dx/dt
−x − 2y
x(0)
−2
dx/dt
2x − 4y
x(0)
1
14)
=
,
=
4)
=
,
=
dy/dt
8x − y
y(0)
−2
dy/dt
2x − 2y
y(0)
−2
dx/dt
2x − 3y
x(0)
1
dx/dt
4x − 2y
x(0)
4
15)
=
,
=
5)
=
,
=
dy/dt
3x + 2y
y(0)
−2
dy/dt
5x + 2y
y(0)
−1
dx/dt
−3x − 5y
x(0)
−2
dx/dt
3x + y
x(0)
−2
16)
=
,
=
6)
=
,
=
dy/dt
5x
+
3y
y(0)
2
dy/dt
−x + 3y
y(0)
1
dx/dt
5x + 41y
x(0)
5
dx/dt
−11x + 2y
x(0)
1
17)
=
,
=
7)
=
,
=
dy/dt
−x
−
3y
y(0)
0
dy/dt
−4x − 7y
y(0)
1
dx/dt
−3x − y
x(0)
1
dx/dt
3x − 2y
x(0)
4
18)
=
,
=
8)
=
,
=
dy/dt
2x
−
y
y(0)
2
dy/dt
5x + y
y(0)
−1
!
13
5
x(0)
2
dx/dt
2x + 2 y
dx/dt
3x − 5y
x(0)
−5
,
=
19)
=
9)
=
,
=
5
y(0)
3
dy/dt
− 13
x
+
y
dy/dt
5x − 3y
y(0)
1
2
2
10)
dx/dt
3x + 2y
=
,
dy/dt
−2x + 3y
x(0)
−1
=
y(0)
−1
dx/dt
−2x + 2y
20)
=
,
dy/dt
−5x − 4y
x(0)
3
=
y(0)
−2
10.. PLANE AUTONOMOUS SYSTEMS
10.
Plane autonomous systems
INL 10.1
1)
2)
3)
4)
5)
6)
7)
8)
Classify the stationary point (origin)
period) if periodic solutions occur.
dx/dt
ax − 2y
=
för a 6= −2.
dy/dt
2x + 2y
dx/dt
19x − 3ay
=
för a 6= 95
22 .
dy/dt
x + (a − 5)y
dx/dt
ax − 3y
=
för a 6= − 94 .
dy/dt
3x + 4y
dx/dt
ax − 2y
=
för a 6= −4.
dy/dt
2x + y
dx/dt
3x − 2y
=
för a 6= − 43 .
dy/dt
2x + ay
dx/dt
ax − 2y
=
för a 6= 2.
dy/dt
(1 − a)x + y
dx/dt
y
=
för alla a.
dy/dt
−x + ay
dx/dt
2x + 7y
=
för a 6= − 49
8 .
dy/dt
−7x + 4ay
9)
10)
dx/dt
dy/dt
dx/dt
dy/dt
=
INL 10.2
1)
2)
3)
4)
5)
6)
21
−ax − 12 y
för a 6= 94 .
9
x
+
y
2
=
3x + 54 y
−5x + ay
för a 6= − 25
12 .
..............................................
ZC 10.1–10.3
of the system. Also, determine the orbital time (the
dx/dt
ax − 12y
11)
=
för a 6= −18.
dy/dt
3x + 2y
dx/dt
−5x − 9y
12)
=
för a 6= − 95 , 0.
dy/dt
a2 x − ay
dx/dt
2ax + 2y
13)
=
för a 6= 6.
dy/dt
−18x − 3y
x + 25 y
dx/dt
för a 6= − 25
14)
=
8 .
dy/dt
− 52 x + 2ay
15)
16)
17)
18)
19)
20)
dx/dt
ax − by
=
för |a| + |b| > 0.
dy/dt
bx + ay
dx/dt
2ax + (a − 2)y
=
för a 6= − 56 .
dy/dt
3x + 4y
dx/dt
3x + ay
=
för a 6= 89 .
dy/dt
5x + (3 − a)y
dx/dt
5ax − 12y
=
för a 6= − 18
5 .
dy/dt
3x + 2y
dx/dt
3x − 5y
=
för a 6= − 100
3 .
dy/dt
20x + ay
dx/dt
ax + 3y
=
för a 6= 56 .
dy/dt
(2 − a)x + 2y
Find all stationary points of the system, and state for each of them whether it is
asymptotically stable, stable or unstable. Also, state the classifications valid for the points
regarded as stationary for the corresponding linearized system.
