Unit I
Differential Equations-II
6.1 METHOD OF VARIATION OF PARAMETERS
Consider a linear differential equation of second order
d2y
dy
+ a1
+ a2 y = φ(x)
...(1)
dx
dx 2
where a1, a2 are functions of ‘x’. If the complimentary function of this equation is known then we
can find the particular integral by using the method known as the method of variation of parameters.
Suppose the complimentary function of the Eqn. (1) is
C.F. = C1 y1 + C2 y2 where C1 and C2 are constants and y1 and y2 are
the complementary solutions of Eqn. (1)
The Eqn. (1) implies that
y1′′+ a1 y1′ + a 2 y1 = 0
...(2)
y2′′ + a1 y2′ + a 2 y2 = 0
...(3)
We replace the arbitrary constants C1, C2 present in C.F. by functions of x, say A, B respectively,
∴
y = Ay1 + By2
...(4)
is the complete solution of the given equation.
The procedure to determine A and B is as follows.
From Eqn. (4)
y′ =
b Ay′ + By ′ g + b A′y + B′y g
1
2
1
2
...(5)
We shall choose A and B such that
A′y1 + B′y2 = 0
...(6)
Thus Eqn. (5) becomes y′1 = Ay1′ + By 2′
...(7)
Differentiating Eqn. (7) w.r.t. ‘x’ again, we have
y″ =
b Ay ′′+ Ay′′ g + b A′y ′ + B′y′ g
1
2
280
1
2
...(8)
281
DIFFERENTIAL EQUATIONS–II
Thus, Eqn. (1) as a consequence of (4), (7) and (8) becomes
A ′ y1′ + B ′ y 2′ = φ(x)
...(9)
Let us consider equations (6) and (9) for solving
A′y1 + B′ y 2 = 0
...(6)
A′y1′ + B′ y 2′ = φ(x)
...(9)
Solving A′ and B′ by cross multiplication, we get
A′ =
bg
Find A and B
A = –
Integrating,
B =
where W =
y1
y1′
bg
– y2 φ x
y φx
, B′ = 1
W
W
z
z
...(10)
y2 φ ( x)
dx + k1
W
y1 φ( x )
dx + k 2
W
y2
= y1 y 2′ – y 2 y1′
y 2′
Substituting the expressions of A and B
y = Ay1 + By2 is the complete solution.
WORKED OUT EXAMPLES
1. Solve by the method of variation of parameters
d2y
+ y = cosec x.
dx 2
Solution. We have
(D2 + 1) y = cosec x
A.E. is
m2 + 1 = 0
⇒
m2 = – 1
⇒
m = ± i
Hence the C.F. is given by
∴
yc = C1 cos x + C2 sin x
...(1)
y = A cos x + B sin x
...(2)
be the complete solution of the given equation where A and B are to be found.
The general solution is
We have
y = Ay1 + By2
y 1 = cos x and y2 = sin x
y1′ = – sin x and y2′ = cos x
W = y1 y2′ − y2 y1′
= cos x . cos x + sin x . sin x = cos2x + sin2x = 1
282
ENGINEERING MATHEMATICS—II
A′ =
bg
– y2 φ x
,
W
B′ =
– sin x ⋅ cosec x
,
1
1
,
= – sin x ⋅
sin x
=
A′ = – 1,
On integrating, we get
A =
B =
bg
y1 φ x
W
cos x ⋅ cosec x
1
1
B′ = cos x ⋅
sin x
B′ =
B′ = cot x
zb g
z
–1 dx + C1 , i.e., A = – x + C1
cot x dx + C2 , i.e., B = log sin x + C2
Hence the general solution of the given Eqn. (2) is
y = (– x + C1) cos x + (log sin x + C2) sin x
i.e.,
y = C1 cos x + C2 sin x – x cos x + sin x log sin x.
