University of Science and Technology of Southern Philippines
ES 208 Differential Equations
MTh 10:30 AM – 12:00 PM
Submitted by:
Dan Lester S. Mier
BSME 2C_M3
Submitted to:
Engr. Dennis E. Ganas
13 DECEMBER 2019
APPLICATIONS OF FIRST ORDER DIFFRENTIAL EQUATION
POPULATION GROWTH
A certain population of bacteria is known to grow at a rate proportional to the
amount present in a culture that provides plentiful food and space. Initially there are
250 bacteria, and after seven hours 800 bacteria are observed in a culture. Find the
expression for the approximate number of bacteria present in the culture at any time
𝑡.
Let 𝑁(𝑡) denote the number of bacteria present in the culture at any time 𝑡. The
growth rate is
𝑑𝑁
𝑑𝑡
, which is proportional to 𝑁. Thus,
𝑑𝑁
𝑑𝑡
= 𝑘𝑁,
(1)
where 𝑘 is a constant of proportionality
Note that (1) can be solved using variable separable
𝑑𝑁
= 𝑘𝑁
𝑑𝑡
𝑑𝑁
= 𝑘𝑑𝑡
𝑑𝑡
ln 𝑁 = 𝑘𝑡 + 𝐶
𝑒 ln 𝐶 = 𝑒 (𝑘𝑡+𝐶)
𝑁 = 𝑒 𝑘𝑡 ∙ 𝑒 𝑐
(Let 𝐶 = 𝑒 𝑐 )
𝑁 = 𝐶𝑒 𝑘𝑡
(2)
At 𝑡 = 0, 𝑁 = 250. Applying this initial condition to (2), we get
250 = 𝐶𝑒 𝑘(0) or 𝐶 = 250
At 𝑡 = 7, 𝑁 = 800. Substituting this condition and solving for 𝑘,
1
800
800 = 250𝑒 𝑘(7) or 𝑘 = 7 ln (250) = 0.166
Now the solution becomes 𝑵 = 𝟐𝟓𝟎𝒆𝟎.𝟏𝟔𝟔𝒕 which is an expression for the approximate
number of bacteria present at any time 𝑡 measured in hours.
 Determine the approximate number of bacteria that will be present in the culture
described in the previous problem after 24 hours.
We require 𝑁 at 𝑡 = 24. Substituting 𝑡 = 24 into 𝑁 = 250𝑒 0.166𝑡 we obtain 𝑁 =
250𝑒 0.166𝑡 = 𝟏𝟑, 𝟒𝟑𝟑.
 In a culture of yeast the amount of active ferment grows at rate proportional to
the amount present. If the amount doubles in 1 hour, how many times the original
3
amount may be anticipated at the end of 2 4 hours?
Let 𝑁(𝑡) denote the amount of yeast present at time 𝑡. Then
𝑑𝑁
𝑑𝑡
= 𝑘𝑁, where 𝑘
is a constant of proportionality. The solution to this equation is given 𝑁 = 𝐶𝑒 𝑘𝑡 . If we
designate the initial amount of yeast as 𝑁𝑜 = 𝐶𝑒 𝑘(0) = 𝐶. We may then rewrite the
solution as 𝑁 = 𝑁𝑜 𝑒 𝑘𝑡 .
After 1 hour, the amount present is 𝑁 = 2𝑁𝑜 ; Applying this condition and solving
for 𝑘, we find 2𝑁𝑜 = 𝑁0 𝑒 𝑘(1) so that 𝑒 𝑘 = 2 and 𝑘 = ln 2 = 0.693. Thus the amount of
yeast present at any time 𝑡 is 𝑁 = 𝑁𝑜 𝑒 0.693𝑡 . After 2.75 hours the amount will be 𝑁 =
𝑁𝑜 𝑒 0.693(2.75) = 6.72𝑁𝑜 . This represents a 𝟔. 𝟕𝟐 fold increase over the original amount.
EXAMPLE
A colony of bacteria is growing exponentially. At time t=0 it has 10 bacteria in it, and
at time t=4 it has 2000. At what time will it have 100,000 bacteria?
𝑁 = 10𝑒 𝑘𝑡
@ t = 4 ; N = 2000
2000 = 10𝑒 4𝑘
2000
= 𝑒 4𝑘
10
ln 200 = 4𝑘
𝑘=
ln 200
= 1.3245793
4
Thus
𝑁 = 10𝑒 1.3245793𝑡
@ N = 100000
100000 = 10𝑒 1.3245793𝑡
ln 10000 = 1.3245793𝑡
𝑡=
ln 10000
1.3245793
𝒕 ≈ 𝟔. 𝟗𝟓
DECOMPOSITION/DECAY

A certain reactive material is known to decay at a rate proportional to the
amount present. If initially there is 100mg of the material present and if after 2 years it
is observed that 5% of the original mass has decayed, find an expression for the mass
at any time 𝑡.
Let 𝑁(𝑡) denote the amount of material present at time 𝑡. The differential
equation governing this system is
𝑑𝑁
𝑑𝑡
= 𝑘𝑁, and its solution is 𝑁 = 𝐶𝑒 𝑘𝑡 .
At 𝑡 = 0, 𝑁 = 100. Applying this initial condition, we get 100 = 𝐶𝑒 𝑘(0) = 𝐶. Thus
the solution is 𝑁 = 100𝑒 𝑘𝑡 .
At 𝑡 = 2, 𝑁 = 100 − 5 = 95 since 5% has decayed. Substituting this condition
in the equation 𝑁 = 100𝑒 𝑘𝑡 and solving for 𝑘, we get 95 = 100𝑒 𝑘(2) or 𝑘 =
1
2
95
ln 100 = −0.0256. (Note: For decay, 𝑘 is negative.)
Thus the amount reactive material present at any time 𝑡 is 𝑵 = 𝟏𝟎𝟎𝒆−𝟎.𝟎𝟐𝟓𝟔𝒕 .
 In the previous problem, determine the time necessary for 10% of the original
mass to decay.
We require 𝑡 when 𝑁 has decayed to 90% of its original mass was 100mg, we
seek the value 𝑡 corresponding to 𝑁 = 90. Substituting 𝑁 = 90 into 𝑁 = 100𝑒 −0.0256𝑡
of the previous problem gives us 90 = 100𝑒 −0.0256𝑡 so that −0.0256𝑡 = ln 0.9 and 𝑡 =
−(ln 0.9)
0.0256
= 𝟒. 𝟏𝟐 years.
 Radium decomposes at a rate proportional to the amount present. If half the
original amount disappears in 1600 years, find the percentage lost in 100years.
Let 𝑅(𝑡) denote the amount of radium present at time 𝑡. It follows that
𝑑𝑅
𝑑𝑡
𝑘𝑡
= 𝑘𝑅
where K is a constant of proportionality. Solving this equation, we get 𝑅 = 𝐶𝑒 . If we
designate the initial amount as 𝑅𝑜 (at 𝑡 = 0) and apply this condition, we find 𝑅𝑜 =
𝐶𝑒𝑘(0) = 𝐶, so that the solution becomes 𝑅 = 𝑅𝑜 𝑒 𝑘𝑡 .
1
Since the half-life of radium is 1600 years, we have the condition 𝑅 = 2 𝑅𝑜 when
1
𝑡 = 1600. Applying this condition to the last equation and then solving for 𝑘 give 2 𝑅𝑜 =
1
ln
1
2
𝑅𝑜 𝑒 𝑘(1600) , from which 2 = 𝑒 𝑘(1600) , and 𝑘 = 1600
= −0.0004332. The amount of radium
present at any time 𝑡 is thus 𝑅 = 𝑅𝑜 𝑒 −0.0004332𝑡 . The amount present after 100 years
will be 𝑅 = 𝑅𝑜 𝑒 −0.0004332(100) = 0.958𝑅𝑜 , so the percent decrease from the initial
amount 𝑅𝑜 is
𝑅𝑜 −0.958𝑅𝑜
𝑅𝑜
∙ 100 = 𝟒. 𝟐%.
EXAMPLE
A zircon sample contains 4000 atoms of the radioactive element 235U. Given that
235U has a half‐ life of 700 million years, how long would it take to decay to 125
atoms?
𝑁 = 4000𝑒 𝑘𝑡
@ t = 700 years;𝑁 =
𝑁0
2
𝑁0
= 𝑁0 𝑒 700𝑘
2
1
= 𝑒 700𝑘
2
ln
𝑘=
1
ln 2
700
1
= 700𝑘
2
= −0.0009902
Thus
𝑁 = 4000𝑒 −0.0009902𝑡
@ N = 125
125 = 4000𝑒 −0.0009902𝑡
ln
125
= − 0.0009902𝑡
4000
125
4000
𝑡=
−0.0009902
ln
𝒕 ≈ 𝟑𝟓𝟎𝟎 𝒎𝒊𝒍𝒍𝒊𝒐𝒏 𝒚𝒆𝒂𝒓𝒔
EXERCISES
 The population of a certain country is known to increase at a rate proportional
to the number of people recently living in the country. If after 2 years the population
has doubled, and after 3 years the population is 20,000, find the number of people
initially living in the country.
𝑁𝒐 = 𝐶𝑒𝑘(0)
𝐶 = 𝑁𝑜
2𝑁𝑜 = 𝑁𝑜 𝑒 𝑘(2)
2
ln( )
1 = 0.3466
𝑘=
2
𝑁 = 𝑁𝑜 𝑒 0.3466𝑡
20,000 = 𝑁𝑜 𝑒 0.3466(3)
𝑵𝒐 = 𝟕𝟎𝟕𝟎. 𝟓𝟏
 A certain value of bacteria grows at a rate that is proportional to the number
present. It is found that the number doubles in 4 hours, how many may be expected
at the end of 12 hours?
𝑁 = 𝐶𝑒 𝑘𝑡
𝑁𝑜 = 𝐶𝑒 𝑘(0)
𝐶 = 𝑁𝑜
𝑁 = 𝑁𝑜 𝑒 𝑘𝑡
2𝑁𝑜 = 𝑁𝑜 𝑒 𝑘(4)
ln 2
𝑘=
= 0.1733
4
𝑁 = 𝑁𝑜 𝑒 0.1733𝑡
𝑁 = 𝑁𝑜 𝑒 0.1733(12)
𝑵=𝟖
 A certain radioactive material is known to decay at a rate proportional to the
amount present. If after 1 hour it is observed that 10% of the material has decayed,
find the half-life of the material.
𝑁 = 𝐶𝑒 𝑘𝑡
100 = 𝐶𝑒𝑘(0)
𝐶 = 100
𝑁 = 100𝑒 𝑘𝑡
90 = 100𝑒 𝑘(1)
90
ln(100)
𝑘=
= −0.1053605
1
𝑁 = 𝑁𝑜 𝑒 −0.1053605𝑡
1
= 𝑁𝑜 𝑒 −0.1053605𝑡
2
𝒕 = 𝟔. 𝟓𝟖 𝒉𝒓𝒔
 A certain radioactive material is known to decay at a rate proportional to the
1
amount present. If initially 2g of the material is present and 0.1% of the original mass
has decayed after 1 week, find an expression for the mass at any time 𝑡.
𝑁 = 𝐶𝑒 𝑘𝑡
1
= 𝐶𝑒𝑘(0)
2
1
𝐶=
2
1 𝑘𝑡
𝑁= 𝑒
2
𝟏 −𝟎.𝟎𝟎𝟏𝒕
𝑵= 𝒆
𝟐
SEATWORK
1. A herd of llamas has 1000 llamas in it, and the population is growing
exponentially. At time 𝑡 = 4 it has 2000 llamas. Write a formula for the number of
llamas at arbitrary time 𝑡.
𝑁 = 𝐶𝑒 𝑘𝑡
1000 = 𝐶𝑒𝑘(0)
𝐶 = 1000
𝑁 = 1000𝑒 𝑘𝑡
2000 = 1000𝑒 𝑘(4)
2000
ln(1000)
𝑘=
= 0.1732868
4
𝑵 = 𝟏𝟎𝟎𝟎𝒆𝟎.𝟏𝟕𝟑𝟐𝟖𝟔𝟖𝒕
2. A colony of bacteria is growing exponentially. At time 𝑡 = 0 it has 10 bacteria in
it, and at time 𝑡 = 4 it has 2000. At what time will it have 100,000 bacteria?
𝑁 = 𝐶𝑒 𝑘𝑡
10 = 𝐶𝑒𝑘(0)
𝐶 = 10
𝑁 = 10𝑒 𝑘𝑡
2000 = 10𝑒 𝑘(4)
2000
)
10 = 1.3245793
𝑘=
4
𝑁 = 10𝑒 1.3245793𝑡
100,000 = 10𝑒 1.3245793𝑡
100,000
ln ( 10 )
𝑡=
1.3245793
𝒕 = 𝟔. 𝟗𝟓
ln(
3. A slow economy caused a company’s annual revenues to drop from $530,000
in 2008 to $386,000 in 2010. If the revenue is following an exponential pattern of
decline, what is the expected revenue in 2012?
𝑁 = 𝐶𝑒 𝑘𝑡
$530,000 = 𝐶𝑒𝑘(0)
𝐶 = $530,000
𝑁 = $530,000𝑒 𝑘𝑡
$386,000 = $530,000𝑒 𝑘(2)
$386,000
ln(
)
$530,000
𝑘=
= −0.1585198
2
𝑁 = $530,000𝑒 −0.1585198(4)
𝑵 = $𝟐𝟖𝟏, 𝟏𝟐𝟒. 𝟓𝟒
4. Suppose a population of insects increase according to the law of exponential
growth. There were 130 insects after the third day of the experiment and 380 insects
after the seventh day. Approximately how many insects were in the original
population?
𝑁 = 𝑁𝑜 𝑒 𝑘𝑡
130 = 𝑁0 𝑒 𝑘(0)
𝑁𝑜 = 130
𝑁 = 130𝑒 𝑘𝑡
380 = 130𝑒 𝑘(4)
380
ln(130)
𝑘=
= 0.2681592
4
𝑁 = 𝑁𝑜 𝑒 0.2681592𝑡
380 = 𝑁𝑜 𝑒 0.2681592(7)
𝑵𝒐 = 𝟓𝟖
5. A zircon sample contains 4000 atoms of the radioactive element 235U. Given
that 235U has a half-life of 700 million years, how long would it take to decay to 125
atoms?
