Differential Equations 1.Solve; dy dy 1 x y xy dx 1 x y xy dx = (1 + x)
Download ReportTranscript Differential Equations 1.Solve; dy dy 1 x y xy dx 1 x y xy dx = (1 + x)
Slide 1
Differential Equations
Slide 2
1.Solve;
dy
dy
1 x y xy
dx
1 x y xy
dx
= (1 + x) + y(1 + x)
= (1 + x)(1 + y)
dy
1 y
(1 x) dx
Integrating
dy
1 y (1 x)dx
log(1 y ) x
x
2
2
Which is the
required solution.
c
Slide 3
2.Solve;
( x y)
2
dy
( x y)
2
dy
a
2
dx
a
2
dx
Put x + y = z
1
dy
dx
dy
dx
dz
dx
dz
z
1
dx
dz
2
z
1 a
dx
2
dx
dx
The given eqn becomes
dz
dz
1
a
2
z
2
a
2
z
2
a z
2
1
z
2
2
2
z a
2
2
dz dx
z a a
2
2
z a
2
2
2
2
a
1 2
2
z
a
dz dx
dz dx
Integrating
Slide 4
2
a
1 z 2 a 2
dz a z
dz dx
1
2
2
a
2
dz dx
z
za
tan x c
a
a
2
1
1
x y
x y a tan
xc
a
1
x y
y a tan
c
a
1
Slide 5
3. Solve 3ex tan y dx + (1 + ex) sec 2y dy = 0
Given equation is
3ex tan y dx + (1 + ex) sec2y dy = 0
3ex tan y dx = – (1 + ex) sec2y dy
3e
x
1 e
3e
2
x
dx
sec y
tan y
x
1 e
dy
Integrating
2
x
dx
sec y
dy
tan y
3 log(1+ex) = – log(tan y) + log c
log(1+ex)3 + log(tan y) = log c
log(1+ex)3(tan y) = log c
(1+ex)3(tan y) = c
Which is the
required solution
Slide 6
4. Solve (x2 – y)dx + (y2 – x)dy = 0, if it passes
through the origin.
The given equation is
(x2 – y)dx + (y2 – x)dy = 0
x2 dx – ydx + y2 dy – xdy = 0
x2 dx + y2 dy = ydx + xdy
Integrating,
x
x
2
2
x
dx y dy ydx xdy
0+0=0+c
c=0
2
dx y dy d ( xy )
2
3
3
Since it passes through
the origin,
y
The required solution is
x
3
3
y
3
xy
3
3
3
xy c
x y 3 xy
3
3
Slide 7
5. Solve:
dy
dy
y cot x 2 cos x
dx
y cot x 2 cos x
dx
It is a linear diff eqn with
P = cot x, Q = 2cosx
P dx
IF e
e
e
cot x dx
log sin x
sin x
cos x
e
sin x dx
The solution is
y I.F Q I.F dx c
y sin x 2 cos x sin x dx c
y sin x sin 2 x dx c
y sin x
y sin x
cos x
2
cos x
2
c
c
Slide 8
6. Solve:
(1 x )
2
dy
(1 x )
2
dy
2 xy x 1 x
dx
2x
1 x
2
y
2
x 1 x
1 x
e
2
P
1 x
IF e
2
2
y
P dx
2x
e
1 x
1 x 2 dx
e
2 x
1 x 2 dx
1
1 x
2
y I.F Q I.F dx c
x
,Q
2
log(1 x )
The solution is
2
It is a linear diff eqn with
2x
2
dx
dx
dy
2 xy x 1 x
y
1
1 x
2
1
1 x
2
x
1
2
1 x 1 x
x
(1 x )
2 3/ 2
2
dx c
dx c
Slide 9
y
y
1
1 x
2
1
1 x
2
x
dx c
(1 x )
1
2 3/ 2
2x
2 (1 x
2 3/ 2
)
3
1
2
1 (1 x )
y
2
3
1 x
2
1
2
1
2
1 (1 x )
y
2
1
1 x
2
2
1
1
y
c
2
2
1 x
1 x
1
2
dx c
c
1
2
c
Slide 10
7. Solve: (1 + y2)dx = (tan–1 y – x )dy
The given equation
(1+y2)dx = (tan–1y – x)dy can be written as
dx
tan
dy
dx
dy
yx
1 y
dy
dx
1
tan
1
1 y
x
1 y
2
y
2
IF e
2
e
x
1 y
tan
1
1 y
1
P dy
tan
1
e
y
2
The solution is
y
2
x (I.F) Q (I.F ) dy c
It is a linear diff eqn in x with
P
1
1 y
2
,Q
tan
1 y 2 dy
1
1 y
-1
y
2
xe
tan y
tan
1
1 y
y
2
-1
e
tan y
dy c
Slide 11
-1
xe
tan y
tan
1
y
1 y
2
-1
e
tan y
dy c.........(1)
Put u = tan – 1 y
du
dy
du
1
-1
1 y
2
-1
1
1 y
xe
2
xe
dy
Eqn (1) becomes
-1
xe
tan y
-1
xe
tan y
xe
ue du c
u
ue e du c
u
u
tan
tan y
-1
tan y
tan y
e
1
tan
1
ue e c
u
ye
y
tan
1
(tan
y
1
u
e
tan
1
y
c
y 1) c
It is the required soln.