2. Solve by the method of variation of parameters
d2 y
+ 4y = 4 tan 2x.
dx 2
Solution. We have
(D2 + 4) y = 4 tan 2x
A.E. is
m2 + 4 = 0
where φ(x) = 4 tan 2x.
i.e.,
i.e.,
m2 = – 4
m = ± 2i
Hence the complementary function is given by
yc = C1 cos 2x + C2 sin 2x
y = A cos 2x + B sin 2x
...(1)
be the complete solution of the given equation where A and B are to be found
We have
Then
y 1 = cos 2x
and y2 = sin 2x
y1′ = – 2 sin 2x
and y2′ = 2 cos 2x
W = y1 y2′ − y2 y1′
= cos 2x . 2 cos 2x + 2 sin 2x . sin 2x
= 2 (cos22x + sin22x)
Also,
= 2
φ(x) = 4 tan 2x
A′ =
bg
– y2 φ x
W
and B′ =
bg
y1φ x
W
283
DIFFERENTIAL EQUATIONS–II
A′ =
– sin 2 x ⋅ 4 tan 2 x
– cos 2 x ⋅ 4 tan 2 x
, B′ =
2
2
A′ =
– 2 sin 2 2 x
, B′ = 2 sin 2x
cos 2 x
On integrating, we get
z
z
z
zl
sin 2 2 x
dx , B = 2 sin 2 x dx
cos 2 x
A = –2
1 − cos2 2 x
dx
cos 2 x
= –2
= −2
= –2
z
q
sec 2 x − cos 2 x dx
RS 1 log bsec 2 x + tan 2 xg – 1 sin 2 xUV
2
T2
W
A = – log (sec 2x + tan 2x) + sin 2x + C1
B = 2
sin 2 x dx
b
g
2 – cos 2 x
+ C2
2
B = – cos 2x + C2
Substituting these values of A and B in Eqn. (1), we get
y = C1 cos 2x + C2 sin 2x – cos 2x log (sec 2x + tan 2x)
which is the required general solution.
=
3. Solve by the method of variation of parameters
d2y
+ a 2 y = sec ax.
dx 2
Solution. We have
(D2 + a2) y = sec ax
A.E. is
⇒ m = ± ai
m2 + a2 = 0
C.F. = yc = C1 cos ax + C2 sin ax
y = A cos ax + B sin ax
...(1)
be the complete solution of the given equation where A and B are to be found.
We have,
y 1 = cos ax,
y2 = sin ax
y1′ = – a sin ax, y2′ = a cos ax
W = y1 y2′ − y2 y1′ = a. Also, φ (x) = sec ax
bg
– y2 φ x
, and
W
– sin ax ⋅ sec ax
,
A′ =
a
A′ =
bg
y1 φ x
W
cos ax ⋅ sec ax
B′ =
a
B′ =
284
ENGINEERING MATHEMATICS—II
A′ =
– tan ax
,
a
A = –
A =
1
a
z
1
a
1
dx + c2
B=
a
x
+ C2
B=
a
B′ =
tan ax dx + C1 ,
b
– log sec ax
2
g+C
z
1
a
Substituting these values of A and B in Eqn. (1), we get
Thus,
y = C1 cos ax + C2 sin ax –
cos ax log (sec ax ) x sin ax
+
⋅
a
a2
4. Solve by the method of variation of parameters
d2y
dy
+2
+ 2y = e–x sec3x.
2
dx
dx
Solution. We have
(D2 + 2D + 2) y = e–x sec3x
A.E. is
i.e.,
m2 + 2m + 2 = 0
m =
–2± 4−8
2
m = –1±i
∴The complementary function (C.F.) is
C.F. = yc = e–x(C1cos x + C2 sin x)
∴
y = C1 e–x cos x + C2 e–x sin x
⇒
y = A e–x cos x + B e–x sin x
Thus
–x
y1 = e
–x
cos x and y2 = e
...(1)
sin x
y1′ = – e– x(sin x + cos x)
y2′ = e–x (cos x – sin x)
W = y1 y2′ – y2 y1′ = e–2x
Also,
φ(x) = e–x sec3x
A′ =
=
bg
– e – x sin x ⋅ e – x sec 3 x
= – tan x sec 2 x
e −2 x
A = –
A = –
and
B′ =
bg
– y2 φ x
– y1 φ x
, B′ =
W
W
z
tan x sec 2 x dx
1
tan 2 x + C1
2
bg
y1 φ x
W
285
DIFFERENTIAL EQUATIONS–II
e – x cos x ⋅ e − x sec 3 x
e −2 x
2
B′ = sec x
=
B =
z
sec 2 x dx
B = tan x + C2
Substituting these values of A and B in Eqn. (1), we get
y =
=
RS – 1 tan x + C UV e cos x + ltan x + C q e
T2
W
1
e bC cos x + C sin x g + e sin x sec x
2
−x
2
1
−x
–x
1
–x
2
sin x
2
2
This is general solution of the given solution.