𝑁 = 𝐶𝑒 𝑘𝑡
1 𝑡
125 = 4000( )700
2
ln
125
1 𝑡
= ln( )700
4000
2
ln
125
𝑡
1
=
ln
4000 700 2
125
ln 4000
1
ln 2
5=
=
𝑡
700
𝑡
700
𝑡 = 5(700)
𝒕 = 𝟑𝟓𝟎𝟎
6. The number of bacteria in a liquid culture is observed to grow at a rate
proportional to the number of cells present. At the beginning of the experiment there
are 10,000 cells and after 3 hours there are 500,000. How many will there be after 1
day of growth if this unlimited growth continues? What is the doubling time of the
bacteria?
𝑁 = 𝐶𝑒 𝑘𝑡
10,000 = 𝐶𝑒 𝑘(0)
𝐶 = 10,000
𝑁 = 10,000𝑒 𝑘𝑡
500,000 = 10,000𝑒 𝑘(3)
𝑘=
500,000
ln( 10,000 )
= 1.3040077
3
𝑁 = 10,000𝑒 1.3040077(24)
𝑵 = 𝟒 × 𝟏𝟎𝟏𝟕 bacteria
ln(
𝑡=
2𝑁𝑜
) ln(2(10,000))
𝑁𝑜
1000
=
= 𝟎. 𝟓𝟑 𝒉𝒓𝒔 𝒐𝒓 𝟑𝟐 𝒎𝒊𝒏.
𝑘
1.3040077
7. Carbon-14 is a radioactive isotope of carbon that has a half-life of 5600 years.
It is used extensively in dating organic material that is tens of thousands of years old.
What fraction of the original amount of Carbon-14 in a sample would be present after
10,000 years?
𝑁=
1
𝑁 𝑒 𝑘𝑡
2 𝑜
1
𝑁 = 𝑁𝑜 𝑒 𝑘(5600)
2
1
ln 1
2 = 1.24 × 10−4
5600
−4
𝑁 = 𝑒 1.24×10 (10,000)
𝑁 = 0.29
𝑵 = 𝟑𝟎%
𝑘=
MIXING (Non-reacting Fluids)
A tank initially holds 𝑉𝑜 gallons of brine that contains 𝑎 lb. of salt. Another brine
solution, containing 𝑏 lb. of salt per gallon, is poured into the tank at the rate of 𝑒
gal/min. while, simultaneously, the well-stirred solution leaves the tank at the rate 𝑓
gal/min. Find a differential equation for the amount of salt in the tank at any time 𝑡.
Let 𝑄 denote the amount (in lbs.) of salt in the tank at any time. The time rate of
change of 𝑄,
𝑑𝑄
𝑑𝑡
, equals the rate at which slat enters the tank minus the rate at which
salt leaves the tank. Salt enters the tank at the rate of 𝑏𝑒 lb/min. To determine the rate
at which salt leaves the tanks, we first calculate the volume of brine in the tank, which
is the initial volume 𝑉𝑜 plus the volume of brine added 𝑒𝑡 minus the volume of brine
removed 𝑓𝑡. Thus, the volume of brine in the tank at any time is 𝑉𝑜 + 𝑒𝑡 − 𝑓𝑡. The
concentration of salt in the tank at any time is then
𝑄
(𝑉𝑜 + 𝑒𝑡 − 𝑓𝑡)
from which it follows that salt leaves in the tank at the rate of
𝑓[
𝑄
] 𝑙𝑏/𝑚𝑖𝑛.
𝑉𝑜 + 𝑒𝑡 − 𝑓𝑡
Thus,
𝑑𝑄
𝑑𝑡
𝑄
= 𝑏𝑒- 𝑓 [𝑉 +𝑒𝑡−𝑓𝑡]
𝑜
𝑑𝑄
𝑓
+[
] 𝑄 = 𝑏𝑒
𝑑𝑡
𝑉𝑜 + 𝑒𝑡 − 𝑓𝑡
At 𝑡 = 0, 𝑄 = 𝑎, so we also have the initial condition 𝑄(0) = 𝑎
EXAMPLES
1. A tank initially holds 100 gal. of brine solution containing 1 lb. of salt. At 𝑡 = 0
another brine solution containing 1 lb. of salt per gallon is poured into the tank at the
rate of 3 gal/min, while the well-stirred mixture leaves the tank at the same rate. Find
the amount of salt in the tank at any time 𝑡.
𝑑𝑄
3
+[
] 𝑄 = (1)(3)
𝑑𝑡
100 + 3𝑡 − 3𝑡
𝑑𝑄
3
+[
]𝑄 = 3
𝑑𝑡
100
𝑄 = 100 + 𝐶𝑒 −0.03𝑡
At 𝑡 = 0, 𝑄 = 1, 𝐶 =?
(Using Linear Eq’n)
1 = 100 + 𝐶𝑒 (−0.03)(0)
𝑸 = 𝟏𝟎𝟎 − 𝟗𝟗𝒆−𝟎.𝟎𝟑𝒕
2. Find the time at which the mixture described in the previous problem contains
2 lb. of salt.
We require 𝑡 when 𝑄 = 2
𝑄 = 100 − 99𝑒 −0.03𝑡
2 = 100 − 99𝑒 −0.03𝑡
𝑒 −0.03𝑡 =
𝑡=−
98
99
1
98
ln
0.03 99
𝒕 = 𝟎. 𝟑𝟑𝟖 𝒎𝒊𝒏.
3. A 50-gal. tank initially contains 10 gal. of fresh water. At 𝑡 = 0, a brine solution
containing 1 lb. of salt per gallon is poured into the tank at the rate of 4 gal/min, while
the well-stirred mixture leaves the tank at the rate of 2 gal/min. Find the amount of time
required for overflow to occur.
Here 𝑎 = 0, 𝑏 = 1, 𝑒 = 4, 𝑓 = 2 and 𝑉𝑜 = 10. Using the formula, the volume of the
brine in the tank at any time 𝑡 is 𝑉𝑜 + 𝑒𝑡 − 𝑓𝑡 = 10 − 2𝑡. We require 𝑡 when 10 + 2𝑡 =
50; Hence, 𝒕 = 𝟐𝟎 𝒎𝒊𝒏.
4. A tank contains 1,500 L of water and 20 kg of dissolved salt. Fresh water is
entering the tank at 15 L/min (the solution stays perfectly mixed), and the solution
drains at a rate of 10 L/min. How much salt is in the tank at t minutes and at 10
minutes?
𝑑𝑄
10
+[
] 𝑄 = (15)(0)
𝑑𝑡
1500 + 15𝑡 − 10𝑡
𝑑𝑄
10
+[
]𝑄 = 0
𝑑𝑡
1500 + 5𝑡
𝑄 = 𝐶(1500 + 5𝑡)−2
@ t = 0; Q = 20 kg
20 = 𝐶(1500)−2
𝐶 = 20(1500)2
𝐶 = 45000000
Thus,
𝑄 = 45000000(1500 + 5𝑡)−2
@ t = 10
𝑄 = 45000000(1500 + 50)−2
𝑄 = 45000000(1550)−2
𝑸 = 𝟏𝟖. 𝟕
EXERCISES:
1. A 120-gallon tank holds purified water. Salt water with 1.5 lbs. of salt per gallon
leaks into the tank at 2 gallons per minute. The mixture in the tank is constantly
(perfectly) mixed, and it flows out of the tank at 3 gallons per minute. Write a model
for the amount of salt in the tank (measured in lbs.) after 𝑡 minutes.
𝑑𝑄
3
+[
] 𝑄 = (1.5)(2)
𝑑𝑡
120 + 2𝑡 − 3𝑡
𝑑𝑄
3
+[
]𝑄 = 3
𝑑𝑡
120 − 𝑡
Using Linear Equation:
𝑄=
3
(120 − 𝑡) + 𝐶(120 − 𝑡)3
2
At 𝑡 = 0, 𝑄 = 0, 𝐶 =?
3
𝑄 = (120 − 0) + 𝐶(120 − 3)3
2
𝐶=−
3
= −𝟏. 𝟎𝟒 × 𝟏𝟎−𝟒
2(120)3
2. A tank has pure water flowing into it at 10L/min. The contents of the tank are
kept thoroughly mixed, and the contents flow out at 10L/min. Initially, the tank contains
10 kg. of salt in 100L of water. How much salt will there be in the tank after 30 minutes?
𝑑𝑄
1
+ 𝑄=0
𝑑𝑡 10
Using Variable Separable
1
𝑄 = 𝐶𝑒 −10𝑡
When 𝐶 = 10
1
𝑄 = 10𝑒 −10𝑡
At 𝑡 = 30 , 𝑄 =?
1
𝑄 = 10𝑒 −10(30)
𝑸 = 𝟎. 𝟒𝟗𝟕𝟗 𝒌𝒈
3. A tank originally contains 100 gal of fresh water. Then water containing
1
2
lb. of
salt per gallon is poured into the tank at a rate of 2gal/min. and the mixture is allowed
to leave at the same rate. What is the amount of salt at any instant? Determine the
amount of salt after 10 minutes.
𝑑𝑄
2
+[
] 𝑄 = (0.5)(2)
𝑑𝑡
100 + 2𝑡 − 2𝑡
𝑑𝑄
1
+ 𝑄=1
𝑑𝑡 50
Using Linear Equation
1
𝑄 = 50 + 𝐶𝑒 −50𝑡
At 𝑡 = 0
1
= 50 + 𝐶𝑒 −50(0)
𝐶 = −50
1
𝑄 = 50 − 50𝑒 −50𝑡
At 𝑡 = 10 𝑚𝑖𝑛.
1
𝑄 = 50 − 50𝑒 −50(10)
𝑸 = 𝟗. 𝟎𝟔 𝒍𝒃𝒔.
1
4. A tank initially holds 80 gal of brine solution containing 2lb. of salt per gallon. At
t=0, another brine solution containing 1lb. of salt per gallon is poured into the tank at
a rate of 4 gal/min, while the well-stirred mixture leaves the tank at the rate of 8gal/min.
Find the amount of salt in the tank at anytime 𝑡. Determine the amount of salt after 3
minutes.
𝑑𝑄
8
+[
] 𝑄 = (1)(4)
𝑑𝑡
80 + 4𝑡 − 8𝑡
𝑑𝑄
8
+[
]𝑄 = 4
𝑑𝑡
80 − 4𝑡
Using Linear Equation
𝑄 = (80 − 4𝑡) + 𝐶(80 − 4𝑡)2
At 𝑡 = 0, 𝑄 = 40𝑙𝑏𝑠.
40 = (80 − 4(0)) + 𝐶(80 − 4(0))2
𝐶=−
1
160
At 𝑡 = 3, 𝑄 =?
𝑄 = (80 − 4(3)) + (−
1
)(80 − 4(3))2
160
𝑸 = 𝟑𝟗. 𝟏 𝒍𝒃𝒔
LINEAR DIFFERENTIAL EQUATION OF ORDER 𝒏
Standard Form of an 𝑛𝑡ℎ order Linear Differential Equation
The general linear ODE of order 𝑛 is
𝒚(𝒏) + 𝑷𝟏(𝒙)𝒚(𝒏 − 𝟏) + ⋯ + 𝒑𝒏(𝒙)𝒚 = 𝒒(𝒙)
(1)
If 𝑞(𝑥) ≠ 0, the equation is non-homogeneous. We then call
𝒚(𝒏) + 𝑷𝟏(𝒙)𝒚(𝒏 − 𝟏) + ⋯ + 𝒑𝒏(𝒙)𝒚 = 𝟎
(2)
the associated homogeneous equation or the reduced equation.
The theory of the 𝑛𝑡ℎ order linear ODE runs parallel to that of the second order
equation. In particular, the general solution to the associated homogeneous equation
(2) is called complementary function or solution, and it has the form
𝒚𝒄 = 𝑪 𝟏 𝒚𝟏 + ⋯ + 𝑪 𝒏 𝒚𝒏
where 𝐶𝑖 are constants
(3)
Where the 𝑦𝑖 are 𝑛 solutions to (2) which are linearly independent, meaning that none
of them can be expressed as a linear combination of the others, i.e., by a relation of
the form (the left side could also be any of the other 𝑦𝑖 );
𝒚𝒏 = 𝒂𝟏 𝒚𝟏 + ⋯ + 𝒂𝒏 − 𝒚𝒏 − 𝟏
where 𝑎𝑖 are constants.
Once the associated homogeneous equation (2) has been solved by finding 𝑛
independent solutions, the solution to the original ODE (1) can be expressed as
𝒚 = 𝒚𝒑 + 𝒚𝒄
(4)
Where 𝑦𝑝 is a particular solution to (1), and 𝑦𝑐 is as in (3)
DIFFERENTIAL OPERATORS
In calculus differentiation is often denoted by the capital letter 𝐷- that is,
𝑑𝑦
𝑑𝑥
=
𝐷𝑦. The symbol 𝐷 is called a differential operator because it transforms a differentiable
function into another function. For example,𝐷(5𝑥 3 − 6𝑥 2 ) = 15𝑥 2 − 12𝑥. Higher-order
derivatives can be expressed in terms of 𝐷 in
𝑑
𝑑𝑦
( )=
𝑑𝑥 𝑑𝑥
𝑑2 𝑦
𝑑𝑥 2
= 𝐷(𝐷𝑦) = 𝐷2 𝑦
𝒅𝒏 𝒚
and, in general, 𝒅𝒙𝒏 = 𝑫𝒏 𝒚.
where 𝑦 represents a sufficiently differentiable function. Polynomial expressions
involving 𝐷, such as 𝐷 + 3, 𝐷2 + 3𝐷 − 4, and5𝑥 3 𝐷3 − 6𝑥 2 𝐷2 + 4𝑥𝐷 + 9, are also
differential operators. In general, we define an 𝑛𝑡ℎ order differential operator or
polynomial operator to be
𝑳 = 𝒂𝒏 (𝒙)𝑫𝒏 + 𝒂𝒏−𝟏 (𝒙)𝑫𝒏−𝟏 + ⋯ + 𝒂𝟏 (𝒙)𝑫 + 𝒂𝟎 (𝒙)
(5)
As a consequence of two basic properties of differentiation,𝐷(𝑐𝑓(𝑥)) = 𝑐𝐷𝑓(𝑥),
𝑐 is a constant and 𝐷{𝑓(𝑥) + 𝑔(𝑥)} = 𝐷𝑓(𝑥) + 𝐷𝑔(𝑥), the differential operator 𝐿
possesses a linearity property; that is, 𝐿 operating on a linear combination of two
differentiable function is the same as the linear combination of 𝐿 operating on the
individual functions. In symbols this means that
𝑳{∝ 𝒇(𝒙) + 𝜷𝒈(𝒙)} =∝ 𝑳𝒇(𝒙) + 𝜷𝑳𝒈(𝒙)
(6)
Where ∝ and 𝛽 are constants. Because of (6) we say that the 𝑛𝑡ℎ order differential
operator 𝐿 is a linear operator.