Slide 12
8. Solve (D2 – 13D + 12)y = e –2x
The given equation is
(D2 – 13D + 12)y = e –2x
The characteristic equation is
p2 – 13p + 12 = 0
(p – 12)(p – 1) = 0
p = 1, 12
The complementary function is
CF = A ex + B e 12x
Particular integral is
PI
1
D 13D 12
2
e
2 x
1
( 2) 13( 2) 12
2
1
4 26 12
1
e
e
e
2 x
2 x
2 x
42
The general solution is
y = CF + PI
y Ae Be
x
12 x
1
42
e
2 x
Slide 13
9. Solve (D2 + 6D + 8)y = e –2x
The given equation is
1
2 x
e
(D2 + 6D + 8)y = e –2x
( D 4)( D 2)
The characteristic equation is
1
2 x
2
p + 6p + 8 = 0
e
( 2 4)( D 2)
(p + 2)(p + 4) = 0
1 2 x
x 2 x
p = – 2, – 4
xe
e
2
2
The complementary function is
CF = A e–2x + B e– 4x
The general solution is
Particular integral is
y = CF + PI
PI
1
D 6D 8
2
e
2 x
y Ae
2 x
Be
4 x
x
2
e
2 x
Slide 14
10. Solve (D2 – 4D + 13)y = e –3x
CF = e2x(Acos3x + B sin3x)
The given equation is
(D2 – 4D + 13)y = e –3x
The characteristic equation is
p2 – 4p + 13 = 0
p
p
( 4)
( 4) 4 1 13
2
2 1
4 16 52
2
PI
4 6i
2
4
36
1
D 4 D 13
1
2
e
3 x
( 3) 4( 3) 13
2
1
9 12 13
e
3 x
e
e
3 x
3 x
34
The general solution is
2
y = CF + PI
2 3i
y e ( A cos 3 x B sin 3 x)
2x
1
34
e
3 x
Slide 15
11. Solve (D2 – 4)y = sin2x
The given equation is
(D2 – 4)y = sin2x
The characteristic equation is
p2 – 4 = 0
(p – 2)(p + 2) = 0
p = –2, 2
The complementary function is
CF = A e –2x + B e 2x
Particular integral is
PI
1
D 4
2
( 2 4)
2
sin 2 x
sin 2 x
8
The general solution is
y = CF + PI
y Ae
sin 2 x
1
2 x
Be
2x
1
4
sin 2 x
Slide 16
12. Solve (D2 – 2D – 3)y = sinx cosx
The given equation is
(D2 – 2D – 3)y = sinx cosx
The characteristic equation is
p2 – 2p – 3 = 0
(p – 3)(p + 1) = 0
p = –1, 3
The complementary function is
CF = A e –x + B e 3x
Particular integral is
PI
sin x cos x
D 2D 3
2
1
2 D 2D 3
sin 2 x
2
2( 2 2 D 3)
sin 2 x
2
sin 2 x
2( 2 D 7)
( 2 D 7) sin 2 x
2( 2 D 7)(2 D 7)
( 2 D 7) sin 2 x
2( 4 D 49)
2
Slide 17
( 2 D 7)
2( 4 D 49)
2
sin 2 x
2 D (sin 2 x) 7 sin 2 x
2( 4( 2 ) 49)
2
4 cos 2 x 7 sin 2 x
2( 16 49)
1
( 4 cos 2 x 7 sin 2 x)
130
The general solution is
y = CF + PI
y Ae
x
Be
3x
1
130
( 4 cos 2 x 7 sin 2 x)
Slide 18
13. Solve (D2 + 14D + 49)y = e–7x + 4
The given equation is
(D2 + 14D + 49)y = e–7x + 4
The characteristic equation is
p2 + 14p + 49 = 0
(p + 7)(p + 7) = 0
p = – 7, – 7
CF = (Ax + B)e–7x
Particular integral is
PI
e
7 x
4
e
7 x
( D 7)
2
x e
2
7 x
x e
0x
( D 7)
4e
7 x
2
2
0x
(0 7 )
2
2
4e
2
4
49
The general solution is
y = CF + PI
D 14 D 49
2
2
y ( Ax B )e
7 x
x e
7 x
2
4
49
Slide 19
14. Solve (D2 – 13D + 12)y = e–2x + 5ex
The given equation is
2 x
x
e
5e