5. Solve by the method of variation of parameters
2
d2y
.
–y =
2
1 + ex
dx
Solution. We have
(D2 – 1) y =
A.E. is
i.e.,
Hence C.F. is
2
1 + ex
m2 – 1 = 0
m2 = 1
⇒m =±1
x
yc = C1 e + C2 e–x
y = Aex + Be–x
...(1)
be the complete solution of the given equation where A and B are to be found
We have
y 1 = ex
y2 = e–x
and
y1′ = ex
y2′ = – e–x
W = y1 y2′ − y2 y1′
also
LM
N
φ(x) =
2
1 + ex
= ex (– e–x) – e–x – e–x.
ex – x = e0 = 1
= –1–1=–2
e–x + x = e0 = 1
– y2 φ x
A′ =
W
2
– e– x ⋅
1 + ex
e− x
1
=
+
=
= x
–2
1 + ex
e 1+ ex
bg
d
A′ =
A =
1
e
z
x
d1 + e i
x
d
1
e 1 + ex
x
i
dx + C1
i
OP
Q
286
ENGINEERING MATHEMATICS—II
Substitute
Hence,
ex = t, then ex dx = dt
A =
=
z b
z LMN
t
dx =
dt
t
1
g dt + C
1 1
1 O
– +
P dt + C ,
t 1+ tQ
t
2
1
1+ t
1
2
using partial fractions.
1
– log t + log (1 + t ) + C1
t
A = – e–x – x + log (1 + ex) + C1
= –
B′ =
bg
y1 φ x
W
ex ⋅
=
B = –
z
2
1 + ex
ex
=–
–2
1 + ex
ex
dx + C2
1 + ex
B = – log (1 + ex) + C2
Substituting these values of A and B in eqn. (1), we get
y =
d
i
d
i
– e − x − x + log 1 + e x + C1 e x + – log 1 + e x + C2 e – x
= C1 ex + C2 e–x – 1 – xex + ex log (1 + ex) – e–x log (1 + ex)
= C1 ex + C2 e–x – 1 – xex + (ex – e–x) log (1 + ex)
This is the complete solution of the given equation.
6. Solve by the method of variation of parameters
y″ + y = tan x.
Solution. We have (D2 + 1) y = tan x
A.E. is
m2 + 1 = 0
i.e.,
m2 = – 1
i.e.,
m = ±i
C.F. is
C.F. = yc = C1cos x + C2 sin x
∴
y = A cos x + B sin x
be the complete solution of the given equation where A and B are to be found
We have
y 1 = cos x
and
y2 = sin x
y1′ = – sin x
y2′ = cos x
W = y1 y2′ − y2 y1′
= cos x . cos x + sin x . sin x = cos2 x + sin2 x = 1
...(1)
287
DIFFERENTIAL EQUATIONS–II
φ (x) = tan x
Also,
bg
– y2 φ x
W
– sin x ⋅ tan x
=
1
A′ =
A′ =
– sin 2 x
cos x
A = –
= –
= –
z
z
zb
sin 2 x
dx + C1
cos x
1 − cos2 x
dx + C1
cos x
g
sec x − cos x dx + C1
A = – [log (sec x + tan x) – sin x] + C1
B′ =
bg
y1 φ x
W
cos x ⋅ tan x
1
B′ = sin x
=
B =
z
sin x dx + C2
B = – cos x + C2
Substitute these values of A and B in Eqn. (1), we get
y = {– log (sec x + tan x) + sin x + C1} cos x + {– cos x + C2} sin x
y = C1 cos x + C2 sin x – cos x log (sec x + tan x)
This is the complete solution.