Any linear differential equation can be expressed in terms of 𝐷 notation. For
example, the differential equation 𝑦 ′′ + 5𝑦 ′ + 6𝑦 = 5𝑥 − 3 can be written as 𝐷2 𝑦 +
5𝐷𝑦 + 6𝑦 = 5𝑥 − 3 or (𝑫𝟐 + 𝟓𝑫 + 𝟔)𝒚 = 𝟓𝒙 − 𝟑.
SUPERPOSITION PRINCIPLE
Let 𝑦1 , 𝑦2 , … 𝑦𝑘 be solutions of the homogeneous 𝑛𝑡ℎ order differential
equations on an interval 𝐼. Then the linear combination
𝒚 = 𝑪𝟏 𝒚𝟏 (𝒙) + 𝑪𝟐 𝒚𝟐 (𝒙) + ⋯ + 𝑪𝒌 𝒚𝒌 (𝒙)
Where the 𝐶𝑖= 1,2,3 … 𝑘 are arbitrary constants, is also a solution on the interval.
HOMOGENEOUS LINEAR DE WITH CONSTANT COEFFICIENTS
AUXILIARY EQUATION
Considering the special case of the second-order equation
𝒂𝒚′′ + 𝒃𝒚′ + 𝒄𝒚 = 𝟎
(7)
Where 𝑎, 𝑏, and 𝑐 are constants. If we try to find a solution of the form 𝑦 = 𝑒 𝑚𝑥 , then
after substitution of 𝑦 ′ = 𝑚𝑒 𝑚𝑥 and 𝑦 ′′ = 𝑚2 𝑒 𝑚𝑥 , equation (7) becomes
𝒂𝒎𝟐 𝒆𝒎𝒙 + 𝒃𝒎𝒆𝒎𝒙 +𝑪𝒆𝒎𝒛 = 𝟎
Or
𝒆𝒎𝒙 (𝒂𝒎𝟐 + 𝒃𝒎 + 𝑪) = 𝟎
We argue that because 𝑒 𝑚𝑥 ≠ 0 for all 𝑥, it is apparent that the only way 𝑦 = 𝑒 𝑚𝑥 can
satisfy the differential equation (7) is when 𝑚 is chosen as a root of the quadratic
equation
𝒂𝒎𝟐 + 𝒃𝒎 + 𝒄 = 𝟎
(8)
This last equation is called the auxiliary equation of the differential equation (7).
Since the two roots of (8) 𝑚1 =
−𝑏+√𝑏 2 −4𝑎𝑐
2𝑎
and 𝑚2 =
−𝑏−√𝑏 2 −4𝑎𝑐
2𝑎
, there will be three
forms of the general solutions of (7) corresponding to the three cases:



𝑚1 and 𝑚2 real and distinct (𝑏 2 − 4𝑎𝑐 > 0),
𝑚1 and 𝑚2 real and equal (𝑏 2 − 4𝑎𝑐 = 0), and
𝑚1 and 𝑚2 conjugate complex numbers (𝑏 2 − 4𝑎𝑐 < 0).
CASE 1: DISTINCT REAL ROOTS
Under the assumption that the auxiliary equation (8) has two unequal real roots 𝑚1
and 𝑚2 , we find the solution 𝑦1 = 𝑒 𝑚1 𝑥 and 𝑦2 = 𝑒 𝑚2 𝑥 . We see that these functions are
linearly independent on (−∞, ∞) and hence form a fundamental set. It follows that the
general solution of (7) on this interval is
𝒚 = 𝑪 𝟏 𝒆𝒎 𝟏 𝒙 + 𝑪 𝟐 𝒆𝒎 𝟐 𝒙
For higher-order equations,
𝒚 = 𝑪 𝟏 𝒆𝒎 𝟏 𝒙 + 𝑪 𝟐 𝒆𝒎 𝟐 𝒙 + ⋯ 𝑪 𝒏 𝒆𝒎 𝒏 𝒙
EXAMPLES:
4. Solve 𝑦 ′′ − 𝑦 ′ − 2𝑦 = 0
(9)
The characteristics (or auxiliary) equation is 𝑚2 − 𝑚 − 2 = 0, which can be
factored into (𝑚 + 1)(𝑚 − 2) = 0. Since the roots 𝑚 = −1 and 𝑚 = 2 are real and
distinct, the solution is
𝒚 = 𝑪𝟏 𝒆−𝒙 + 𝑪𝟐 𝒆𝟐𝒙
𝑑3 𝑦
𝑑2 𝑦
𝑑𝑦
2. Solve 𝑑𝑥 3 + 2 𝑑𝑥 2 − 5 𝑑𝑥 − 6𝑦 = 0
We rewrite the equation as (𝑚3 + 2𝑚2 − 5𝑚 − 6) = (𝑚 − 2)(𝑚 + 1)(𝑚 + 3) =
0. Then the characteristic roots are 2, −1, and −3. They are real and distinct, so the
solution is
𝒚 = 𝑪𝟏 𝒆𝟐𝒙 + 𝑪𝟐 𝒆−𝒙 + 𝑪𝟑 𝒆−𝟑𝒙
3. Solve (3𝐷3 + 5𝐷 2 − 2𝐷)𝑦 = 0
1
The auxiliary equation is 3𝑚3 + 5𝑚2 − 2𝑚 = 0 and its roots are 𝑚 = 0, −2, 3.
Therefore the solution is
𝟏
𝒚 = 𝑪𝟏 + 𝑪𝟐 𝒆−𝟐𝒙 + 𝑪𝟑 𝒆𝟑𝒙
4. Solve 𝑦 ′′ − 36𝑦 = 0
The auxiliary equation is 𝑚2 − 36 = 0 and its roots are 𝑚 = 6, −6.
Therefore the solution is
𝒚 = 𝑪𝟏 𝒆𝟔𝒙 + 𝑪𝟐 𝒆−𝟔𝒙
CASE 2: CONJUGATE COMPLEX (IMAGINARY) ROOTS
If 𝑚1 and 𝑚2 are complex, then we can write 𝑚1 = ∝ +𝑖𝛽 and 𝑚2 = ∝ −𝑖𝛽, where
∝ and 𝛽 > 0 are real and 𝑖 2 = −1. Formally, there is no difference between this case
and case 1, and hence
𝒚 = 𝑪𝟏 𝒆(∝+𝒊𝜷)𝒙 + 𝑪𝟐 𝒆(∝−𝒊𝜷)𝒙
However, in practice we prefer to work with real functions instead of complex
exponentials. To this end we use the Euler’s Formula:
𝒆𝒊𝜽 = 𝒄𝒐𝒔𝜽 + 𝒊𝒔𝒊𝒏𝜽
where 𝜃 is any real number. It follows from this formula that
𝒆𝒊𝜷𝒙 = 𝒄𝒐𝒔𝜷𝒙 + 𝒊𝒔𝒊𝒏𝜷𝒙 and 𝒆−𝒊𝜷𝒙 = 𝒄𝒐𝒔𝜷𝒙 − 𝒊𝒔𝒊𝒏𝜷𝒙
(10)
where we have used cos(−𝛽𝑥) = cos 𝛽𝑥 and sin(−𝛽𝑥) = − sin 𝛽𝑥. Note that by first
adding and then substituting the two equations in (10), we obtain, respectively,
𝒆𝒊𝜷𝒙 + 𝒆−𝒊𝜷𝒙 = 𝟐 𝐜𝐨𝐬 𝜷𝒙 and 𝒆𝒊𝜷𝒙 − 𝒆−𝒊𝜷𝒙 = 𝟐𝒊 𝐬𝐢𝐧 𝜷𝒙
But 𝒚𝟏 = 𝒆𝒂𝒙 (𝒆𝒊𝜷𝒙 + 𝒆−𝒊𝜷𝒙 ) = 𝟐𝒆𝒂𝒙 𝐜𝐨𝐬 𝜷𝒙 and 𝒚𝟐 = 𝒆𝒂𝒙 (𝒆𝒊𝜷𝒙 − 𝒆−𝒊𝜷𝒙 ) = 𝟐𝒊𝒆𝒂𝒙 𝐬𝐢𝐧 𝜷𝒙
Hence, the last two results show that 𝑒 𝑎𝑥 cos 𝛽𝑥 and 𝑒 𝑎𝑥 sin 𝛽𝑥 are real solutions of (7).
Moreover, these solutions form a fundamental set on(−∞, ∞). Consequently, the
general solution is
𝒚 = 𝑪𝟏 𝒆𝒂𝒙 𝒄𝒐𝒔 𝜷𝒙 + 𝑪𝟐 𝒆𝒂𝒙 𝒔𝒊𝒏 𝜷𝒙
EXAMPLES:
1. Solve 𝑦 ′′ + 4𝑦 ′ + 7𝑦 = 0
Rewriting into auxiliary equations, 𝑚2 + 4𝑚 + 7 = 0, using the quadratic formula,
the roots are found to be 𝑚 = −2 ± √3𝑖. The corresponding general solution is
𝒚 = 𝑪𝟏 𝒆−𝟐𝒕 𝐜𝐨𝐬 √𝟑𝒙 + 𝑪𝟐 𝒆−𝟐𝒕 𝐬𝐢𝐧 √𝟑𝒙
2. Solve 3𝑦 ′′ + 2𝑦 ′ + 𝑦 = 0
1
1
The auxiliary equation is 3𝑚2 + 2𝑚 + 1 = 0 and the roots are 𝑚 = − 3 ± 3 √2𝑖. Thus,
the general solution becomes
𝟏
𝒚 = 𝑪𝟏 𝒆−𝟑𝒙 𝒄𝒐𝒔
𝟏
𝟏
𝟏
√𝟐𝒙 + 𝑪𝟐 𝒆−𝟑𝒙 𝒔𝒊𝒏 √𝟐𝒙
𝟑
𝟑
3. Solve (𝐷3 − 3𝐷2 + 9𝐷 + 13)𝑦 = 0
The auxiliary equation is 𝑚3 − 3𝑚2 + 9𝑚 = 13 and the roots are 𝑚 = −1, 2 ± 3𝑖.
Hence, the general solution of the differential equation is
𝒚 = 𝑪𝟏 𝒆−𝒙 + 𝑪𝟐 𝒆𝟐𝒙 𝒄𝒐𝒔 𝟑𝒙 + 𝑪𝟑 𝒆𝟐𝒙 𝒔𝒊𝒏 𝟑𝒙
4. Solve 𝑦 ′′ + 4𝑦 = 0
The auxiliary equation is 𝑚2 + 4 = 0 and the roots are 𝑚 = ±2𝑖. Hence the
general solution is
𝒚 = 𝑪𝟏 𝐜𝐨𝐬 𝟐𝒙 + 𝑪𝟐 𝐬𝐢𝐧 𝟐𝒙
CASE 3A: REPEATED REAL ROOTS
When 𝑚1 = 𝑚2 , we necessarily obtain only one exponential solution, 𝑦1 = 𝑒 𝑚𝑥 .