2
dx/dt
x + x − 2xy
dx/dt
14x − 2x2 − xy
dx/dt
30x − 3x2 + xy
=
7)
=
13)
=
dy/dt
xy + 5y
dy/dt
16y − 2y 2 − xy
dy/dt
60y − 3y 2 + 4xy
dx/dt
x + xy − 3x2
dx/dt
6x + x2 + 3xy
dx/dt
2 − x2 + y
=
8)
=
14)
=
dy/dt
4y − 2xy − y 2
dy/dt
4xy − 12y
dy/dt
2x2 − 2xy
dx/dt
xy − 2y − 4
dx/dt
20x − xy − 2x2
dx/dt
y − xy + y 2
=
9)
=
15)
=
dy/dt
4y 2 − x2
dy/dt
16y − y 2 − xy
dy/dt
5x + xy − 2x2
dx/dt
4x − xy
dx/dt
5x − x2 − xy
dx/dt
4x − 2xy
=
10)
=
16)
=
dy/dt
5xy − y − y 2
dy/dt
xy − 2y
dy/dt
3y − xy − y 2
2
dx/dt
7x − xy − x2
dx/dt
−3x + x2 − xy
dx/dt
x + 3y 2 − 4
=
11)
=
17)
=
dy/dt
xy − 5y
dy/dt
−5y + xy
dy/dt
3xy − x2
dx/dt
3x − x2 − xy
dx/dt
x + x2 − 2xy
dx/dt
2x − 2xy
=
12)
=
18)
=
dy/dt
−2xy + 4y
dy/dt
xy − y
dy/dt
x2 − 2xy 2 − 3y
Block 4 – Fourier series, Partial DE
11.
Fourier series
INL 11.1
...........................................................................
ZC 11.1–11.3
Find the Fourier series, the Fourier-cosinus series and the Fourier-sinus series of the function
f , and sketch the corresponding graphs in the interval Iskiss . Finally, use the series to
determine the sums of the two chosen of the following series:
∞
∞
∞
∞
X
X
X
X
1
(−1)n−1
1
(−1)n−1
,
S
=
,
S
=
.
,
S2 =
S1 =
3
4
2n − 1
(2n − 1)2
n2
n2
n=1
n=1
n=1
n=1
1) f (x) = 2 − x2 /2
Df = (0, 2]
Isketch = [−6, 6]
S3 and S4
6) f (x) = x2 /3 − 1
Df = (0, 3)
Isketch = [−9, 9]
S3 and S4
11) f (x) = −(x + 2)
Df = (−2, 0]
Isketch = [−6, 6]
S1 and S2
16) f (x) = −(x + 2)x
Df = (−1, 0)
Isketch = [−3, 3]
S1 and S2
2) f (x) = 2x + 1
Df = [0, 1)
Isketch = [−3, 3]
S1 and S2
7) f (x) = −(2x + 1)
Df = [−2, 0]
Isketch = [−6, 6]
S1 and S2
12) f (x) = 3x + 1
Df = [−1, 0]
Isketch = [−3, 3]
S1 and S2
17) f (x) = 3 − x
Df = [0, 3]
Isketch = [−9, 9]
S1 and S2
3) f (x) = −x(4 + x)
Df = (−2, 0)
Isketch = [−6, 6]
S3 and S4
8) f (x) = (4 − x)x
Df = [0, 2)
Isketch = [−6, 6]
S2 and S3
13) f (x) = 1 + x2 /9
Df = (0, 3]
Isketch = [−9, 9]
S3 and S4
18) f (x) = x2 − 1
Df = (0, 1]
Isketch = [−3, 3]
S3 and S4
4) f (x) = x + 3
Df = [−3, 0]
Isketch = [−9, 9]
S1 and S2
9) f (x) = 3 − 2x
Df = [0, 1)
Isketch = [−3, 3]
S1 and S2
14) f (x) = 2 − x/2
Df = [0, 2)
Isketch = [−6, 6]
S1 and S2
19) f (x) = x + 2
Df = [−2, 0)
Isketch = [−6, 6]
S1 and S2
5) f (x) = x − 1
Df = (0, 1]
Isketch = [−3, 3]
S1 and S2
10) f (x) = x(x + 6)
Df = (−3, 0)
Isketch = [−9, 9]
S3 and S4
15) f (x) = (x + 3)2
Df = [−3, 0]
Isketch = [−9, 9]
S3 and S4
20) f (x) = (x − 3)2 + 3
Df = (0, 3)
Isketch = [−9, 9]
S2 and S3
12.