(D2 – 13D + 12)y = e–2x + 5ex
4 26 12 (1 12)( D 1)
The characteristic equation is
2 x
x
e
5
xe
2
p – 13p + 12 = 0
42
11
(p – 12)(p – 1) = 0
2 x
x
e
5 xe
p = 12, 1
42
11
CF = A e12x + B ex
The general solution is
Particular integral is
2 x
x
e
5e
y = CF + PI
PI
D 13D 12
2
e
2 x
D 13D 12
2
y Ae
12 x
5e
x
( D 12)( D 1)
Be
x
e
2 x
42
5 xe
11
x
Slide 20
15. Solve (D2 + 4D + 13)y = cos3x
The given equation is
(D2 + 4D + 13)y = cos 3x
The characteristic equation is
p2 + 4p + 13 = 0
p
p
4
4 4 1 13
2
CF = e–2x(Acos3x + Bsin3x)
Particular integral is
PI
2 1
4 16 52
2
p
p
4
36
2
4 6i
2
2 3i
1
D 4 D 13
1
2
3 4 D 13
2
1
4D 4
cos 3 x
cos 3 x
cos 3 x
( D 1)
4( D 1)( D 1)
cos 3 x
Slide 21
( D 1)
4( D 1)
2
cos 3 x
D (cos 3 x) cos 3 x
4( 3 1)
2
3 sin 3 x cos 3 x
40
1
(3 sin 3 x cos 3 x)
40
The general solution is
y = CF + PI
y e ( A cos 3 x B sin 3 x)
2x
1
40
(3 sin 3 x cos 3 x)
Slide 22
16. Solve: (D2 – 1)y = cos 2x – 2sin 2x
The given equation is
(D2 – 1)y = cos 2x – 2sin 2x
The characteristic equation is
p2 – 1 = 0
(p – 1)(p + 1) = 0
p = 1, – 1
CF = A ex + B e–x
The particular integral is
PI
1
D 1
2
cos 2 x
D 1
2
2
cos 2 x
2 1
2
cos 2 x
sin 2 x
D 1
2
2
2
sin 2 x
2 1
2
sin 2 x
5
5
1
2
cos 2 x sin 2 x
5
5
The general solution is
(cos 2 x 2 sin 2 x)
y = CF + PI
y Ae Be
x
x
1
5
( 2 sin 2 x cos 2 x)
Slide 23
17. Solve (D2 – 4D + 1)y = x2
The given equation is
Particular integral is
2
2
(D – 4D + 1)y = x
1
2
PI 2
x
The characteristic equation is
D 4D 1
p2 – 4p + 1 = 0
= {1 + (D2 – 4D)}–1 x2
p
( 4)
( 4) 4 1 1
2
2 1
4 16 4
4 12
2
42 3
2
2 3
2
CF Ae
( 2 3 ) x
Be
( 2 3 ) x
= {1 – (D2 – 4D) + (D2 – 4D)2 – } x2
= {1 – D2 + 4D + 16D2} x2
= x2 – 2 + 4 (2x) + 16(2)
= x2 + 30 + 8x
The general solution is
y = CF + PI
y Ae
( 2 3 ) x
Be
( 2 3 ) x
x 8 x 30
2
Slide 24
18. Solve (D2 + 3D – 4)y = x2
1
2
The given equation is
PI
x
2
D
3D
2
2
(D + 3D – 4)y = x
41
4
The characteristic equation is
1
2
1
D 3D
2
p2 + 3p – 4 = 0
1
x
4
4
(p + 4)(p – 1) = 0
2
2
2
1
D 3D D 3D
2
p = 1, – 4
.. x
1
4
4
4
x
–4x
CF = Ae + Be
2
2
2
1
D 3D 9 D
The particular integral is
PI
1
D 3D 4
2
x
2
1
4
4
.. x
16
2
1
13D
12 D 2
1
x
4
16
16
Slide 25
1 2 13
12
x
(
2
)
(
2
x
)
4
16
16
1 2 13 3
x
x
4
8
2
1 2 3
13
x x
4
2
8
The general solution is
y = CF + PI
y Ae Be
x
4 x
1 2 3
13
x x
4
2
8
Slide 26
19. In a chemical reaction the rate of conversion of a
substance at time t is proportional to the quantity of the
substance still untransformed at that instant. At the end of
one hour, 60grams remain and at the end of 4 hours 21
grams. How many grams of the first substance was there
initially?