𝑏
From the quadratic formula we find that 𝑚1 = − 2𝑎. Since the only way to have 𝑚1 =
𝑚2 is to have 𝑏 2 − 4𝑎𝑐 = 0. It follows that a second solution of the equation is
𝒆𝟐𝒎𝟏 𝒙
𝒎𝟏 𝒙
𝒚𝟐 = 𝒆
∫ 𝟐𝒎 𝒙 𝒅𝒙 = 𝒆𝒎𝟏 𝒙 ∫ 𝒅𝒙 = 𝒙𝒆𝒎𝟏 𝒙
𝒆 𝟏
The general solution is then
𝒚 = 𝑪𝟏 𝒆𝒎𝟏 𝒙 + 𝑪𝟐 𝑿𝒆𝒎𝟐 𝒙
For higher-order equations,
𝒚 = 𝑪𝟏 𝒆𝒎𝟏 𝒙 + 𝑪𝟐 𝑿𝒆𝒎𝟐 𝒙 + 𝑪𝟑 𝑿𝟐 𝒆𝒎𝟑 𝒙 + ⋯ + 𝑪𝒌 𝑿𝒌−𝟏 𝒆𝒎𝒌𝒙
EXAMPLES:
1. Solve 𝑦 (5) − 3𝑦 (4) + 3𝑦 (3) − 𝑦 ′′ = 0
The characteristic equation is 𝑚5 − 3𝑚4 + 3𝑚3 − 𝑚2 = 𝑚2 (𝑚 − 1)3 = 0, which
has roots 𝑚 = 0, 0, 1, 1, 1. Hence, the general solution is
𝒚 = 𝑪𝟏 + 𝑪𝟐 𝑿 + 𝑪𝟑 𝒆𝒙 + 𝑪𝟒 𝑿𝒆𝒙 + 𝑪𝟓 𝑿𝟐 𝒆𝒙
𝑑4 𝑦
𝑑3 𝑦
𝑑2 𝑦
2. Solve 𝑑𝑥 4 + 2 𝑑𝑥 3 + 𝑑𝑥 2 = 0
The auxiliary equation is 𝑚4 + 2𝑚3 + 𝑚2 = 0, with roots 𝑚 = 0, 0, −1, −1. Hence
the desired equation is
𝒚 = 𝑪𝟏 + 𝑪𝟐 𝑿 + 𝑪𝟑 𝒆−𝒙 + 𝑪𝟒 𝑿𝒆−𝒙
3. Solve (𝐷3 − 4𝐷2 + 4𝐷)𝑦 = 0
The auxiliary equation is 𝑚3 − 4𝑚2 + 4𝑚 = 0 with roots 𝑚 = 0, 2 ,2. Thus, the
desired solution
𝒚 = 𝑪𝟏 + 𝑪𝟐 𝒆𝟐𝒙 + 𝑪𝟑 𝑿𝒆𝟐𝒙
4. Solve 4𝑦 ′′ + 14𝑦 ′ + 12 = 0
3
The auxiliary equation is 4𝑚2 + 14𝑚′ + 12 = 0 with roots 𝑚 = −2, − 2. Thus, the
desired solution
𝟑
𝒚 = 𝑪𝟏 𝒆−𝟐𝒙 + 𝑪𝟐 𝒆−𝟐𝒙
CASE 3B: REPEATED CONJUGATE COMPLEX ROOTS
If 𝑚 = 𝑎 ± 𝛽𝑖 are complex conjugate roots each appears 𝑘 times, then the general
solution is,
𝒚 = 𝑪𝟏 𝒆𝒂𝒙 𝒄𝒐𝒔 𝜷𝒙 + 𝑪𝟐 𝒆𝒂𝒙 𝒔𝒊𝒏 𝜷𝒙 + 𝑪𝟑 𝑿𝒆𝒂𝒙 𝒄𝒐𝒔 𝜷𝒙 + 𝑪𝟒 𝑿𝒆𝒂𝒙 𝒔𝒊𝒏 𝜷𝒙
+ 𝑪𝒌 𝑿𝒌−𝟏 𝒆𝒂𝒕 𝒄𝒐𝒔 𝜷𝒙 + 𝑪𝒌 𝑿𝒌−𝟏 𝒆𝒂𝒕 𝒔𝒊𝒏 𝜷𝒙
EXAMPLES:
1. Solve 𝑦 (4) + 𝑦 (3) + 8𝑦 ′′ + 8𝑦 ′ + 4𝑦 = 0
The characteristic equation is 𝑚4 + 4𝑚3 + 8𝑚2 + 8𝑚 + 4 = (𝑚2 + 2𝑚 + 2)2 = 0,
which has roots 𝑚 = −1 ± 𝑖, −1 ± 𝑖. Hence, the general solution is
𝒚 = 𝑪𝟏 𝒆−𝒙 𝒄𝒐𝒔 𝒙 + 𝑪𝟐 𝒆−𝒙 𝒔𝒊𝒏 𝒙 + 𝑪𝟑 𝑿𝒆−𝒙 𝒄𝒐𝒔 𝒙 + 𝑪𝟒 𝑿𝒆−𝒙 𝒔𝒊𝒏𝒙
𝑑4 𝑦
𝑑2 𝑦
2. Solve 16 𝑑𝑥 4 + 24 𝑑𝑥 2 + 9𝑦 = 0
The auxiliary equation is 16𝑚4 + 24𝑚2 + 9 = (4𝑚2 + 3)2 = 0 and the roots are 𝑚 =
1
1
± 2 √3𝑥, ± 2 √3𝑥. The general solution becomes
𝒚 = 𝑪𝟏 𝒄𝒐𝒔
𝟏
𝟏
𝟏
𝟏
√𝟑𝒙 + 𝑪𝟐 𝒔𝒊𝒏 √𝟐𝒙 + 𝑪𝟑 𝑿𝒄𝒐𝒔 √𝟑𝒙 + 𝑪𝟒 𝑿𝒔𝒊𝒏 √𝟐𝒙
𝟐
𝟑
𝟐
𝟑
3. Solve 𝑦 ′′ − 8𝑦 ′ + 17𝑦 = 0
The auxiliary equation is 𝑚2 − 8𝑚 + 17 = 0 and the roots are 𝑚 = 4 ± 𝑖. The
general solution becomes
𝒚 = 𝑪𝟏 𝒆𝟒𝒙 𝐜𝐨𝐬 𝒙 + 𝑪𝟐 𝒆𝟒𝒙 𝐬𝐢𝐧 𝒙
EXERCISES:
𝑑3 𝑦
𝑑2 𝑦
𝑑𝑦
1. 𝑑𝑥 3 + 4 𝑑𝑥 3 − 3 𝑑𝑥 − 18𝑦 = 0
𝑚3 + 4𝑚2 − 3𝑚 − 18 = 0
(𝑚 − 2)(𝑚 + 3)(𝑚 + 3) = 0
𝑚 = 2, −3, −3
𝒚 = 𝑪𝟏 𝒆𝟐𝒙 + 𝑪𝟐 𝒆−𝟑𝒙 + 𝑪𝟑 𝑿𝒆−𝟑𝒙
2. 𝑄 (5) + 5𝑄 (4) + 10𝑄 (3) + 10𝑄 ′′ + 5𝑄 ′ + 𝑄 = 0
𝑚5 + 5𝑚4 + 10𝑚3 + 10𝑚2 + 5𝑚 + 1 = 0
(𝑚 + 1)5
𝑚 = −1, −1, −1, −1, −1
𝒚 = 𝑪𝟏 𝒆−𝒙 + 𝑪𝟐 𝑿𝒆−𝒙 + 𝑪𝟑 𝑿𝟐 𝒆−𝒙 + 𝑪𝟒 𝑿𝟑 𝒆−𝒙 + 𝑪𝟓 𝑿𝟒 𝒆−𝒙
3. 𝑦 4 + 8𝑦 ′′ + 16𝑦 = 0
𝑚4 + 8𝑚2 + 16 = 0
(𝑚2 + 4)(𝑚2 + 4) = 0
𝑚 = ±2𝑖, ±2𝑖
𝒚 = 𝑪𝟏 𝒄𝒐𝒔 𝟐𝒙 + 𝑪𝟐 𝒔𝒊𝒏 𝟐𝒙 + 𝑪𝟑 𝑿𝒄𝒐𝒔 𝟐𝒙 + 𝑪𝟒 𝑿𝒔𝒊𝒏𝟐𝒙
4. Find the general solution of a sixth-order linear homogeneous differential
equation for 𝑦(𝑥) with real coefficients of its characteristic equation has roots 5𝑖 and
−5𝑖, each with multiplicity three.
𝒚 = 𝑪𝟏 𝒄𝒐𝒔 𝟓𝒙 + 𝑪𝟐 𝒔𝒊𝒏 𝟓𝒙 + 𝑪𝟑 𝑿𝒄𝒐𝒔 𝟓𝒙 + 𝑪𝟒 𝑿𝒔𝒊𝒏𝟓𝒙 + 𝑪𝟓 𝑿𝟐 𝒄𝒐𝒔 𝟓𝒙 + 𝑪𝟔 𝑿𝟐 𝒔𝒊𝒏 𝟓𝒙
5. Find a homogeneous differential equation with constant coefficients whose
general solution is 𝑦 = 𝐶1 + 𝐶2 𝑋 + 𝐶3 𝑒 8𝑥
Roots: 𝑚 = 0, 0 ,8
(𝑚)(𝑚)(𝑚 − 8) = 0
𝑚3 − 8𝑚2 = 0
𝒚(𝟑) − 𝟖𝒚′′ = 𝟎
6. Find the general solution to a fourth-order linear homogeneous differential
equation for 𝑥(𝑡) with real coefficients if two particular solutions are 3𝑒 2𝑡 and 6𝑡 2 𝑒 −𝑡 .
𝑚 = 2, −1, −1, −1
𝒚 = 𝑪𝟏 𝒆𝟐𝒕 + 𝑪𝟐 𝒆−𝒕 + 𝑪𝟑 𝒕𝒆−𝒕 + 𝑪𝟒 𝒕𝟐 𝒆−𝒕
7. Solve the given boundary-value problem:𝑦 ′′ − 10𝑦 ′ + 25𝑦 = 0
𝑦(0) = 1, 𝑦(1) = 0
𝑚2 − 10𝑚 + 25 = 0
𝑚 = 5,5
𝑦 = 𝐶1 𝑒 5𝑥 + 𝐶2 𝑋𝑒 5𝑥
1 = 𝐶1 𝑒 5(0) + 𝐶2 (0)𝑒 5(0)
𝐶1 = 1
0 = (1)𝑒 5(1) + 𝐶2 (1)𝑒 5(1)
𝐶2 = −1
𝒚 = 𝒆𝟓𝒙 − 𝒙𝒆𝟓𝒙
NON-HOMOGENEOUS DIFFERENTIAL EQUATION WITH
CONSTANT COEFFICIENTS
GENERAL SOLUTION – NON-HOMOGENEOUS EQUATIONS
Let 𝑦𝑝 be any particular solution of the non-homogeneous linear 𝑛𝑡ℎ order
differential equation on an interval 𝐼, and let 𝑦1 , 𝑦2 … 𝑦𝑝 be a fundamental set of
solutions of the associated homogeneous differential equation. Then the general
solution of the equation on the interval is
𝒚 = 𝑪𝟏 𝒚𝟏 (𝒙) + 𝑪𝟐 𝒚𝟐 (𝒙) + ⋯ + 𝑪𝒏 𝒚𝒏 (𝒙) + 𝒚𝒑
Where 𝐶𝑖 , 𝑖 = 1,2, … 𝑛 are arbitrary constants.
The linear combination 𝑦𝑐 = 𝐶1 𝑦1 (𝑥) + 𝐶2 𝑦2 (𝑥) + ⋯ + 𝐶𝑛 𝑦𝑛 (𝑥) which is the
general solution of a homogeneous linear differentia equation is called the
complementary function. In other words, to solve a non-homogeneous linear
differential equation, we first solve the associated homogeneous equation and then
find any particular solution of the non-homogeneous equation. The general solution of
the non-homogeneous equation is then
𝒚 = 𝒚𝒄 + 𝒚𝒑
SOLUTION BY METHOD OF UNDETERMINED FUNCTION
A. EQUATIONS WITH EXPONENTIAL RIGHT SIDE
EXAMPLES:
1. Solve 𝑦 ′ − 5𝑦 = 𝑒 2𝑥
I. 𝑦𝑐 = 𝐶1 𝑒 5𝑥
II. 𝑦𝑝 = 𝐴𝑜 𝑒 2𝑥
𝑦 ′ = 2𝐴𝑜 𝑒 2𝑥
Substitute to equation (1)
2𝐴𝑜 𝑒 2𝑥 − 5𝐴𝑜 𝑒 2𝑥 = 𝑒 2𝑥
(2−5)𝐴𝑜 𝑒 2𝑥
𝑒 2𝑥
=
(1)
𝑒 2𝑥
𝑒 2𝑥
(2 − 5)𝐴𝑜 = 1
−3𝐴𝑜 = 1
1
𝐴𝑜 = − 3
𝟏
𝒚 = 𝑪𝟏 𝒆𝟓𝒙 − 𝟑 𝒆𝟐𝒙
2. Solve 𝑦 ′′ − 7𝑦 ′ = 6𝑒 6𝑥
I. 𝑦𝑐 = 𝐶1 + 𝐶2 𝑒 7𝑥
II. 𝑦𝑝 = 𝐴𝑜 𝑒 6𝑥
𝑦 ′ = 6𝐴𝑜 𝑒 6𝑥
𝑦 ′′ = 36𝐴𝑜 𝑒 6𝑥
Substitute to equation (1)
(1)
36𝐴𝑜 𝑒 6𝑥 − 7(6𝐴𝑜 𝑒 6𝑥 ) = 6𝑒 6𝑥
−6𝐴𝑜 𝑒 6𝑥 = 6𝑒 6𝑥
𝐴𝑜 = −1
𝒚 = 𝑪𝟏 + 𝑪𝟐 𝒆𝟕𝒙 − 𝒆𝟔𝒙
3. Solve
𝑑3 𝑄
𝑑𝑡 3
I.
II.
𝑑2 𝑄
𝑑𝑄
− 5 𝑑𝑄2 + 25 𝑑𝑡 − 125𝑄 = −60𝑒 7𝑡
(1)
𝑄𝑐 = 𝐶1 𝑒 5𝑡 + 𝐶2 cos 5𝑡 + 𝐶3 sin 5𝑡
𝑄𝑝 = 𝐴𝑜 𝑒 7𝑡
15
𝑄 ′ = 7𝐴𝑜 − 37
𝑄 ′′ = 49𝐴𝑜 𝑒 7𝑡
𝑄 ′′′ = 343𝐴𝑜 𝑒 7𝑡
Substitute to equation (1)
343𝐴𝑜 𝑒 7𝑡 − 5(49𝐴𝑜 𝑒 7𝑡 ) + 25(7𝐴𝑜 𝑒 7𝑡 ) − 125𝐴𝑜 𝑒 7𝑡 = −60𝑒 7𝑡
148𝐴𝑜 𝑒 7𝑡 = −60𝑒 7𝑡
15
𝐴𝑜 = − 37
𝟏𝟓
𝑸 = 𝑪𝟏 𝒆𝟓𝒕 + 𝑪𝟐 𝐜𝐨𝐬 𝟓𝒕 + 𝑪𝟑 𝐬𝐢𝐧 𝟓𝒕 − 𝟑𝟕 𝒆𝟕𝒕
1. Solve 𝑦 ′′ − 36𝑦 = 4𝑒 2𝑥
𝑦𝑐 = 𝐶1 𝑒 6𝑥 + 𝐶2 𝑒 −6𝑥
𝑦𝑝 = 𝐴0 𝑒 2𝑥
𝑦′ = 2𝐴0 𝑒 2𝑥
𝑦′′ = 4𝐴0 𝑒 2𝑥
Substitute
4𝐴0 𝑒 2𝑥 − 36𝐴0 𝑒 2𝑥 = 4𝑒 2𝑥
−32𝐴0 = 4
𝐴0 = −
1
8
𝟏
𝒚 = 𝑪𝟏 𝒆𝟔𝒙 + 𝑪𝟐 𝒆−𝟔𝒙 − 𝒆𝟐𝒙
𝟖
B. EQUATIONS WITH CONSTANT RIGHT SIDE
1. Solve 𝑦 ′ − 5𝑦 = 8
We assume a particular solution of the form 𝑦𝑝 = 𝐴𝑜 where 𝐴𝑜 is a
constant to be determined. The general solution of the associated homogeneous
equation is to be 𝑦𝑐 = 𝐶1 𝑒 5𝑥 . Since any non-zero constant 𝐴𝑜 is linearly independent
of 𝑒 5𝑥 , there is no need to modify 𝑦𝑝 .
Substituting 𝑦𝑝 into the non-homogeneous equation and that noting that
𝑦′𝑝 = 0, we get
0 − 5𝐴𝑜 = 8
8
𝐴𝑜 = −
5
The general solution to the non-homogeneous equation is 𝑦 = 𝑦𝑐 + 𝑦𝑝
𝟖
𝒚 = 𝑪𝟏 𝒆𝟓𝒙 −
𝟓
2. Solve 𝑦 ′′ − 𝑦 ′ − 2𝑦 = 7
We assume a particular solution of the form 𝑦𝑝 = 𝐴𝑜 where 𝐴𝑜 is a
constant to be determined. The general solution of the associated homogeneous
equation is to be 𝑦𝑐 = 𝐶1 𝑒 −𝑥 + 𝐶2 𝑒 2𝑥 . Since 𝑦𝑝 cannot be obtained from 𝑦𝑐 by any
choice of 𝐶1 , there is no need to modify it.