7. Solve by the method of variation of parameters
d2y
dy
–2
+ 2y = ex tan x.
2
dx
dx
Solution. We have (D2 – 2D + 2) y = ex tan x
A.E. is
m2 – 2m + 2 = 0
i.e.,
m =
2± 4−8
2
m = 1±i
Therefore C.F. is
yc = ex (C1 cos x + C2 sin x)
∴
y = ex (A cos x + B sin x)
be the complete solution of the given equation
...(1)
288
ENGINEERING MATHEMATICS—II
where A and B are to be found
y 1 = ex cos x
We have
and
y2 = ex sin x
. y1′ = ex (cos x – sin x), y2′ = ex (sin x + cos x)
W = y1 y2′ − y2 y1′ = e2x Also, φ(x)= ex tan x
A′ =
=
A′ =
bg
– y2 φ x
W
– e x sin x ⋅ e x tan x
e2 x
– sin 2 x
cos x
A = –
= –
z
zb
sin 2 x
dx = –
cos x
z
g
1 − cos2 x
dx
cos x
sec x − cos x dx
A = – log (sec x + tan x) + sin x + C1
B′ =
bg
y1 ⋅ φ x
W
e x cos x ⋅ e x tan x
e2 x
B′ = sin x
=
B =
z
sin x dx + C2
B = – cos x + C2
Substituting these values of A and B in Eqn. (1), we get
y = {– log (sec x + tan x) + sin x + C1}ex cos x + {– cos x + C2} ex sin x
y = ex (C1 cos x + C2 sin x) – ex cos x log (sec x + tan x)
This is the complete solution of the given equation.
8. Solve by the method of variation of parameters
d2y
dy
–2
+ y = ex log x.
2
dx
dx
Solution. We have (D2 – 2D + 1) y = ex log x
A.E. is
i.e.,
i.e.,
m2 – 2m + 1 = 0
(m – 1)2 = 0
m = 1, 1
Hence C.F. is
yc = (C1 + C2 x) ex = C1 ex + C2 x ex
∴
y = Aex + Bxex
...(1)
289
DIFFERENTIAL EQUATIONS–II
y 1 = ex and y2 = xex
We have
y1′ = ex,
y2′ = xex + ex
W = y1 y2′ − y2 y1′ = xe2x + e2x – xe2x = e2x
Also
φ(x) = ex log x
A′ =
bg
– y2 φ x
,
W
B′ =
– xe x ⋅ e x log x e x ⋅ e x log x
=
e2 x
e2 x
A′ = – x . log x,
bg
– y1 φ x
W
=
A= –
z
log x ⋅ x dx
LM
N
x2
–
2
Integrating both these terms by parts, we get
A = – log x ⋅
z
B′ = log x
OP
Q
x2 1
⋅ dx + C1
2 x
– x 2 log x x 2
+
+ C1 ,
A=
2
4
Substituting these values of A and B in Eqn. (1), we get
Thus,
2
x
1
y=
bC + C x g e
x
–
x 2 e x log x x 2 e x
+
+ x 2 log x e x – x 2 e x
2
4
=
bC + C x g e
x
+
x 2 log x ⋅ e x 3 2 x
– x e
2
4
y=
2
1
1
b
2
g
C1 + C2 x e x +
y″ + 4y′ + 4y = 4 +
b
m2 + 4m + 4
(m + 2)2
m
yc
=
=
=
=
g
x 2e x
2 log x – 3 .