Let x grams of the substance remain after t hours.
dx
x
dx
k x Where k is constant of proportionality
dt
dt
dx
k dt Integrating
x
dx
x
k
dt
log x = – k t + c1
xe
kt c1
x = c e– k t …….(1)
When t = 1, x = 60
60 = c e–k ……….(2)
Slide 27
x = c e– k t …….(1)
When t = 4, x = 21
taking log
3log c = 4log60 – log21
21 = c e–4k ……….(3)
= 4(1.7782) – 1.3222
When t = 0, x = c
= 7.1128 – 1.3222
From (2) e – k = 60/c
= 5.7906
Sub in (3) we get
4
4
60
60
21 c
3
c
c
c
3
60
21
4
log c = 5.7906/3 = 1.9302
c = antilog 1.9302
= 85.15 85
Hence initially there was
85gms of the substance.
Slide 28
20. For a postmortem report, a doctor requires to know
approximately the time of death of the deceased. He
records the first temperature at 10.00am to be 93.4F.
After 2 hours he finds the temperature to be 91.4F. If the
room temperature is 72F, estimate the time of death.
(Assume that the normal temperature of a human body to
be 98.6F). Given
19.4
26.6
log e
21.4
0.0426 and log e
0.00945
21.4
Let T be the temperature of the body at any time t
The temperature difference is T – 72
By Newton’s law of cooling,
dT
dt
T 72
dT
k (T 72)
dt
Where k is constant of proportionality
Slide 29
dT
T 72
k dt
When, t = 120, T = 91.4F
Integrating
91.4F = 72 + 21.4 e– 120k
dT
T 72 k dt
21.4e– 120k = 91.4 – 72
log (T – 72) = – k t + c1
T 72 e
21.4e– 120k = 19.4
kt c1
e
120 k
21.4
T = 72 + c e– k t …….(1)
120k log
Initially, t = 0, T = 93.4F
19.4
21.4
93.4F = 72 + c e0
k
1
120
c = 93.4 – 72 = 21.4F
T = 72 + 21.4 e–k t ……(2)
19.4
k
1
120
log
19.4
21.4
( 0.0426) 3.55 10
4
Slide 30
When T = 98.6F, t = t1
98.6F = 72 + 21.4 e–k t1
21.4 e–k t1 = 98.6 – 72 = 26.6
e
kt1
26.6
t1
21.4
kt1 log e
t1
t1
1
k
26.6
21.4
26.6
log e
1
3.55 10
4
t1
10
4
(0.0945)
3.55
945
266 min
3.55
21.4
4hours 26min before the
first recorded temperature
(0.0945)
The approximate time of
death = 10 – 4:26 = 5:34am
Slide 31
21. The number of bacteria in a yeast culture grows at a rate
which is proportional to the number present. If the
population of a colony of yeast bacteria triples in 1hour.
Show that the number of bacteria at the end of five hours will
be 35 times of the population at initial time.
Let x be the number of bacteria present in the yeast
culture at any time t
dx
dt
dx
x
k dt
dx
x
k x
dt
x
dx
k
Integrating
dt
log x = k t + c1
Where k is constant of proportionality
kt c1
xe
x = c e k t …….(1)
When t = 0, x = x0
x0 = c e0
c = x0
Slide 32
Sub in eqn (1) we get
x = x0 e k t………(2)
Given x = 3x0, when t = 1
3x0 = x0 ek
e k = 3……..(3)
When t = 5, let x = x1
x1 = x0 e 5k
= x0 (ek)5 = x0 35
= 35 times of the population at initial time
Slide 33
22. The sum of Rs. 1000 is compounded continuously, the
nominal rate of interest being four percent per annum. In
how many years will the amount be twice the original
principal? (loge2 = 0.6931)
Let x be the principal
dx
dt
dx
dt
dx
4% of x
0.04 x
0.04 dt
Integrating
dx
x
xe
0.04 t c1
x = c e 0.04 t …….(1)
When t = 0, x = 1000
x
log x = 0.04 t + c1
0.04 dt
1000 = c e0
c = 1000
x = 1000e 0.04t……..(2)
Slide 34
x = 1000 e 0.04 t …….(2)
When x = 2000, t = ?
2000 = 1000 e0.04t
e0.04t = 2
0.04t = loge2
0.04t = 0.6931
t
0.6931
17.32 17
0.04
In 17 years the amount will be twice the
original principal
Slide 35
23. Radium disappears at a rate proportional to the
amount present. If 5% of the original amount disappears
in 50years, how much will remain at the end of 100years.
(Take A0 as the initial amount.)