Substituting 𝑦𝑝 and its derivatives (all of which are zero) into the nonhomogeneous differential equation, we get
0 − 0 − 2𝐴𝑜 = 7
7
𝐴𝑜 = −
2
The general solution to the non-homogeneous equation is
𝟕
𝒚 = 𝑪𝟏 𝒆−𝒙 + 𝑪𝟐 𝒆𝟐𝒙 −
𝟐
3. Solve
𝑑3 𝑥
𝑑𝑡 3
𝑑2 𝑥
𝑑𝑥
+ 5 𝑑𝑡 2 + 26 𝑑𝑡 − 150𝑥 = 30
We assume a particular solution of the form 𝑥𝑝 = 𝐴𝑜 . The general
solution of the associated homogeneous equation is to be 𝑥𝑐 = 𝐶1 𝑒 3𝑡 +
𝑒 −4𝑡 (𝐶2 𝑐𝑜𝑠√34𝑡 + 𝐶3 𝑠𝑖𝑛√34𝑡). Since 𝑥𝑝 cannot be obtained from 𝑥𝑐 no matter how the
arbitrary constants 𝐶1 through 𝐶3 are chosen, there is no need to modify 𝑥𝑝 .
Substituting 𝑥𝑝 into the given differential equation, we get
0 + 5(0) + 26(0) − 150𝐴𝑜 = 30
1
𝐴𝑜 = −
5
The general solution is then
𝒙 = 𝑪𝟏 𝒆𝟑𝒕 + 𝒆−𝟒𝒕 (𝑪𝟐 𝒄𝒐𝒔√𝟑𝟒𝒕 + 𝑪𝟑 𝒔𝒊𝒏√𝟑𝟒𝒕) −
𝟏
𝟓
4. Solve 𝑦 ′′ − 8𝑦 ′ + 17𝑦 = 442
𝑦𝑐 = 𝐶1 𝑒 4𝑥 cos 𝑥 + 𝐶2 𝑒 4𝑥 sin 𝑥
𝑦𝑝 = 𝐴0
Substitute
0 − 0 + 17𝐴0 = 442
𝐴0 = 26
𝒚 = 𝑪𝟏 𝒆𝟒𝒙 𝐜𝐨𝐬 𝒙 + 𝑪𝟐 𝒆𝟒𝒙 𝐬𝐢𝐧 𝒙 + 𝟐𝟔
C. EQUATIONS WITH POLYNOMIAL RIGHT SIDE
1. Solve 𝑦 ′ − 5𝑦 = 3𝑥 + 1
Since the right side of the differential equation is a first-degree
polynomial, we try a general first-degree polynomial as a particular solution. We
assume 𝑦𝑝 = 𝐴1 𝑥 + 𝐴𝑜 , where the coefficients 𝐴1 and 𝐴𝑜 must be determined. The
solution to the associated homogeneous equation is shown to be 𝑦𝑐 = 𝐶1 𝑒 5𝑥 . Since no
part of 𝑦𝑝 solves the homogeneous equation, there is no need to modify 𝑦𝑝 .
Substituting 𝑦𝑝 into the the given non-homogeneous equation and noting
that 𝑦′𝑝 = 𝐴1 , we get
𝐴1 − 5(𝐴1 𝑥 + 𝐴𝑜 ) = 3𝑥 + 1
Equating the coefficients of like power of 𝑥, we obtain
−5𝐴1 = 3
𝐴1 − 5𝐴𝑜 = 1
3
𝐴1 = − 5
3
− 5 − 5𝐴𝑜 = 1
8
𝐴𝑜 = − 25
𝟑
𝟖
𝒚 = 𝑪𝟏 𝒆𝟓𝒙 − 𝒙 −
𝟓
𝟐𝟓
2. Solve 𝑦 ′′ − 𝑦 ′ − 2𝑦 = 4𝑥 2
We assume a particular solution of the form 𝑦𝑝 = 𝐴1 𝑥 2 + 𝐴1 𝑥 + 𝐴𝑜 . The
general solution to the associated homogeneous differential equation is found to be
𝑦𝑐 = 𝐶1 𝑒 −𝑥 + 𝐶2 𝑒 2𝑥 . Since 𝑦𝑝 and 𝑦𝑐 have no terms in common except perhaps for a
multiplicative constant, there is no need to modify 𝑦𝑝 .
Differentiating𝑦𝑝 , we get 𝑦′𝑝 = 2𝐴2 𝑥 + 𝐴1 and 𝑦′′𝑝 = 2𝐴2 . Substituting these
derivatives to the given differential equation, we get
2𝐴2 − (2𝐴2 𝑥 + 𝐴1 ) − 2(𝐴1 𝑥 2 + 𝐴1 𝑥 + 𝐴𝑜 ) = 4𝑥 2
Equating the coefficients of like powers of 𝑥, we obtain
−2𝐴2 = 4
−2𝐴2 𝑥 + 2𝐴1 = 0
2𝐴2 − 𝐴1 − 2𝐴𝑜 = 0
𝐴2 = −2
𝐴1 = 2
𝐴𝑜 = −3
Then 𝑦𝑝 = −2𝑥 2 + 2𝑥 − 3, and the general solution is
𝒚 = 𝑪𝟏 𝒆−𝒙 + 𝑪𝟐 𝒆𝟐𝒙 − 𝟐𝒙𝟐 + 𝟐𝒙 − 𝟑
3. Determine the form of a particular solution to
𝑑4 𝑦
𝑑𝑥 4
= 12𝑥 2 − 60
The complementary solution is 𝑦𝑐 = 𝐶1 + 𝐶2 𝑥 + 𝐶3 𝑥 2 + 𝐶4 𝑥 3 . Since the
right side of the given differential equation is a second-degree polynomial, we try, as
a particular solution, the general second-degree polynomial 𝐴𝑜 + 𝐴1 𝑥 + 𝐴1 𝑥 2 . But this
is a part of 𝑦𝑐 for suitable choices of 𝐶1 through 𝐶4 , and so must be modified. To do so,
we multiply 𝑥 4 , the smallest positive integral power of 𝑥 that eliminated any duplication
of 𝑦𝑐 . The result is a proper particular solution
𝒚𝒑 = 𝑨𝒐 𝒙𝟒 + 𝑨𝟏 𝒙𝟓 + 𝑨𝟐 𝒙𝟔
5. Find 𝑦𝑝 of 𝑦′′ − 4𝑦′ − 12𝑦 = 2𝑡 3 − 𝑡 + 3
Once, again we will generally want the complementary solution in hand first, but
again we’re working with the same homogeneous differential equation (you’ll
eventually see why we keep working with the same homogeneous problem) so we’ll
again just refer to the first example.
For this example, the right side is a cubic polynomial. For this we will need the
following guess for the particular solution.
𝑦𝑝 = 𝐴𝑡 3 + 𝐵𝑡 2 + 𝐶𝑡 + 𝐷
𝑦′ = 3𝐴𝑡 2 + 2𝐵𝑡 + 𝐶
𝑦′′ = 6𝐴𝑡 + 2𝐵
6𝐴𝑡 + 2𝐵 − 4(3𝐴𝑡 2 + 2𝐵𝑡 + 𝐶) − 12(𝐴𝑡 3 + 𝐵𝑡 2 + 𝐶𝑡 + 𝐷) = 2𝑡 3 − 𝑡 + 3
−12𝐴𝑡 3 + (−12𝐴 − 12𝐵)𝑡 2 + (6𝐴 − 8𝐵 − 12𝐶)𝑡 + 2𝐵 − 4𝐶 − 12𝐷 = 2𝑡 3 − 𝑡 + 3
𝑡3 : −12𝐴 = 2 ⇒ 𝐴 = −1/6
𝑡2 : −12𝐴 − 12𝐵 = 0 ⇒ 𝐵 = 1/6
𝑡1 : 6𝐴 − 8𝐵 − 12𝐶 = −1 ⇒ 𝐶 = −1/9
𝑡0 : 2𝐵 − 4𝐶 − 12𝐷 = 3 ⇒ 𝐷 = −5/27
𝟏
𝟏
𝟏
𝟓
𝒚 𝒑 = − 𝒕𝟑 + 𝒕𝟐 − 𝒕 −
𝟔
𝟔
𝟗
𝟐𝟕
Below is the table of particular solutions for any given right side:
Form of g(x)
Guess for a Particular Solution
1 (any constant)
A
𝟓𝒙 + 𝟕
𝐴𝑥 + 𝐵
2
𝟐
𝟑𝒙 − 𝟐
𝐴𝑥 + 𝐵𝑥 + 𝐶
𝒔𝒊𝒏 𝟒𝒙
𝒄𝒐𝒔 𝟒𝒙
𝐴 cos 4𝑥 + 𝐵 sin 4𝑥
𝐴 cos 4𝑥 + 𝐵 sin 4𝑥
𝒆𝟓𝒙
(𝟗𝒙 − 𝟐)𝒆𝟓𝒙
𝒙𝟐 𝒆𝟓𝒙
𝒆𝟑𝒙 𝒔𝒊𝒏 𝟒𝒙
𝐴𝑒 5𝑥
(𝐴𝑥 + 𝐵)𝑒 5𝑥
(𝐴𝑥 2 + 𝐵𝑥 + 𝐶)𝑒 5𝑥
𝐴𝑒 3𝑥 cos 4𝑥 + 𝐵𝑒 3𝑥 sin 4𝑥
(𝐴𝑥 2 + 𝐵𝑥 + 𝐶) cos 4𝑥 + (𝐸 2 + 𝐹𝑥
+ 𝐺) sin 4𝑥
3𝑥
(𝐴𝑥 + 𝐵)𝑒 cos 4𝑥 + (𝐶𝑥 + 𝐸)𝑒 3𝑥 sin 4𝑥
𝟓𝒙𝟐 𝒔𝒊𝒏 𝟒𝒙
𝒙𝒆𝟑𝒙 𝒄𝒐𝒔 𝟒𝒙
(𝟓𝒙 + 𝟕) + 𝒔𝒊𝒏 𝟒𝒙
EXERCISES:
𝑑2 𝑦
(𝐴𝑥 + 𝐵) + (𝐶 cos 4𝑥 + 𝐷𝑠𝑖𝑛 4𝑥)
𝑑𝑦
1. Solve 𝑑𝑥 2 − 4 𝑑𝑥 + 𝑦 = 3𝑥 − 4
𝑚2 − 4𝑚 + 1 = 0
𝑚 = 2 ± √3
𝑦𝑐 = 𝐶1 𝑒 (2+√3)𝑥 + 𝐶2 𝑒 (2−√3)𝑥
𝑦𝒑 = 𝐴1 𝑥 + 𝐴𝑜
𝑦 ′ = 𝐴1
𝑦 ′′ = 0
0 − 4𝐴1 + 𝐴1 𝑥 + 𝐴𝑜 = 3𝑥 − 4
𝐴1 𝑥 = 3𝑥
−4𝐴1 + 𝐴𝑜 = −4
𝐴1 = 3
−12 + 𝐴𝑜 = −4
𝐴𝑜 = 8
𝒚 = 𝑪𝟏 𝒆(𝟐+√𝟑)𝒙 + 𝑪𝟐 𝒆(𝟐−√𝟑)𝒙 + 𝟑𝒙 + 𝟖
2. Solve 2𝑦 ′ − 5𝑦 = 2𝑥 2 − 5
2𝑚 − 5 = 0
5
𝑚=2
5
𝑦𝑐 = 𝐶1 𝑒 2𝑥
𝑦𝑝 = 𝐴2 𝑥 2 + 𝐴1 𝑥 + 𝐴𝑜
𝑦 ′ = 2𝐴2 𝑥 + 𝐴1
2(2𝐴2 𝑥 + 𝐴1 ) − 5(𝐴2 𝑥 2 + 𝐴1 𝑥 + 𝐴𝑜 ) = 2𝑥 2 − 5
4𝐴2 𝑥 + 2𝐴1 − 5𝐴2 𝑥 2 − 5𝐴1 𝑥 − 5𝐴𝑜 = 2𝑥 2 − 5
−5𝐴2 𝑥 2 = 2𝑥 2
4𝐴2 𝑥 − 5𝐴1 𝑥 = 0
2𝐴1 − 5𝐴𝑜 = −5
2
8
𝐴2 = − 5
𝐴1 = − 25
109
𝐴𝑜 = 125
𝟓
𝟐
𝟖
𝟏𝟎𝟗
𝒚 = 𝑪 𝟏 𝒆𝟐 𝒙 − 𝒙 𝟐 −
𝒙+
𝟓
𝟐𝟓
𝟏𝟐𝟓
𝑑2 𝑦
𝑑𝑦
3. Solve 𝑑𝑥 2 + 4 𝑑𝑥 + 8𝑦 = 8𝑥 4 + 16𝑥 3 − 12𝑥 2 − 24𝑥 − 6
𝑚2 + 4𝑚 + 8 = 0
𝑚 = −2 ± 2𝑖
𝑦𝑐 = 𝐶1 𝑒 −2𝑥 cos 2𝑥 + 𝐶2 𝑒 −2𝑥 sin 2𝑥
𝑦𝑝 = 𝐴4 𝑥 4 + 𝐴3 𝑥 3 + 𝐴2 𝑥 2 + 𝐴1 𝑥 + 𝐴𝑜
𝑦 ′ = 4𝐴4 𝑥 3 + 3𝐴3 𝑥 2 + 2𝐴2 𝑥 + 𝐴1
𝑦 ′′ = 12𝐴4 𝑥 2 + 6𝐴3 𝑥 + 2𝐴2
12𝐴4 𝑥 2 + 6𝐴3 𝑥 + 2𝐴2 + 4(4𝐴4 𝑥 3 + 3𝐴3 𝑥 2 + 2𝐴2 𝑥 + 𝐴1 )
+ 8(𝐴4 𝑥 4 + 𝐴3 𝑥 3 + 𝐴2 𝑥 2 + 𝐴1 𝑥 + 𝐴𝑜 ) = 8𝑥 4 + 16𝑥 3 − 12𝑥 2 − 24𝑥 − 6
8𝐴4 𝑥 4 = 8𝑥 4
𝐴4 = 1
16𝐴4 𝑥 3 + 8𝐴3 𝑥 3 = 16𝑥 3
𝐴3 = 0
12𝐴4 𝑥 2 + 12𝐴3 𝑥 2 + 8𝐴2 𝑥 2 = −12𝑥 2
𝐴2 = −3
6𝐴3 𝑥 + 8𝐴2 𝑥 + 8𝐴1 𝑥 = −24𝑥
2𝐴2 + 4𝐴1 + 8𝐴0 = −6
𝐴1 = 0
𝐴0 = 0
−𝟐𝒙
−𝟐𝒙
𝒚 = 𝑪𝟏 𝒆
𝐜𝐨𝐬 𝟐𝒙 + 𝑪𝟐 𝒆
𝐬𝐢𝐧 𝟐𝒙 + 𝒙𝟒 − 𝟑𝒙𝟐
EQUATIONS WHOSE RIGHT SIDE CONTAINS SINES AND COSINES
1. Solve 𝑦 ′ − 5𝑦 = sin 𝑥
The complementary solution is found to be 𝑦𝑐 = 𝐶1 𝑒 5𝑥 . We assume a particular
solution of the form 𝑦𝑝 = 𝐴𝑜 sin 𝑥 + 𝐵0 cos 𝑥, which needs no modification because 𝑦𝑝
and 𝑦𝑐 have no terms in common except for a multiplicative constant.