Partial differential equations
...........................................
ZC 12.1–12.5
INL 12.1
2) The temperature u of an ideal rod of length
3 is assumed to obey the partial differential
equation u00xx = 71 u0t , 0 < x < 3, t >
0. The left endpoint of the rod is held at
temperature 0 while the right is insulated from
the environment. At time 0, the temperature is
4 degrees in the interior of the rod. Determine
equation u00xx = 15 u0t , 0 < x < 8, t > 0. The
endpoints of the rod are insulated from the
environment. At time 0, the temperature is x
degrees in the interior of the rod. Determine u
for 0 < x < 8, t > 0.
22
12.. PARTIAL DIFFERENTIAL EQUATIONS
u for 0 < x < 3, t > 0.
equation u00xx = 2u0t , 0 < x < 5, t > 0. The
endpoints of the rod are held at temperature 0.
At time 0, the temperature is 3x degrees in the
interior of the rod. Determine u for 0 < x < 5,
t > 0.
23
equation u00xx = 81 u0t , 0 < x < 6, t > 0.
The left endpoint of the rod is insulated from
the environment, while the right is held at
temperature 0. At time 0, the temperature is 3
for 0 < x < 6, t > 0.
11)
12)
for 0 < x < 7, t > 0.
The temperature u of an ideal rod of length
environment. At time 0, the temperature is 2x
for 0 < x < 9, t > 0.
The temperature u of an ideal rod of length
equation u00xx = 3u0t , 0 < x < 2, t > 0. The
At time 0, the temperature is x degrees in the
t > 0.
At time 0, the temperature is 5 degrees in the
t > 0.
00
0
equation uxx = 3ut , 0 < x < 4, t > 0.
for
0 < x < 8, t > 0.
temperature 0. At time 0, the temperature is
2x degrees in the interior of the rod. Determine
u for 0 < x < 4, t > 0.
At time 0, the temperature is 2 degrees in the
t > 0.
the environment. At time 0, the temperature is 15) The temperature u of an ideal rod of length
x degrees in the interior of the rod. Determine
u for 0 < x < 9, t > 0.
equation u00 = 4u0 , 0 < x < 5, t > 0. The
xx
t
environment. At time 0, the temperature is 3x
for 0 < x < 5, t > 0.
environment. At time 0, the temperature is x+3 16) The temperature u of an ideal rod of length
for 0 < x < 2, t > 0.
equation u00xx = 4u0t , 0 < x < 7, t >
2x degrees in the interior of the rod. Determine
u for 0 < x < 7, t > 0.
the environment. At time 0, the temperature is 17) The temperature u of an ideal rod of length
u for 0 < x < 7, t > 0.
24
environment. At time 0, the temperature is 2−x
for 0 < x < 6, t > 0.
At time 0, the temperature is 4x degrees in the
t > 0.
equation u00xx = 2u0t , 0 < x < 3, t > 20) The temperature u of an ideal rod of length
temperature 0. At time 0, the temperature is x
u for 0 < x < 3, t > 0.
for 0 < x < 4, t > 0.