Let A be the amount of radium
disappears at any time t.
dA
dt
A
dA
A
k A Where k is constant of proportionality
dt
k dt Integrating
dA
A
dA
k
dt
log A = –k t + c1
Ae
kt c1
A ce
kt
Initially, t = 0 and A = A0
A0 = c e0 c = A0
Slide 36
A A0 e
kt
When t = 50years, A = 95% of A0 = 0.95A0
0.95A0 = A0 e –50k
e – 50k = 0.95
When t = 100 years, A = A1
A1 = A0 e – 100k = A0 (e – 50k)2
= A0 (0.95)2 = 0.9025A0
Amount of radium at the end of 100years = 0.9025A0
Slide 37
Differential Equations
Slide 2
1.Solve;
dy
dy
1 x y xy
dx
1 x y xy
dx
= (1 + x) + y(1 + x)
= (1 + x)(1 + y)
dy
1 y
(1 x) dx
Integrating
dy
1 y (1 x)dx
log(1 y ) x
x
2
2
Which is the
required solution.
c
Slide 3
2.Solve;
( x y)
2
dy
( x y)
2
dy
a
2
dx
a
2
dx
Put x + y = z
1
dy
dx
dy
dx
dz
dx
dz
z
1
dx
dz
2
z
1 a
dx
2
dx
dx
The given eqn becomes
dz
dz
1
a
2
z
2
a
2
z
2
a z
2
1
z
2
2
2
z a
2
2
dz dx
z a a
2
2
z a
2
2
2
2
a
1 2
2
z
a
dz dx
dz dx
Integrating
Slide 4
2
a
1 z 2 a 2
dz a z
dz dx
1
2
2
a
2
dz dx
z
za
tan x c
a
a
2
1
1
x y
x y a tan
xc
a
1
x y
y a tan
c
a
1
Slide 5
3. Solve 3ex tan y dx + (1 + ex) sec 2y dy = 0
Given equation is
3ex tan y dx + (1 + ex) sec2y dy = 0
3ex tan y dx = – (1 + ex) sec2y dy
3e
x
1 e
3e
2
x
dx
sec y
tan y
x
1 e
dy
Integrating
2
x
dx
sec y
dy
tan y
3 log(1+ex) = – log(tan y) + log c
log(1+ex)3 + log(tan y) = log c
log(1+ex)3(tan y) = log c
(1+ex)3(tan y) = c
Which is the
required solution
Slide 6
4. Solve (x2 – y)dx + (y2 – x)dy = 0, if it passes
through the origin.
The given equation is
(x2 – y)dx + (y2 – x)dy = 0
x2 dx – ydx + y2 dy – xdy = 0
x2 dx + y2 dy = ydx + xdy
Integrating,
x
x
2
2
x
dx y dy ydx xdy
0+0=0+c
c=0
2
dx y dy d ( xy )
2
3
3
Since it passes through
the origin,
y
The required solution is
x
3
3
y
3
xy
3
3
3
xy c
x y 3 xy
3
3
Slide 7
5. Solve:
dy
dy
y cot x 2 cos x
dx
y cot x 2 cos x
dx
It is a linear diff eqn with
P = cot x, Q = 2cosx
P dx
IF e
e
e
cot x dx
log sin x
sin x
cos x
e
sin x dx
The solution is
y I.F Q I.F dx c
y sin x 2 cos x sin x dx c
y sin x sin 2 x dx c
y sin x
y sin x
cos x
2
cos x
2
c
c
Slide 8
6. Solve:
(1 x )
2
dy
(1 x )
2
dy
2 xy x 1 x
dx
2x
1 x
2
y
2
x 1 x
1 x
e
2
P
1 x
IF e
2
2
y
P dx
2x
e
1 x
1 x 2 dx
e
2 x
1 x 2 dx
1
1 x
2
y I.F Q I.F dx c
x
,Q
2
log(1 x )
The solution is
2
It is a linear diff eqn with
2x
2
dx
dx
dy
2 xy x 1 x
y
1
1 x
2
1
1 x
2
x
1
2
1 x 1 x
x
(1 x )
2 3/ 2
2
dx c
dx c
Slide 9
y
y
1
1 x
2
1
1 x
2
x
dx c
(1 x )
1
2 3/ 2
2x
2 (1 x
2 3/ 2
)
3
1
2
1 (1 x )
y
2
3
1 x
2
1
2
1
2
1 (1 x )
y
2
1
1 x
2
2
1
1
y
c
2
2
1 x
1 x
1
2
dx c
c
1
2
c
Slide 10
7. Solve: (1 + y2)dx = (tan–1 y – x )dy
The given equation
(1+y2)dx = (tan–1y – x)dy can be written as
dx
tan
dy
dx
dy
yx
1 y
dy
dx
1
tan
1
1 y
x
1 y
2
y
2
IF e
2
e
x
1 y
tan
1
1 y
1
P dy
tan
1
e
y
2
The solution is
y
2
x (I.F) Q (I.F ) dy c
It is a linear diff eqn in x with
P
1
1 y
2
,Q
tan
1 y 2 dy
1
1 y
-1
y
2
xe
tan y
tan
1
1 y
y
2
-1
e
tan y
dy c
Slide 11
-1
xe
tan y
tan
1
y
1 y
2
-1
e
tan y
dy c.........(1)
Put u = tan – 1 y
du
dy
du
1
-1
1 y
2
-1
1
1 y
xe
2
xe
dy
Eqn (1) becomes
-1
xe
tan y
-1
xe
tan y
xe
ue du c
u
ue e du c
u
u
tan
tan y
-1
tan y
tan y
e
1
tan
1
ue e c
u
ye
y
tan
1
(tan
y
1
u
e
tan
1
y
c
y 1) c
It is the required soln.