4
e –2x
⋅
x
Solution. We have (D2 + 4D + 4) y = 4 +
C.F. is
x
2
9. Solve by the method of variation of parameters
A.E. is
i.e.,
z
e –2 x
x
0
0
– 2, – 2
e–2x (C1 + C2 x)
x⋅
1
dx + C2
x
B = x log x – x + C2
RS – x log x + x + C UV e + b x log x − x + C g xe
4
T 2
W
2
y=
B = log x ⋅ x –
290
ENGINEERING MATHEMATICS—II
y = C1e–2x + C2x e–2x
∴
y = Ae–2x + Bxe–2x
...(1)
be the complete solution of the given equation where A and B are to be found
We have
y 1 = e–2x
y 2 = xe–2x
and
y1′ = – 2e–2x
y2′ = e–2x (1 – 2x)
W = y1 y2′ − y2 y1′
= e–2x · e–2x (1 – 2x) + xe–2x · 2 · e–2x
W = e–4x
A′ =
Also,
bg
– y2 φ x
W
FG
H
– xe −2 x ⋅ 4 +
=
e −2 x
x
e −4 x
IJ
K
A = –
z
LM
N
d4 xe
A = − 4x ⋅
2x
i
bg
F
GH
e –2 x ⋅ 4 +
=
F
GH
+ 1 dx + C1 ,
OP
Q
e2 x
e2x
– x + C1 ,
− 4⋅
2
4
A = – 2xe2x + e2x – x + C1
B =
=
z
z FGH
e –2 x
x
e –4 x
2x
4+
B′ = e
A′ = – (4xe2x + 1) ,
On integrating, we get
e –2 x
x
– y1 φ x
B′ =
W
φ (x) = 4 +
e –2 x
x
F
GH
e2 x 4 +
4e2 x +
I
JK
I
JK
I
JK
e –2 x
dx + C2
x
IJ
K
1
dx + C2
x
B = 2e2x + log x + C2
Substituting these values of A and B in Eqn. (1), we get
y = (– 2x e2x + e2x – x + C1) e–2x + (2e2x + log x + C2) xe–2x
= (C1 + C2x) e–2x – 2x + 1 – xe–2x + 2x + xe–2x log x
y = (C1 + C2x) e–2x + 1 + xe–2x(log x – 1).
10. Solve by the method of variation of parameters
y″ + 2y′ + 2y = e–x sec3x.
Solution. We have (D2 + 2D + 2) y = e–x sec3x
A.E.. is
m2 + 2m + 2 = 0
–2± 4−8
i.e.,
m =
∴ C.F. is
yc = e–x
∴
= –1± i
2
(C1cos x + C2 sin x)
y = Ae–x cos x + Be–x sin x
...(1)
be the complete solution of the given equation where A and B are functions of x to be found
291
DIFFERENTIAL EQUATIONS–II
We have
y 1 = e–x cos x
y 2 = e–x sin x
and
y1′ = – e–x (sin x + cos x)
y2′ = e–x (cos x – sin x)
W = y1 y2′ − y2 y1′
= e–2x
Also,
φ(x) = e–x sec3x
A′ =
bg
– y2 φ x
W
−x
B′ =
−x
–e
A = –
A = –
z
3
B′ =
tan x sec 2 x dx + C1 ,
tan 2 x
+ C1 ,
2
B =
Substituting these values of A and B in Eqn. (1), we get
e
z
−x
sec 2 x dx + C2
B = tan x + C2
F – tan x + C I e cos x + btan x + C g e sin x
GH 2
JK
e tan x sin x
+ e tan x sin x
e bC cos x + C sin x g –
2
2
y =
W
cos x ⋅ e − x sec 3 x
e– 2 x
2
B′ = sec x
sin x ⋅ e sec x
e– 2 x
A′ = – tan x sec2x,
A′ =
bg
– y1 φ x
−x
1
–x
2
–x
=
–x
–x
1
2
Thus,
b
g
e – x tan x sin x
2
This is complete solution of the given equation.
–x
y = e C1 cos x + C2 sin x +
EXERCISE 6.1
Solve the following equations by the method of variation of parameters:
1.
d2y
+ y = tan2 x.
dx 2
2.
d2y
+ y = x sin x.
dx 2
3.
d2y
dy
– 5⋅
+ 6 y = e4 x.