INL 12.2
1) An ideal string of normal length 3 can vibrate in
a xu-plane. The string is in its both ends (the
positions 0 and 3) held fixed at displacement
level 0. At time 0, the string is in motion such
that no point of the string is displaced but has
the velocity −3. Find the displacement u of the
string for 0 < x < 3, t > 0, where it is assumed
that the displacement obeys the wave equation
1 00
u00xx = 49
utt .
level 0. At time 0, the string is in rest such
that interior points of the string are displaced
according to x. Find the displacement u of the
1 00
u00xx = 25
utt .
2. Find the displacement u of the string for
0 < x < 7, t > 0, where it is assumed
1 00
u00xx = 36
utt .
the velocity x. Find the displacement u of the
1 00
utt .
u00xx = 64
according to 2x. Find the displacement u of the
u00xx = 19 u00tt .
the velocity 2. Find the displacement u of the
1 00
u00xx = 16
utt .
1 00
u00xx = 81
utt .
1 00
u00xx = 36
utt .
according to −x. Find the displacement u of
the string for 0 < x < 6, t > 0, where it is
assumed that the displacement obeys the wave
1 00
equation u00xx = 64
utt .
25
u00xx = 91 u00tt .
positions 0 and 7) held fixed at displacement 16) An ideal string of normal length 5 can vibrate in
−2. Find the displacement u of the string
1 00
utt .
u00xx = 25
for 0 < x < 5, t > 0, where it is assumed
1 00
utt .
u00xx = 25
positions 0 and 2) held fixed at displacement 17) An ideal string of normal length 3 can vibrate in
u00xx = 41 u00tt .
1 00
u00xx = 81
utt .
according to −x. Find the displacement u of
1 00
u00xx = 81
utt .
equation u00xx = 14 u00tt .
the velocity x. Find the displacement u of the
1 00
00
uxx = 49 utt .
1 00
utt .
u00xx = 36
level 0. At time 0, the string is in rest such 20) An ideal string of normal length 4 can vibrate in
according to −3x. Find the displacement u of
1 00
00
equation uxx = 16 utt .
1 00
u00xx = 49
utt .
INL 12.3
26
1) En rektangulär, värmeledande platta med
sidlängderna 5 och 2 är placerad i ett koordinatsystem på så vis att två av sidorna tangerar
de positiva koordinataxlarna. Den sida som
tangerar y-axeln är av den kortare längden och
hålls vid temperaturen 0 grader, medan dess
motstående sida hålls isolerad från omgivningen. Den sida som tangerar x-axeln hålls vid
temperaturen 0 grader, medan dess motstående
sida hålls vid temperaturen 9 grader. Bestäm
den statiska temperaturfördelningen u i det
inre av plattan om fördelningen antas lyda den
partiella differentialekvationen u00xx + u00yy = 0.
tangerar y-axeln är av den kortare längden
och hålls isolerad från omgivningen, medan
dess motstående sida hålls vid temperaturen
0 grader. Den sida som tangerar x-axeln
hålls isolerad, medan dess motstående sida
hålls vid temperaturen 7 grader. Bestäm den
statiska temperaturfördelningen u i det inre av
plattan om fördelningen antas lyda den partiella
differentialekvationen u00xx + u00yy = 0.
dess motstående sida hålls vid temperaturen 0
grader. Den sida som tangerar x-axeln hålls vid
och hålls vid temperaturen 0 grader, medan
dess motstående sida hålls isolerad från omgivningen. Den sida som tangerar x-axeln
hålls isolerade från omgivningen. Den sida
som tangerar x-axeln hålls isolerad, medan dess
motstående sida hålls vid temperaturen 2 − x
grader. Bestäm den statiska temperaturfördelningen u i det inre av plattan om fördelningen
antas lyda den partiella differentialekvationen
u00xx + u00yy = 0.
de positiva koordinataxlarna. De sidor som
är parallella med y-axeln är de kortare och
hålls vid temperaturen 0 grader. Den sida
som tangerar x-axeln hålls vid temperaturen
0 grader, medan dess motstående sida hålls
vid temperaturen 3 grader.