Substituting 𝑦𝑝 and its derivative into the given differential equation
𝐴𝑜 cos 𝑥 − 𝐵0 sin 𝑥 − 5(𝐴𝑜 sin 𝑥 + 𝐵0 cos 𝑥) = sin 𝑥
(−5𝐴𝑜 − 𝐵0 ) sin 𝑥 + (𝐴𝑜 − 5𝐵0 ) cos 𝑥 = 1(sin 𝑥) + 0(cos 𝑥)
Equating the coefficients of like terms, we obtain the system
−5𝐴𝑜 − 𝐵0 = 1 and 𝐴𝑜 − 5𝐵0 = 0
5
1
From which we find 𝐴𝑜 = − 26 and 𝐵0 = − 26. Then the general solution is
𝒚 = 𝑪𝟏 𝒆𝟓𝒙 −
𝟓
𝟏
𝐬𝐢𝐧 𝒙 −
𝐜𝐨𝐬 𝒙
𝟐𝟔
𝟐𝟔
2. Solve 𝑦 ′′ − 7𝑦 ′ = 48 sin 4𝑥 + 84 cos 4𝑥
The complementary solution is found to be 𝑦𝑐 = 𝐶1 + 𝐶2 𝑒 7𝑥 . We assume a
particular solution of the form 𝑦𝑝 = 𝐴𝑜 sin 4𝑥 + 𝐵0 cos 4𝑥, which needs no modification.
Substituting 𝑦′𝑝 = 4𝐴𝑜 cos 4𝑥 − 4𝐵0 sin 4𝑥 and 𝑦′′𝑝 = −16𝐴𝑜 sin 4𝑥 − 16𝐵𝑜 cos 4𝑥
into the given differential equation and rearranging give
(−16𝐴𝑜 + 28𝐵𝑜 ) sin 4𝑥 + (−28𝐴𝑜 − 16𝐵𝑜 ) cos 4𝑥 = 48 sin 4𝑥 + 84 cos 4𝑥
Equating coefficients of like terms and solving the resulting system, we get 𝐴𝑜 =
−3 and 𝐵𝑜 = 0. The general solution is then
𝒚 = 𝑪𝟏 + 𝑪𝟐 𝒆𝟕𝒙 − 𝟑 𝐬𝐢𝐧 𝟒𝒙
3. Solve
𝑑3 𝑄
𝑑𝑡 3
𝑑2 𝑄
𝑑𝑄
− 5 𝑑𝑡 2 + 25 𝑑𝑡 − 125𝑄 = 504 cos 2𝑡 − 651 sin 2𝑡
The complementary solution is found to be 𝑄𝑐 = 𝐶1 𝑒 5𝑡 + 𝐶2 cos 5𝑡 + 𝐶3 sin 5𝑡. We
assume a particular solution of the form 𝑄𝑝 = 𝐴𝑜 sin 2𝑡 + 𝐵0 cos 2𝑡, which needs no
modification.
Substituting 𝑄𝑝 into the given differential equation and rearranging
(−105𝐴𝑜 − 42𝐵𝑜 ) sin 2𝑡 + (42𝐴𝑜 − 105𝐵𝑜 ) cos 2𝑡 = −651 sin 2𝑡 + 504 cos 2𝑡
Equating coefficients of like terms and solving the resulting system, we find 𝐴𝑜 =
7 and 𝐵𝑜 = −2, so the general solution is
𝒚 = 𝑪𝟏 𝒆𝟓𝒕 + 𝑪𝟐 𝐜𝐨𝐬 𝟓𝒕 + 𝑪𝟑 𝐬𝐢𝐧 𝟓𝒕 + 𝟕 𝐬𝐢𝐧 𝟐𝒕 − 𝟐 𝐜𝐨𝐬 𝟐𝒕
4. Solve 𝑦′′ − 4𝑦′ − 12𝑦 = 𝑠𝑖𝑛(2𝑡)
𝑦𝑝 = 𝐴𝑠𝑖𝑛(2𝑡)
𝑦 ′ = 2𝐴 cos 2𝑡
𝑦 ′′ = −4𝐴 sin 2𝑡
Subtitute
−4𝐴𝑠𝑖𝑛(2𝑡) − 4(2𝐴𝑐𝑜𝑠(2𝑡)) − 12(𝐴𝑠𝑖𝑛(2𝑡)) = 𝑠𝑖𝑛(2𝑡)
−16𝐴𝑠𝑖𝑛(2𝑡) − 8𝐴𝑐𝑜𝑠(2𝑡) = 𝑠𝑖𝑛(2𝑡)
cos(2𝑡) : −8𝐴 = 0 ⇒ 𝐴 = 0
𝑠𝑖𝑛(2𝑡): −16𝐴 = 1 ⇒ 𝐴 = −
𝒚 = 𝑪𝟏 𝒆𝟔𝒙 + 𝑪𝟐 𝒆−𝟐𝒙 −
1
16
𝟏
𝐬𝐢𝐧 𝟐𝒕
𝟏𝟔
EQUATIONS WHOSE RIGHT SIDE IS THE PRODUCT OF A
POLYNOMIAL AND EXPONENTIAL
1. Solve 𝑦 ′ − 5𝑦 = 𝑥𝑒 2𝑥
Since the right side of this equation is the product of a first-degree polynomial
and an exponential, we assume a particular solution of the same form- a general firstdegree polynomial times an exponential. We try𝑦𝑝 = (𝐴1 𝑥 + 𝐴0 )𝑒 2𝑥 . The solution to
the associated homogeneous differential equation is found to be 𝑦𝑐 = 𝐶1 𝑒 5𝑥 . Since 𝑦𝑝
and 𝑦𝑐 have no terms in common, there is no need to modify 𝑦𝑝 .
Differentiating 𝑦𝑝 we get, 𝑦′𝑝 = 𝐴1 𝑒 2𝑥 + 2(𝐴1 𝑥 + 𝐴0 )𝑒 2𝑥 . Substituting these
values to the given differential equation, we obtain
𝐴1 𝑒 2𝑥 + 2(𝐴1 𝑥 + 𝐴0 )𝑒 2𝑥 − 5(𝐴1 𝑥 + 𝐴0 )𝑒 2𝑥 = 𝑥𝑒 2𝑥
−3𝐴1 𝑥 + (𝐴1 − 3𝐴𝑜 ) = 𝑥
Equating coefficients of like powers,
−3𝐴1 = 1
𝐴1 − 3𝐴𝑜 = 0
1
1
𝐴1 = − 3
𝐴𝑜 = − 9
𝟏
𝟏
𝒚 = 𝑪𝟏 𝒆𝟓𝒙 − 𝒙𝒆𝟐𝒙 − 𝒆𝟐𝒙
𝟑
𝟗
2. Solve 𝑦 ′′ − 7𝑦 ′ = (3 − 36𝑥)𝑒 4𝑥
We try 𝑦𝑝 = (𝐴1 𝑥 + 𝐴0 )𝑒 4𝑥 , a first-degree polynomial times an exponential, as a
particular solution. The complementary solution is known to be 𝑦𝑐 = 𝐶1 + 𝐶2 𝑒 7𝑥 . Since
𝑦𝑝 and 𝑦𝑐 have no term in common, there is no need to modify 𝑦𝑝 .
Substituting 𝑦′𝑝 = (4𝐴1 𝑥 + 𝐴1 + 4𝐴0 )𝑒 4𝑥 and 𝑦 ′′𝑝 = (16𝐴1 𝑥 + 8𝐴1 + 16𝐴0 )𝑒 4𝑥
into the differential equation and simplifying, we obtain
(−12𝐴1 )𝑥 + (𝐴1 − 12𝐴𝑜 ) = −36𝑥 + 3
Equating coefficients of like powers of 𝑥 yields a system of two equations from
which we find 𝐴1 = 3 and 𝐴𝑜 = 0. The general solution is then
𝒚 = 𝑪𝟏 + 𝑪𝟐 𝒆𝟕𝒙 + 𝟑𝒙𝒆𝟒𝒙
3. Solve
𝑑2 𝑥
𝑑𝑡 2
𝑑𝑥
+ 4 𝑑𝑡 + 8𝑥 = (29𝑡 3 + 30𝑡 2 − 52𝑡 − 20)𝑒 3𝑡
The complementary solution of the preceding two problems is found to be 𝑥𝑐 =
𝐶1 𝑒
cos 2𝑡 + 𝐶2 𝑒 −2𝑡 sin 2𝑡, and we try a particular solution of the form 𝑥𝑝 = (𝐴3 𝑡 3 +
𝐴2 𝑡 2 + 𝐴1 𝑡 + 𝐴𝑜 )𝑒 3𝑡 . This trial needs no modification.
Substituting 𝑥𝑝 , 𝑥′𝑝 = (3𝐴3 𝑡 3 + 3𝐴3 𝑡 2 + 3𝐴2 𝑡 2 + 2𝐴2 𝑡 + 3𝐴1 𝑡 + 𝐴1 + 3𝐴𝑜 )𝑒 3𝑡 and
𝑥′′𝑝 = (9𝐴3 𝑡 2 + 18𝐴3 𝑡 2 + 9𝐴2 𝑡 2 + 6𝐴3 𝑡 + 12𝐴2 𝑡 + 9𝐴1 𝑡 + 2𝐴2 + 6𝐴1 + 9𝐴𝑜 )𝑒 3𝑡
into
the given differential equation yields, after simplification,
(29𝐴3 )𝑡 3 + (30𝐴3 + 29𝐴2 )𝑡 2 + (6𝐴3 + 20𝐴2 + 29𝐴1 )𝑡 + (2𝐴2 + 10𝐴1 + 29𝐴𝑜 )
= 29𝑡 3 + 30𝑡 2 − 52𝑡 − 20
We find that 𝐴1 = 1, 𝐴2 = 0, 𝐴0 = 0. The general solution is thus
𝒙 = 𝑪𝟏 𝒆−𝟐𝒕 𝐜𝐨𝐬 𝟐𝒕 + 𝑪𝟐 𝒆−𝟐𝒕 𝐬𝐢𝐧 𝟐𝒕 + (𝒕𝟑 − 𝟐𝒕)𝒆𝟑𝒕
−2𝑡
4. Solve 𝑦′′ − 4𝑦′ − 12𝑦 = 𝑡𝑒 4𝑡
𝑦𝑝 = 𝐴𝑡𝑒 4𝑡 + 𝐵𝑒 4𝑡
𝑦 ′ = 4𝐴𝑡𝑒 4𝑡 + 𝐴𝑒 4𝑡 + 4𝐵𝑒 4𝑡
𝑦 ′ = 16𝐴𝑡𝑒 4𝑡 + 8𝐴𝑒 4𝑡 + 16𝐵𝑒 4𝑡
𝑒 4𝑡 (16𝐴𝑡 + 16𝐵 + 8𝐴) − 4(𝑒 4𝑡 (4𝐴𝑡 + 4𝐵 + 𝐴)) − 12(𝑒 4𝑡 (𝐴𝑡 + 𝐵)) = 𝑡𝑒 4𝑡
(16𝐴 − 16𝐴 − 12𝐴)𝑡𝑒 4𝑡 + (16𝐵 + 8𝐴 − 16𝐵 − 4𝐴 − 12𝐵)𝑒 4𝑡 = 𝑡𝑒 4𝑡
−12𝐴𝑡𝑒 4𝑡 + (4𝐴 − 12𝐵)𝑒 4𝑡 = 𝑡𝑒 4𝑡
1
12
1
𝑒 4𝑡 : 4𝐴 − 12𝐵 = 0 ⇒ 𝐵 = −
36
𝑡𝑒 4𝑡 : −12𝐴 = 1 ⇒ 𝐴 = −
𝒚 = 𝑪𝟏 𝒆𝟔𝒙 + 𝑪𝟐 𝒆−𝟐𝒙 −
𝟏
(𝟑𝒕 + 𝟏)𝒆𝟒𝒕
𝟑𝟔
EQUATIONS WHOSE RIGHT SIDE CONTAINS A PRODUCT
INVOLVING SINES AND COSINES
1. Solve 𝑦 ′ + 6𝑦 = 3𝑒 2𝑥 sin 3𝑥
The complementary solution is 𝑦𝑐 = 𝐶1 𝑒 −6𝑥 . Since the right side of the nonhomogeneous differential equation is the product of an exponential and a sine, we try
a particular solution of the form 𝑦𝑝 = 𝐴𝑜 𝑒 2𝑥 sin 3𝑥 + 𝐵𝑜 𝑒 2𝑥 cos 3𝑥.
Since 𝑦𝑝 has no term in common with 𝑦𝑐 , there is no need to modify 𝑦𝑝 .