Slide 12
8. Solve (D2 – 13D + 12)y = e –2x
The given equation is
(D2 – 13D + 12)y = e –2x
The characteristic equation is
p2 – 13p + 12 = 0
(p – 12)(p – 1) = 0
p = 1, 12
The complementary function is
CF = A ex + B e 12x
Particular integral is
PI
1
D 13D 12
2
e
2 x
1
( 2) 13( 2) 12
2
1
4 26 12
1
e
e
e
2 x
2 x
2 x
42
The general solution is
y = CF + PI
y Ae Be
x
12 x
1
42
e
2 x
Slide 13
9. Solve (D2 + 6D + 8)y = e –2x
The given equation is
1
2 x
e
(D2 + 6D + 8)y = e –2x
( D 4)( D 2)
The characteristic equation is
1
2 x
2
p + 6p + 8 = 0
e
( 2 4)( D 2)
(p + 2)(p + 4) = 0
1 2 x
x 2 x
p = – 2, – 4
xe
e
2
2
The complementary function is
CF = A e–2x + B e– 4x
The general solution is
Particular integral is
y = CF + PI
PI
1
D 6D 8
2
e
2 x
y Ae
2 x
Be
4 x
x
2
e
2 x
Slide 14
10. Solve (D2 – 4D + 13)y = e –3x
CF = e2x(Acos3x + B sin3x)
The given equation is
(D2 – 4D + 13)y = e –3x
The characteristic equation is
p2 – 4p + 13 = 0
p
p
( 4)
( 4) 4 1 13
2
2 1
4 16 52
2
PI
4 6i
2
4
36
1
D 4 D 13
1
2
e
3 x
( 3) 4( 3) 13
2
1
9 12 13
e
3 x
e
e
3 x
3 x
34
The general solution is
2
y = CF + PI
2 3i
y e ( A cos 3 x B sin 3 x)
2x
1
34
e
3 x
Slide 15
11. Solve (D2 – 4)y = sin2x
The given equation is
(D2 – 4)y = sin2x
The characteristic equation is
p2 – 4 = 0
(p – 2)(p + 2) = 0
p = –2, 2
The complementary function is
CF = A e –2x + B e 2x
Particular integral is
PI
1
D 4
2
( 2 4)
2
sin 2 x
sin 2 x
8
The general solution is
y = CF + PI
y Ae
sin 2 x
1
2 x
Be
2x
1
4
sin 2 x
Slide 16
12. Solve (D2 – 2D – 3)y = sinx cosx
The given equation is
(D2 – 2D – 3)y = sinx cosx
The characteristic equation is
p2 – 2p – 3 = 0
(p – 3)(p + 1) = 0
p = –1, 3
The complementary function is
CF = A e –x + B e 3x
Particular integral is
PI
sin x cos x
D 2D 3
2
1
2 D 2D 3
sin 2 x
2
2( 2 2 D 3)
sin 2 x
2
sin 2 x
2( 2 D 7)
( 2 D 7) sin 2 x
2( 2 D 7)(2 D 7)
( 2 D 7) sin 2 x
2( 4 D 49)
2
Slide 17
( 2 D 7)
2( 4 D 49)
2
sin 2 x
2 D (sin 2 x) 7 sin 2 x
2( 4( 2 ) 49)
2
4 cos 2 x 7 sin 2 x
2( 16 49)
1
( 4 cos 2 x 7 sin 2 x)
130
The general solution is
y = CF + PI
y Ae
x
Be
3x
1
130
( 4 cos 2 x 7 sin 2 x)
Slide 18
13. Solve (D2 + 14D + 49)y = e–7x + 4
The given equation is
(D2 + 14D + 49)y = e–7x + 4
The characteristic equation is
p2 + 14p + 49 = 0
(p + 7)(p + 7) = 0
p = – 7, – 7
CF = (Ax + B)e–7x
Particular integral is
PI
e
7 x
4
e
7 x
( D 7)
2
x e
2
7 x
x e
0x
( D 7)
4e
7 x
2
2
0x
(0 7 )
2
2
4e
2
4
49
The general solution is
y = CF + PI
D 14 D 49
2
2
y ( Ax B )e
7 x
x e
7 x
2
4
49
Slide 19
14. Solve (D2 – 13D + 12)y = e–2x + 5ex
The given equation is
2 x
x
e
5e
(D2 – 13D + 12)y = e–2x + 5ex
4 26 12 (1 12)( D 1)
The characteristic equation is
2 x
x
e
5
xe
2
p – 13p + 12 = 0
42
11
(p – 12)(p – 1) = 0
2 x
x
e
5 xe
p = 12, 1
42
11
CF = A e12x + B ex
The general solution is
Particular integral is
2 x
x
e
5e
y = CF + PI
PI
D 13D 12
2
e
2 x
D 13D 12
2
y Ae
12 x