Bestäm den
vid temperaturen 2x − 1 grader. Bestäm den
sidlängderna 5 och 3 är placerad i ett koordinatsystem på så vis att två av sidorna tangerar de
positiva koordinataxlarna. De sidor som är parallella med y-axeln är de kortare och hålls vid
temperaturen 0 grader. Den sida som tangerar
x-axeln hålls isolerad, medan dess motstående
vid temperaturen 3 − x grader. Bestäm den
sidlängderna 3 och 2 är placerad i ett koordinatsystem på så vis att två av sidorna tangerar 10) En rektangulär, värmeledande platta med
sidlängderna 9 och 4 är placerad i ett koordinatär parallella med y-axeln är de kortare och
system på så vis att två av sidorna tangerar
11)
12)
13)
14)
27
hålls vid temperaturen 0 grader. Den sida 15) A rectangular, heat conductive plate with side
lengths 7 and 5 is located in a coordinate system
such that two of the sides are tangential to the
positive coordinate axes. The side tangential
vid temperaturen 2 grader.
Bestäm den
to the y-axis is held at the temperature of
0 degrees, while its opposite side is is kept
00
00
insulated from the environment. The side
differentialekvationen uxx + uyy = 0.
that is tangential to the x-axis is held at the
En rektangulär, värmeledande platta med
temperature of 0 degrees, while its opposite
sidlängderna 6 och 5 är placerad i ett koordinatside is held at the temperature of 6 degrees.
system på så vis att två av sidorna tangerar de
Determine the static temperature distribution
positiva koordinataxlarna. De sidor som är paru in the interior of the plate if the distribution
allella med y-axeln är de kortare och hålls vid
is assumed to obey the partial differential
temperaturen 0 grader. Den sida som tangerar
equation u00xx + u00yy = 0.
sida hålls vid temperaturen 8 grader. Bestäm 16) En rektangulär, värmeledande platta med
sidlängderna 9 och 6 är placerad i ett koordinatden statiska temperaturfördelningen u i det
sidlängderna 5 och 4 är placerad i ett koordinatdess motstående sida hålls vid temperaturen
0 grader. Den sida som tangerar x-axeln
som tangerar x-axeln hålls isolerad, medan dess
motstående sida hålls vid temperaturen 2x + 3
grader. Bestäm den statiska temperaturfördelningen u i det inre av plattan om fördelningen 17) En rektangulär, värmeledande platta med
sidlängderna 7 och 4 är placerad i ett koordinatantas lyda den partiella differentialekvationen
system på så vis att två av sidorna tangerar de
u00xx + u00yy = 0.
positiva koordinataxlarna. De sidor som är parEn rektangulär, värmeledande platta med
allella med y-axeln är de kortare och hålls vid
sidlängderna 7 och 3 är placerad i ett koordinattemperaturen 0 grader. Den sida som tangerar
dess motstående sida hålls isolerad från ompartiella differentialekvationen u00xx + u00yy = 0.
givningen. Den sida som tangerar x-axeln
hålls isolerad, medan dess motstående sida 18) En rektangulär, värmeledande platta med
sidlängderna 8 och 6 är placerad i ett koordinathålls vid temperaturen 5 grader. Bestäm den
sidlängderna 8 och 4 är placerad i ett koordinat0 grader, medan dess motstående sida hålls
vid temperaturen 3x − 2 grader. Bestäm den
grader. Den sida som tangerar x-axeln hålls vid 19) En rektangulär, värmeledande platta med
sidlängderna 6 och 3 är placerad i ett koordinatsida hålls vid temperaturen 3 grader. Bestäm
28
dess motstående sida hålls isolerad från omgivningen. Den sida som tangerar x-axeln
sidlängderna 8 och 7 är placerad i ett koordinat-
grader. Den sida som tangerar x-axeln hålls vid

Assigned problems for MAA134 Differential Equations

Transcription

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