2
dx
dx
4.
d2y
+ 4 y = 4 sec2 2x.
dx 2
5. (D2 + D) y = x cos x.
b
g
Ans. y = C1 cos x + C2 sin x + sin x log sec x + tan x – 2
LMAns. y = C cos x + C sin x + 1 x sin x – 1 x cos xOP
2
4
N
Q
LMAns. y = C e + C e + 1 e OP
2
N
Q
Ans. y = C cos 2 x + C sin 2 x + sin 2 x log bsec 2 x + tan 2 x g − 1
LMAns. y = C + C e + 1 x bsin x − cos xg + cos x + 2 sin xOP
2
N
Q
2
1
2
2x
1
1
2
–x
1
2
3x
2
4x
292
ENGINEERING MATHEMATICS—II
LMAns. y = C cos ax + C sin ax – 1 x cos ax + 1 sin ax log sin ax OP
a
a
N
Q
6. (D2 + a2) y = cosec ax.
1
2
Ans. y = C1 cos x + C2 sin x + x cos x − sin x + sin x log sec x
7. (D2 + 1) y = sec x tan x.
LMAns. y = bC + C xge
N
8. (D2 + 2D + 1) y = e–x log x.
9. (D2 – 3D + 2) y =
2
1
–x
2
1
.
1 + e– x
+
g OPQ
b
1 2 −x
2 log x − 3
x e
4
d
i
d
Ans. y = C1e x + C2 e 2 x − xe x + e x log 1 + e x + e 2 x log 1 + e – x
10. (D2 – 6D + 9) y =
e3x
.
x2
LMAns. y = bC
N
13. (D2 + 1) y =
1
.
1+ sin x
1
1
ex
.
x
15. (D2 + 6D + 9) y =
e –3 x
.
x5
+ C2 x e 3 x +
2
3x
2
1
14. (D2 – 2D + 1) y =
6.2
g
Ans. y = C1 + C2 x e 3 x – e 3 x log x
g 18 d2 x − 4 x + 3i e OPQ
Ans. y = C cos x + C sin x + sin x log bsec x + tan xg + logcos x − 1
Ans. y = C cos x + C sin x – x cos x + sin x log b1 + sin x g – 1
11. (D2 – 2D + 1) y = x2 e3x.
12. (D2 + 1) y = log cos x.
b
i
2
b
g
Ans. y = C1 + C2 x e x + xe x log x
LMAns. y = bC
N
1
g
+ C2 x e –3 x +
1 –3 –3 x
x e
12
OP
Q
SOLUTION OF CAUCHY’S HOMOGENEOUS LINEAR EQUATION AND
LENGENDRE’S LINEAR EQUATION
A linear differential equation of the form
xn
n −1
n −2
dny
y
y
dy
n −1 d
n− 2 d
a
x
a
x
+
⋅
+
+ ⋅⋅⋅ + an −1 x ⋅
+ an y = φ( x )
2
1
dx
dx n
dx n −1
dx n − 2
...(1)
Where a1, a2, a3 ...an are constants and φ(x) is a function of x is called a homogeneous linear
differential equation of order n.
The equation can be transformed into an equation with constant coefficients by changing the
independent variable x to z by using the substitution x = ez or z = log x
Now
z = log x ⇒
dz 1
=
dx x
293
DIFFERENTIAL EQUATIONS–II
dy dy dz
=
⋅
=
dx dz dx
Consider
∴
x
1 dy
⋅
x dz
dy
dy
= Dy
=
dx
dz
d
.
dz
Differentiating w.r.t. ‘x’ we get,
where D =
x
d 2 y dy
+
⋅1 =
dx 2 dx
d2y
dx 2
x
i.e.,
d 2 y dz
⋅
dz 2 dx
=
d 2 y 1 dy
⋅ –
dz 2 x dx
=
1 d 2 y 1 dy
⋅
– ⋅
x dz 2 x dz
i.e.,
x2
d 2 y dy
d2y
−
=
dz 2 dz
dx 2
i.e.,
x2
d2y
= (D2 – D) y = D (D – 1) y
dx 2
Similarly,
x3
d 3y
= D (D – 1) (D – 2) y
dx 3
.............................................................