5e
x
( D 12)( D 1)
Be
x
e
2 x
42
5 xe
11
x
Slide 20
15. Solve (D2 + 4D + 13)y = cos3x
The given equation is
(D2 + 4D + 13)y = cos 3x
The characteristic equation is
p2 + 4p + 13 = 0
p
p
4
4 4 1 13
2
CF = e–2x(Acos3x + Bsin3x)
Particular integral is
PI
2 1
4 16 52
2
p
p
4
36
2
4 6i
2
2 3i
1
D 4 D 13
1
2
3 4 D 13
2
1
4D 4
cos 3 x
cos 3 x
cos 3 x
( D 1)
4( D 1)( D 1)
cos 3 x
Slide 21
( D 1)
4( D 1)
2
cos 3 x
D (cos 3 x) cos 3 x
4( 3 1)
2
3 sin 3 x cos 3 x
40
1
(3 sin 3 x cos 3 x)
40
The general solution is
y = CF + PI
y e ( A cos 3 x B sin 3 x)
2x
1
40
(3 sin 3 x cos 3 x)
Slide 22
16. Solve: (D2 – 1)y = cos 2x – 2sin 2x
The given equation is
(D2 – 1)y = cos 2x – 2sin 2x
The characteristic equation is
p2 – 1 = 0
(p – 1)(p + 1) = 0
p = 1, – 1
CF = A ex + B e–x
The particular integral is
PI
1
D 1
2
cos 2 x
D 1
2
2
cos 2 x
2 1
2
cos 2 x
sin 2 x
D 1
2
2
2
sin 2 x
2 1
2
sin 2 x
5
5
1
2
cos 2 x sin 2 x
5
5
The general solution is
(cos 2 x 2 sin 2 x)
y = CF + PI
y Ae Be
x
x
1
5
( 2 sin 2 x cos 2 x)
Slide 23
17. Solve (D2 – 4D + 1)y = x2
The given equation is
Particular integral is
2
2
(D – 4D + 1)y = x
1
2
PI 2
x
The characteristic equation is
D 4D 1
p2 – 4p + 1 = 0
= {1 + (D2 – 4D)}–1 x2
p
( 4)
( 4) 4 1 1
2
2 1
4 16 4
4 12
2
42 3
2
2 3
2
CF Ae
( 2 3 ) x
Be
( 2 3 ) x
= {1 – (D2 – 4D) + (D2 – 4D)2 – } x2
= {1 – D2 + 4D + 16D2} x2
= x2 – 2 + 4 (2x) + 16(2)
= x2 + 30 + 8x
The general solution is
y = CF + PI
y Ae
( 2 3 ) x
Be
( 2 3 ) x
x 8 x 30
2
Slide 24
18. Solve (D2 + 3D – 4)y = x2
1
2
The given equation is
PI
x
2
D
3D
2
2
(D + 3D – 4)y = x
41
4
The characteristic equation is
1
2
1
D 3D
2
p2 + 3p – 4 = 0
1
x
4
4
(p + 4)(p – 1) = 0
2
2
2
1
D 3D D 3D
2
p = 1, – 4
.. x
1
4
4
4
x
–4x
CF = Ae + Be
2
2
2
1
D 3D 9 D
The particular integral is
PI
1
D 3D 4
2
x
2
1
4
4
.. x
16
2
1
13D
12 D 2
1
x
4
16
16
Slide 25
1 2 13
12
x
(
2
)
(
2
x
)
4
16
16
1 2 13 3
x
x
4
8
2
1 2 3
13
x x
4
2
8
The general solution is
y = CF + PI
y Ae Be
x
4 x
1 2 3
13
x x
4
2
8
Slide 26
19. In a chemical reaction the rate of conversion of a
substance at time t is proportional to the quantity of the
substance still untransformed at that instant. At the end of
one hour, 60grams remain and at the end of 4 hours 21
grams. How many grams of the first substance was there
initially?
Let x grams of the substance remain after t hours.
dx
x
dx
k x Where k is constant of proportionality
dt
dt
dx
k dt Integrating
x
dx
x
k
dt
log x = – k t + c1
xe
kt c1
x = c e– k t …….(1)
When t = 1, x = 60
60 = c e–k ……….(2)
Slide 27
x = c e– k t …….(1)
When t = 4, x = 21
taking log
3log c = 4log60 – log21
21 = c e–4k ……….(3)
= 4(1.7782) – 1.3222
When t = 0, x = c
= 7.1128 – 1.3222
From (2) e – k = 60/c
= 5.7906
Sub in (3) we get
4
4
60
60
21 c
3
c
c
c
3
60
21
4
log c = 5.7906/3 = 1.9302
c = antilog 1.9302
= 85.15 85
Hence initially there was
85gms of the substance.