Substituting
𝑦𝑝 and
𝑦′𝑝 = 2𝐴𝑜 𝑒 2𝑥 sin 3𝑥 + 3𝐴𝑜 𝑒 2𝑥 cos 3𝑥 + 2𝐵𝑜 𝑒 2𝑥 cos 3𝑥 −
3𝐵𝑜 𝑒 2𝑥 sin 3𝑥 into the differential equation yields
2𝐴𝑜 𝑒 2𝑥 sin 3𝑥 + 3𝐴𝑜 𝑒 2𝑥 cos 3𝑥 + 2𝐵𝑜 𝑒 2𝑥 cos 3𝑥 − 3𝐵𝑜 𝑒 2𝑥 sin 3𝑥 + 6𝐴𝑜 𝑒 2𝑥 sin 3𝑥
+ 6𝐵𝑜 𝑒 2𝑥 cos 3𝑥 = 3𝑒 2𝑥 sin 3𝑥
which may be rearranged to
(8𝐴𝑜 − 3𝐵𝑜 ) sin 3𝑥 + (3𝐴𝑜 + 8𝐵𝑜 ) cos 3𝑥 = 3 sin 3𝑥 + 0 cos 3𝑥
24
9
from which we find that 𝐴𝑜 = 73 and 𝐵𝑜 = − 73. The general solution is the
𝟐𝟒 𝟐𝒙
𝟗 𝟐𝒙
𝒚 = 𝑪𝟏 𝒆−𝟔𝒙 +
𝒆 𝐬𝐢𝐧 𝟑𝒙 −
𝒆 𝐜𝐨𝐬 𝟑𝒙
𝟕𝟑
𝟕𝟑
2. Solve 𝑦 ′′ + 6𝑦 ′ + 9𝑦 = 16𝑒 −𝑥 cos 2𝑥
The complementary solution is 𝑦𝑐 = 𝐶1 𝑒 −3𝑥 + 𝐶2 𝑥𝑒 −3𝑥 . We try a particular
solution of the form 𝑦𝑝 = 𝐴𝑜 𝑒 −𝑥 sin 2𝑥 + 𝐵𝑜 𝑒 −𝑥 cos 2𝑥, which needs no modification
because it has no term in common in 𝑦𝑐 .
Substituting 𝑦𝑝 and 𝑦′𝑝 = (−𝐴𝑜 − 2𝐵𝑜 )𝑒 −𝑥 sin 2𝑥 + (2𝐴𝑜 − 𝐵𝑜 )𝑒 −𝑥 cos 2𝑥, and
𝑦′′𝑝 = (−3𝐴𝑜 + 4𝐵𝑜 )𝑒 −𝑥 sin 2𝑥 + (−4𝐴𝑜 − 3𝐵𝑜 )𝑒 −𝑥 cos 2𝑥 into the given differential
equation and rearranging yield
[(−3𝐴𝑜 + 4𝐵𝑜 )𝑒 −𝑥 sin 2𝑥 + (−4𝐴𝑜 − 3𝐵𝑜 )𝑒 −𝑥 cos 2𝑥]
+ [(−𝐴𝑜 − 2𝐵𝑜 )𝑒 −𝑥 sin 2𝑥 + (2𝐴𝑜 − 𝐵𝑜 )𝑒 −𝑥 cos 2𝑥]
+ 9[𝐴𝑜 𝑒 −𝑥 sin 2𝑥 + 𝐵𝑜 𝑒 −𝑥 cos 2𝑥] = 16𝑒 −𝑥 cos 2𝑥
(−8𝐵𝑜 ) sin 2𝑥 + (8𝐴𝑜 ) cos 2𝑥 = 16 cos 2𝑥
By equating coefficients of like terms, we find 𝐴𝑜 = 2 and 𝐵𝑜 = 0. The general solution
is then
𝒚 = 𝑪𝟏 𝒆−𝟑𝒙 + 𝑪𝟐 𝒙𝒆−𝟑𝒙 + 𝟐𝒆−𝒙 𝐬𝐢𝐧 𝟐𝒙
3. Solve
𝑑3 𝑄
𝑑𝑡 3
𝑑2 𝑄
𝑑𝑄
− 5 𝑑𝑡 2 + 25 𝑑𝑡 − 125𝑄 = 5000 cos 2𝑡
The complementary solution is 𝑄𝑐 = 𝐶1 𝑒 5𝑡 + 𝐶2 cos 5𝑡 + 𝐶3 sin 5𝑡. We try a
particular solution of the form 𝑄𝑝 = 𝐴𝑜 𝑒 −𝑡 sin 2𝑡 + 𝐵𝑜 𝑒 −𝑡 cos 2𝑡, which needs no
modification. Since
𝑄′𝑝 = (𝐴𝑜 − 2𝐵𝑜 )𝑒 −𝑡 sin 2𝑡 + (2𝐴𝑜 − 𝐵𝑜 )𝑒 −𝑡 cos 2𝑡
𝑄′′𝑝 = (−3𝐴𝑜 + 4𝐵𝑜 )𝑒 −𝑡 sin 2𝑡 + (−4𝐴𝑜 − 3𝐵𝑜 )𝑒 −𝑡 cos 2𝑡
𝑄′′′𝑝 = (11𝐴𝑜 + 2𝐵𝑜 )𝑒 −𝑡 sin 2𝑡 + (−2𝐴𝑜 + 11𝐵𝑜 )𝑒 −𝑡 cos 2𝑡
And the given differential equation becomes, after substitution and simplification,
(−124𝐴𝑜 − 68𝐵𝑜 ) sin 2𝑡 + (68𝐴𝑜 − 124𝐵𝑜 )𝑐𝑜𝑠 2𝑡 = 5000 cos 2𝑡.
By
equating
coefficients of like terms and solving the resulting system, we find 𝐴𝑜 = 17 and 𝐵𝑜 =
−31, and the general solution is
𝑸 = 𝑪𝟏 𝒆𝟓𝒕 + 𝑪𝟐 𝐜𝐨𝐬 𝟓𝒕 + 𝑪𝟑 𝐬𝐢𝐧 𝟓𝒕 + 𝟏𝟕𝒆−𝒕 𝐬𝐢𝐧 𝟐𝒕 − 𝟑𝟏𝒆−𝒕 𝐜𝐨𝐬 𝟐𝒕
4. Solve 𝑦′′ − 4𝑦′ − 12𝑦 = 481𝑒 −3𝑡 cos 6𝑡
𝑦𝑐 = 𝐴𝑒 −3𝑡 cos 6𝑡 + 𝐵𝑒 −3𝑡 sin 6𝑡
𝑦 ′ = −3𝐴𝑒 −3𝑡 cos 6𝑡 − 6𝐴𝑒 −3𝑡 sin 6𝑡 − 3𝐵𝑒 −3𝑡 sin 6𝑡 + 6𝐵𝑒 −3𝑡 cos 6𝑡
𝑦 ′′ = −27 𝐴𝑒 −3𝑡 cos 6𝑡 + 36𝐴𝑒 −3𝑡 sin 6𝑡 − 36𝐵𝑒 −3𝑡 cos 6𝑡 − 27𝐵𝑒 −3𝑡 sin 6𝑡
Substitute
(−27𝐴 − 60𝐵)𝑒 −3𝑡 cos 6𝑡 + (60𝐴 − 27𝐵)𝑒 −3𝑡 sin 6𝑡 = 481𝑒 −3𝑡 cos 6𝑡
𝐴 = −3
20
𝐵=−
3
𝒚 = 𝑪𝟏 𝒆𝟔𝒙 + 𝑪𝟐 𝒆−𝟐𝒙 − 𝟑𝒆−𝟑𝒕 𝐜𝐨𝐬 𝟔𝒕 −
𝟐𝟎 −𝟑𝒕
𝒆 𝐬𝐢𝐧 𝟔𝒕
𝟑
VARIATION OF PARAMETERS
Consider the second-order differential equation
𝒂(𝒙)𝒚′′ + 𝒃(𝒙)𝒚′ + 𝒄(𝒙)𝒚 = 𝒇(𝒙)
(1)
The method of undetermined coefficients works only when the coefficients 𝑎, 𝑏
and 𝑐 are constants and the right-hand term 𝑓(𝑥) is a special form. If these restrictions
do not apply to a given non-homogeneous linear differential equation, then a more
powerful method of determining a particular solution is needed: the method known as
variation of parameters.
The first step is to obtain the general solution of the corresponding homogeneous
equation, which will have the form
𝒚𝒄 = 𝑪 𝟏 𝒚𝟏 + 𝑪 𝟐 𝒚𝟐
where 𝑦1 and 𝑦2 are known functions.
The next step is to vary the parameters: that is, to replace the constants 𝐶1 and
𝐶2 by (as yet unknown) functions 𝑉1 and 𝑉2 to obtain the form of a particular solution 𝑦
of the given non-homogeneous equation:
Ŷ = 𝑽𝟏 𝒚 𝟏 + 𝑽𝟐 𝒚 𝟐
The goal is to determine these functions 𝑉1 and 𝑉2. Then, since the functions 𝑦1
and 𝑦2 are already known, the expression above for 𝑦 yields a particular solution of
the non-homogeneous equation. Combining 𝑦 with 𝑦𝑐 then gives the general solutions
of the non-homogeneous differential equation.
Since there are two unknowns to be determined, 𝑉1 and 𝑉2, two equations or
conditions are required to obtain a solution. One of these conditions will naturally be
satisfying the given differential equation. But another condition will be imposed first.
Since 𝑦 will be substituted into equation (1), its derivative must be evaluated, the first
derivative of 𝑦 is
Ŷ′ = 𝑽𝟏 𝒚′ 𝟏 + 𝑽′ 𝟏 𝒚𝟏 + 𝑽𝟐 𝒚′ 𝟐 + 𝑽′𝟐 𝒚𝟐
Now, to simplify the rest of the process – and to produce the first condition on 𝑉1
and 𝑉2 – set
𝑽′𝟏 𝒚𝟏 + 𝑽′𝟐 𝒚𝟐 = 𝟎
This will always be the first condition in determining 𝑉1 and 𝑉2; the second
condition will be the satisfaction of the given differential equation.
For third-order differential equation
𝒅𝟑 𝒚 𝒅𝟐 𝒚 𝒅𝒚
+
+
+ 𝒃𝒚 = 𝒇(𝒙)
𝒅𝒙𝟑 𝒅𝒙𝟐 𝒅𝒙
If 𝑦𝑐 = 𝐶1 𝑦1 + 𝐶2 𝑦2 + 𝐶3 𝑦3 then set 𝑦𝑝 = 𝑉1 𝑦1 + 𝑉2 𝑦2 + 𝑉3 𝑦3 so that
𝑽′𝟏 𝒚𝟏 + 𝑽′𝟐 𝒚𝟐 + 𝑽′𝟑 𝒚𝟑 = 𝟎
𝑽′𝟏 𝒚′𝟏 + 𝑽′𝟐 𝒚′𝟐 + 𝑽′𝟑 𝒚′𝟑 = 𝟎
𝑽′𝟏 𝒚′′𝟏 + 𝑽′𝟐 𝒚′′𝟐 + 𝑽′𝟑 𝒚′′𝟑 = 𝒇(𝒙)
Solve for the 𝑉′1, 𝑉′2 , and 𝑉′3 by integration. Then substitute the values to
𝒚 𝒑 = 𝑽𝟏 𝒚 𝟏 + 𝑽𝟐 𝒚 𝟐 + 𝑽𝟑 𝒚 𝟑
Finally, the general solution is
𝑦 = 𝑪 𝟏 𝒚 𝟏 + 𝑪 𝟐 𝒚 𝟐 + 𝑪 𝟑 𝒚 𝟑 + 𝑽𝟏 𝒚 𝟏 + 𝑽𝟐 𝒚 𝟐 + 𝑽𝟑 𝒚 𝟑
EXAMPLES:
1. Solve the differential equation 𝑦 ′′ + 𝑦 = tan 𝑥
Because the characteristics equation 𝑚2 + 1 = 0 has solutions 𝑚 = ±𝑖, the
homogeneous (or complementary) solution is