Slide 28
20. For a postmortem report, a doctor requires to know
approximately the time of death of the deceased. He
records the first temperature at 10.00am to be 93.4F.
After 2 hours he finds the temperature to be 91.4F. If the
room temperature is 72F, estimate the time of death.
(Assume that the normal temperature of a human body to
be 98.6F). Given
19.4
26.6
log e
21.4
0.0426 and log e
0.00945
21.4
Let T be the temperature of the body at any time t
The temperature difference is T – 72
By Newton’s law of cooling,
dT
dt
T 72
dT
k (T 72)
dt
Where k is constant of proportionality
Slide 29
dT
T 72
k dt
When, t = 120, T = 91.4F
Integrating
91.4F = 72 + 21.4 e– 120k
dT
T 72 k dt
21.4e– 120k = 91.4 – 72
log (T – 72) = – k t + c1
T 72 e
21.4e– 120k = 19.4
kt c1
e
120 k
21.4
T = 72 + c e– k t …….(1)
120k log
Initially, t = 0, T = 93.4F
19.4
21.4
93.4F = 72 + c e0
k
1
120
c = 93.4 – 72 = 21.4F
T = 72 + 21.4 e–k t ……(2)
19.4
k
1
120
log
19.4
21.4
( 0.0426) 3.55 10
4
Slide 30
When T = 98.6F, t = t1
98.6F = 72 + 21.4 e–k t1
21.4 e–k t1 = 98.6 – 72 = 26.6
e
kt1
26.6
t1
21.4
kt1 log e
t1
t1
1
k
26.6
21.4
26.6
log e
1
3.55 10
4
t1
10
4
(0.0945)
3.55
945
266 min
3.55
21.4
4hours 26min before the
first recorded temperature
(0.0945)
The approximate time of
death = 10 – 4:26 = 5:34am
Slide 31
21. The number of bacteria in a yeast culture grows at a rate
which is proportional to the number present. If the
population of a colony of yeast bacteria triples in 1hour.
Show that the number of bacteria at the end of five hours will
be 35 times of the population at initial time.
Let x be the number of bacteria present in the yeast
culture at any time t
dx
dt
dx
x
k dt
dx
x
k x
dt
x
dx
k
Integrating
dt
log x = k t + c1
Where k is constant of proportionality
kt c1
xe
x = c e k t …….(1)
When t = 0, x = x0
x0 = c e0
c = x0
Slide 32
Sub in eqn (1) we get
x = x0 e k t………(2)
Given x = 3x0, when t = 1
3x0 = x0 ek
e k = 3……..(3)
When t = 5, let x = x1
x1 = x0 e 5k
= x0 (ek)5 = x0 35
= 35 times of the population at initial time
Slide 33
22. The sum of Rs. 1000 is compounded continuously, the
nominal rate of interest being four percent per annum. In
how many years will the amount be twice the original
principal? (loge2 = 0.6931)
Let x be the principal
dx
dt
dx
dt
dx
4% of x
0.04 x
0.04 dt
Integrating
dx
x
xe
0.04 t c1
x = c e 0.04 t …….(1)
When t = 0, x = 1000
x
log x = 0.04 t + c1
0.04 dt
1000 = c e0
c = 1000
x = 1000e 0.04t……..(2)
Slide 34
x = 1000 e 0.04 t …….(2)
When x = 2000, t = ?
2000 = 1000 e0.04t
e0.04t = 2
0.04t = loge2
0.04t = 0.6931
t
0.6931
17.32 17
0.04
In 17 years the amount will be twice the
original principal
Slide 35
23. Radium disappears at a rate proportional to the
amount present. If 5% of the original amount disappears
in 50years, how much will remain at the end of 100years.
(Take A0 as the initial amount.)
Let A be the amount of radium
disappears at any time t.
dA
dt
A
dA
A
k A Where k is constant of proportionality
dt
k dt Integrating
dA
A
dA
k
dt
log A = –k t + c1
Ae
kt c1
A ce
kt
Initially, t = 0 and A = A0
A0 = c e0 c = A0
Slide 36
A A0 e
kt
When t = 50years, A = 95% of A0 = 0.95A0
0.95A0 = A0 e –50k
e – 50k = 0.95
When t = 100 years, A = A1
A1 = A0 e – 100k = A0 (e – 50k)2
= A0 (0.95)2 = 0.9025A0
Amount of radium at the end of 100years = 0.9025A0
Slide 37