𝑦𝑐 = 𝐶1 cos 𝑥 + 𝐶2 sin 𝑥
Replacing 𝐶1 and 𝐶2 by 𝑉1 and 𝑉2 produces
𝑦𝑝 = 𝑉1 cos 𝑥 + 𝑉2 sin 𝑥
The resulting system of equation is
𝑉′1 𝑐𝑜𝑠 𝑥 + 𝑉′2 𝑠𝑖𝑛 𝑥 = 0
(1)
−𝑉′1 𝑠𝑖𝑛 𝑥 + 𝑉′2 𝑐𝑜𝑠 𝑥 = tan 𝑥
(2)
Multiplying the (1) by sin 𝑥 and (2) by cos 𝑥 the adding these two equations, we obtain
𝑉′1 sin 𝑥 𝑐𝑜𝑠 𝑥 + 𝑉′2 𝑠𝑖𝑛2 𝑥 = 0
−𝑉′1 sin 𝑥 𝑐𝑜𝑠 𝑥 + 𝑉′2 𝑐𝑜𝑠 2 𝑥 = sin 𝑥
_______________________________
𝑉 ′ 2 (𝑠𝑖𝑛2 𝑥 + 𝑐𝑜𝑠 2 𝑥) = sin 𝑥
𝑉 ′ 2 = 𝑠𝑖𝑛𝑥
(3)
From (1)
𝑉′1 𝑐𝑜𝑠 𝑥 + (sin 𝑥)𝑠𝑖𝑛 𝑥 = 0
sin2 𝑥
𝑉′1 = −
cos 𝑥
=
cos2 𝑥−1
cos 𝑥
= cos 𝑥 − sec 𝑥
(4)
Integrating (4)
𝑉1 = ∫(cos 𝑥 − sec 𝑥)𝑑𝑥
= sin 𝑥 − ln(sec 𝑥 + tan 𝑥)
Integrating (3)
𝑉2 = ∫ sin 𝑥 𝑑𝑥
= −cos 𝑥
So that
𝑦𝑝 = 𝑉1 cos 𝑥 + 𝑉2 sin 𝑥
= [sin 𝑥 − ln(sec 𝑥 + tan 𝑥)] cos 𝑥 + [− cos 𝑥 sin 𝑥]
= 𝑠𝑖𝑛𝑥 cos 𝑥 − cos 𝑥 ln(sec 𝑥 + tan 𝑥) − sin 𝑥 cos 𝑥
= − cos 𝑥 ln(sec 𝑥 + tan 𝑥)
Finally, the general solution is
𝒚 = 𝑪𝟏 𝐜𝐨𝐬 𝒙 + 𝑪𝟐 𝐬𝐢𝐧 𝒙 − 𝐜𝐨𝐬 𝒙 𝐥𝐧(𝐬𝐞𝐜 𝒙 + 𝐭𝐚𝐧 𝒙)
2. Solve 𝑦 ′ − 5𝑦 = 𝑒 2𝑥
The complementary solution is 𝑦𝑐 = 𝐶1 𝑒 5𝑥 , so we assume 𝑦𝑝 = 𝑉1 𝑒 5𝑥 . Here 𝑦1 =
𝑒 5𝑥 and 𝑓(𝑥) = 𝑒 2𝑥 , and it follows that
𝑉′1 𝑒 5𝑥 = 𝑒 2𝑥 or 𝑉′1 = 𝑒 −3𝑥
Integrating, we get
𝑉1 = ∫ 𝑒 −3𝑥 𝑑𝑥
1
𝑉1 = − 𝑒 −3𝑥
3
𝟏
𝒚 = 𝑪𝟏 𝒆𝟓𝒙 − 𝒆−𝟑𝒙
𝟑
3. Solve 𝑦 ′′ − 7𝑦 = −3𝑥
The complementary solution is found to be 𝑦𝑐 = 𝐶1 + 𝐶2 𝑒 7𝑥 . We assume 𝑦𝑝 =
𝑉1 + 𝑉2 𝑒 7𝑥 . It follows that
𝑉′1 + 𝑉′2 𝑒 7𝑥 = 0
𝑉 ′1 (0) + 𝑉 ′ 2 (7𝑒 7𝑥 ) = −3𝑥
3
3
The solution to this set of equation is 𝑉′1 = 7 𝑥 and 𝑉′2 = − 7 𝑥𝑒 −7𝑥
Integrating both will produce
3
𝑉1 = ∫ 𝑥 𝑑𝑥
7
3
= 14 𝑥
Thus,
3
𝑉2 = ∫ − 𝑥𝑒 −7𝑥 𝑑𝑥
and
3
2
7
−7𝑥
= 49 𝑥𝑒
3
3
3
+ 343 𝑒 −7𝑥
3
𝑦𝑝 = 14 𝑥 2 + (49 𝑥𝑒 −7𝑥 + 343 𝑒 −7𝑥 )𝑒 −7𝑥
3 2
3
3
𝑥 + 𝑥+
14
49
343
𝟑
𝟑
𝟑
𝒚 = 𝑪𝟏 + 𝑪𝟐 𝒆𝟕𝒙 +
𝒙𝟐 +
𝒙+
𝟏𝟒
𝟒𝟗
𝟑𝟒𝟑
′′′
′
4. Solve 𝑦 + 𝑦 = sec 𝑥
The complementary solution is found to be 𝑦𝑐 = 𝐶1 + 𝐶2 cos 𝑥 + 𝐶3 𝑠𝑖𝑛𝑥 . So we
assume a particular solution of the form 𝑦𝑝 = 𝑉1 + 𝑉2 cos 𝑥 + 𝑉3 sin 𝑥. Here 𝑦1 = 1 and
𝑦2 = cos 𝑥, 𝑦3 = sin 𝑥 and 𝑓(𝑥) = sec 𝑥. Thus,
𝑉′1 + 𝑉′2 cos 𝑥 + 𝑉′3 sin 𝑥 = 0
(1)
′ (0)
𝑉1
− 𝑉′2 sin 𝑥 + 𝑉′3 cos 𝑥 = 0
(2)
′
𝑉 1 (0) + 𝑉′2 cos 𝑥 + 𝑉′3 sin 𝑥 = sec 𝑥
(3)
Combining (1) and (3) yields
𝑦𝑝 =
𝑉′1 = sec 𝑥
(4)
Dividing (2) by cos 𝑥 gives
𝑉′3 = −𝑉 ′ 2 tan 𝑥
Substituting (5) to (2) and simplifying obtains
𝑉′2 = −1
Substituting (6) to (5) thus yields
𝑉′3 = − tan 𝑥
Integrating (4), (6), and (7) obtain
𝑉1 = ln(sec 𝑥 + tan 𝑥)
𝑉2 = −𝑥
𝑉3 = ln(cos 𝑥)
(5)
(6)
(7)
Finally, the general solution is
𝒚 = 𝑪𝟏 + 𝑪𝟐 𝐜𝐨𝐬 𝒙 + 𝑪𝟑 𝐬𝐢𝐧 𝒙 + 𝐥𝐧(𝐬𝐞𝐜 𝒙 + 𝐭𝐚𝐧 𝒙) − 𝒙𝒄𝒐𝒔 𝒙 + 𝐥𝐧(𝐜𝐨𝐬 𝒙) 𝐬𝐢𝐧 𝒙
5. Solve 𝑦 ′ − 5𝑦 = sin 𝑥
𝑦𝑐 = 𝐶1 𝑒 5𝑥
𝑦𝑝 = 𝑉1 𝑒 5𝑥
𝑉 ′1 𝑒 5𝑥 = sin 𝑥
𝑉 ′1 = (sin 𝑥)𝑒 −5𝑥
𝑉1 = ∫ sin 𝑥 ∙ 𝑒 −5𝑥 𝑑𝑥
5
1
= (− 26 sin 𝑥 − 26 cos 𝑥)𝑒 −5𝑥
Using integration by parts twice
5
1
𝑦𝑝 = − 26 sin 𝑥 − 26 cos 𝑥
Finally, the general solution is
𝒚 = 𝑪𝟏 𝒆𝟓𝒙 −
6. Solve 𝑦 ′ − 5𝑦 = 𝑒 2𝑥
𝑦𝑐 = 𝐶1 𝑒 5𝑥
𝑦𝑝 = 𝑉1 𝑒 5𝑥
𝑉 ′1 𝑒 5𝑥 = 𝑒 2𝑥
𝑉 ′1 = 𝑒 −3𝑥
𝑉1 = ∫ 𝑒 −3𝑥 𝑑𝑥
1
= − 3 𝑒 −3𝑥
1
1
𝑦𝑝 = − 3 𝑒 −3𝑥 ∙ 𝑒 5𝑥 = − 3 𝑒 2𝑥
𝟏
𝒚 = 𝑪𝟏 𝒆𝟓𝒙 − 𝟑 𝒆𝟐𝒙
7. Solve 𝑦 ′ − 5𝑦 = 3𝑥 + 1
𝑦𝑐 = 𝐶1 𝑒 5𝑥
𝑦𝑝 = 𝑉1 𝑒 5𝑥
Solving for 𝑉1
𝑉′1 𝑦1 = 𝑓(𝑥)
𝑉 ′1 𝑒 5𝑥 = 3𝑥 + 1
𝑉 ′1 = (3𝑥 + 1)𝑒 −5𝑥
𝑉1 = ∫(3𝑥 + 1)𝑒 −5𝑥 𝑑𝑥
𝟓
𝟏
𝐬𝐢𝐧 𝒙 −
𝐜𝐨𝐬 𝒙
𝟐𝟔
𝟐𝟔
3
8
5
25
𝑉1 = (− 𝑥 −
)𝑒 −5𝑥
Therefore
3
8
𝑦𝑝 = − 5 𝑥 − 25
𝟑
𝟖
𝒚 = 𝑪𝟏 𝒆𝟓𝒙 − 𝟓 𝒙 − 𝟐𝟓
8. Solve 𝑦 ′′ − 2𝑦 ′ + 𝑦 =
𝑒𝑥
𝑥
𝑦𝑐 = 𝐶1 𝑒 𝑥 + 𝐶2 𝑥𝑒 𝑥
𝑉′1 𝑒 𝑥 + 𝑉′2 𝑥𝑒 𝑥 = 0
(1)
−1(𝑉 ′1 𝑒 𝑥 + 𝑉 ′ 2 (𝑥𝑒 𝑥 + 𝑒 𝑥 ) =
𝑒𝑥
𝑥
(2)
Multiplying equation (2) with -1 & adding both equations:
𝑉′2 𝑥𝑒 𝑥 − 𝑉 ′ 2 (𝑥𝑒 𝑥 + 𝑒 𝑥 ) =
′
𝑥
−𝑉 2 𝑥𝑒 = −
𝑉
′
𝑒𝑥
𝑥
𝑒𝑥
𝑥
1
2
=𝑥
𝑉2 = ln 𝑥
Using equation (1) and substituting 𝑉 ′ 2 :
𝑉′1 𝑒 𝑥 + 𝑉′2 𝑥𝑒 𝑥 = 0
1
𝑉′1 𝑒 𝑥 + (𝑥)𝑥𝑒 𝑥 = 0
𝑉′1 𝑒 𝑥 + 𝑒 𝑥 = 0
𝑉′1 = −1
𝑉1 = −𝑥
So that,
𝑦𝑝 = 𝑉1 𝑦1 +𝑉2 𝑦2
𝑦𝑝 = −𝑥𝑒 𝑥 + ln 𝑥(𝑥𝑒 𝑥 )
𝑦𝑝 = −𝑥𝑒 𝑥 + 𝑥𝑒 𝑥 ln(𝑥)
𝒚 = 𝑪𝟏 𝒆𝒙 + 𝑪𝟐 𝒙𝒆𝒙 − 𝒙𝒆𝒙 + 𝒙𝒆𝒙 𝐥𝐧(𝒙)
9. Solve 𝑦 ′′ − 36𝑦 = 4𝑒 2𝑥
𝑦𝑐 = 𝐶1 𝑒 6𝑥 + 𝐶2 𝑒 −6𝑥
𝑦𝑝 = 𝑉1 𝑒 6𝑥 + 𝑉2 𝑒 −6𝑥
𝑉1 ′𝑒 6𝑥 + 𝑉2 ′𝑒 −6𝑥 = 0
6𝑉1′𝑒
6𝑥
(1)
− 6𝑉2 ′𝑒 −6𝑥 = 4𝑒 2𝑥
(2)
Multiply eqn (1) by 6 then add eqn (2)
12𝑉1 ′𝑒 6𝑥 = 4𝑒 2𝑥
1
𝑉1 ′ = 𝑒 −4𝑥
3
1
𝑉1′ = ∫ 𝑒 −4𝑥 𝑑𝑥
3
𝑉1 = −
1 −4𝑥
𝑒
12
1
𝑉2′ = − 𝑒 8𝑥
3
1
𝑉2′ = − ∫ 𝑒 8𝑥 𝑑𝑥
3
𝑉2 = −
1 8𝑥
𝑒
24
Thus
𝑦𝑝 = −
𝑦𝑝 = −
1 −4𝑥 6𝑥
1 8𝑥 −6𝑥
𝑒 𝑒 −
𝑒 𝑒
12
24
1 2𝑥
1
1
𝑒 − 𝑒 2𝑥 = − 𝑒 2𝑥
12
24
8
𝟏
𝒚 = 𝑪𝟏 𝒆𝟔𝒙 + 𝑪𝟐 𝒆−𝟔𝒙 − 𝒆𝟐𝒙
𝟖
USING WRONSKIAN METHOD
𝑦2 𝑓(𝑥)
𝑦1 𝑓(𝑥)
𝑦 = −𝑦1 [∫
𝑑𝑥 + 𝐶1 ] + 𝑦2 [∫
𝑑𝑥 + 𝐶2 ]
𝑊(𝑦1 𝑦2 )
𝑊(𝑦1 𝑦2 )
where
𝑊(𝑦1 𝑦2 ) = 𝑦1 𝑦′2 − 𝑦′1 𝑦2 ≠ 0
1. Solve 𝑦 ′′ − 2𝑦 ′ + 𝑦 =
𝑥
𝑦𝑐 = 𝐶1 𝑒 + 𝐶2 𝑥𝑒
𝑦1 = 𝑒 𝑥
𝑒𝑥
𝑥
𝑥
𝑦2 = 𝑥𝑒 𝑥
𝑦′1 = 𝑒 𝑥
𝑦′2 = 𝑥𝑒 𝑥 + 𝑒 𝑥
𝑊(𝑦1 𝑦2 ) = 𝑒 𝑥 (𝑥𝑒 𝑥 + 𝑒 𝑥 ) − (𝑒 𝑥 ∙ 𝑥𝑒 𝑥 )
= 𝑒 2𝑥
𝑦 = −𝑒 𝑥 [∫
𝑒𝑥
𝑥𝑒 𝑥 ( 𝑥 )
𝑒𝑥
𝑒𝑥( 𝑥 )
𝑑𝑥 + 𝐶1 ] + 𝑥𝑒 𝑥 [∫ 2𝑥 𝑑𝑥 + 𝐶2 ]
𝑒 2𝑥
𝑒
𝑥 (𝑥
𝑥
)
𝑦 = −𝑒
+ 𝐶1 + 𝑥𝑒 (ln 𝑥 + 𝐶2 )
𝑥
𝑦 = −𝐶1 𝑒 − 𝑥𝑒 𝑥 + 𝐶2 𝑥𝑒 𝑥 + 𝑥𝑒 𝑥 ln 𝑥
Let −𝐶1 = 𝐶1
𝒚 = 𝑪𝟏 𝒆𝒙 − 𝒙𝒆𝒙 + 𝑪𝟐 𝒙𝒆𝒙 + 𝒙𝒆𝒙 𝐥𝐧 𝒙
2. Solve 𝑦 ′′ − 36𝑦 = 4𝑒 2𝑥
𝑦𝑐 = 𝐶1 𝑒 6𝑥 + 𝐶2 𝑒 −6𝑥
𝑤 = 6𝑒 6𝑥 𝑒 −6𝑥 + 6𝑒 6𝑥 𝑒 −6𝑥
𝑤 = 12
𝑒 −6𝑥 (4𝑒 2𝑥 )
𝑒 6𝑥 (4𝑒 2𝑥 )
𝑑𝑥 + 𝐶1 ] + 𝑒 −6𝑥 [∫
𝑑𝑥 + 𝐶2 ]
12
12
𝑒 −4𝑥
𝑒 8𝑥
𝑦 = −𝑒 6𝑥 [∫
𝑑𝑥 + 𝐶1 ] + 𝑒 −6𝑥 [∫
𝑑𝑥 + 𝐶2 ]
3
3
𝑦 = −𝑒 6𝑥 [∫
𝑦 = −𝑒 6𝑥 (
1 −4𝑥
1
𝑒
+ 𝐶1 ) + 𝑒 −6𝑥 (− 𝑒 8𝑥 + 𝐶2 )
12
24
𝑦 = 𝐶1 𝑒 6𝑥 + 𝐶2 𝑒 −6𝑥 −
1 2𝑥
1
𝑒 − 𝑒 2𝑥
12
24
𝟏
𝒚 = 𝑪𝟏 𝒆𝟔𝒙 + 𝑪𝟐 𝒆−𝟔𝒙 − 𝒆𝟐𝒙
𝟖