.............................................................
xn
dny
= D (D – 1) ... (D – n + 1) y
dx n
n
dy 2 d 2 y
n d y
,x
x
⋅⋅⋅⋅⋅
⋅⋅⋅⋅
in Eqn. (1), it reduces to a linear
dx
dx 2
dx n
differential equation with constant coefficient can be solved by the method used earlier.
Also, an equation of the form,
Substituting these values of x
bax + bg
n
⋅
b
dny
+ a1 ax + b
dx n
g
n −1
⋅
d n −1 y
+ ... any = ( x )
dx n −1
...(2)
where a1, a2 .....an are constants and φ (x) is a function of x is called a homogeneous linear differential
equation of order n. It is also called “Legendre’s linear differential equation”.
This equation can be reduced to a linear differential equation with constant coefficients by using
the substitution.
ax + b = ez or z = log (ax + b)
As above we can prove that
bax + bg ⋅ dydx
= a Dy
294
ENGINEERING MATHEMATICS—II
bax + bg
2
bax + bg
n
d2y
= a2 D (D – 1) y
dx 2
.............................................................
.............................................................
⋅
dny
= an D (D – 1)(D – 2) ..... (D – n + 1) y
dx n
The reduced equation can be solved by using the methods of the previous section.
⋅
WORKED OUT EXAMPLES
dy
d2y
− 2x
− 4y = x4.
2
dx
dx
Solution. The given equation is
1. Solve x 2
d2y
dy
− 2x
− 4 y = x4
2
dx
dx
Substitute
x = ez or z = log x
x2
So that
x
dy
= Dy,
dx
x2
...(1)
d2y
= D (D – 1) y
dx 2
The given equation reduces to
D (D – 1) y – 2Dy – 4y = (ez)4
[D (D – 1) – 2D – 4] y = e4z
i.e.,
(D2 – 3D – 4) y = e4z
...(2)
which is an equation with constant coefficients
A.E. is
i.e.,
∴
C.F. is
m2 – 3m – 4 = 0
(m – 4) (m + 1) = 0
m = 4, –1
C.F. = C1e4z + C2e–z
P.I. =
=
1
e4 z
3
4
D − D−
2
b4 g
1
2
bg
–3 4 −4
=
1
ze 4 z
2D − 3
=
1
ze 4 z
( 2) 4 − 3
=
1 4z
ze
5
bg
e4 z
D→ 4
Dr = 0
D→ 4
295
DIFFERENTIAL EQUATIONS–II
∴ The general solution of (2) is
y = C.F. + P.I.
y = C1 e4z + C2 e–z +
1 4z
ze
5
Substituting ez = x or z = log x, we get
–1
4
y = C1 x + C2 x +
d i
1
log x x 4
5
C2 x 4
log x
+
5
x
4
y = C1 x +
is the general solution of the Eqn. (1).
d2y
dy
− 3x
+ 4y = (x + 1)2.
dx
dx 2
Solution. The given equation is
2
2. Solve x
d2y
dy
− 3x
+ 4 y = (x + 1)2
dx
dx 2
Substituting
x = ez or z = log x
x2
x
Then
dy
= Dy,
dx
x2
...(1)
d2y
= D (D – 1) y
dx 2
∴ Eqn. (1) reduces to
D (D – 1) y – 3 Dy + 4y = (ez + 1)2
i.e.,
(D2 – 4D + 4) y = e2z + 2ez + 1
which is a linear equation with constant coefficients.
A.E. is
i.e.,
∴
m2 – 4m + 4 = 0
(m – 2)2 = 0
m = 2, 2
C.F. = (C1 + C2z) e2z
P.I. =
=
b
1
D−2
e
gd
2
e2 z
+
2z
i
+ 2e z + 1
2e z
+
...(2)
e0z
b D – 2g b D – 2 g b D – 2 g
2
2
2
= P.I.1 + P.I.2 + P.I.3
P.I.1 =
=
=
e2 z
b D – 2g
(D → 2)
2
e2 z
b2 – 2 g
ze 2 z
2 D–2
b
(Dr = 0)
2
g
(D → 2)
296
ENGINEERING MATHEMATICS—II
=
P.I.1 =
P.I.2 =
=
ze 2 z
2 2−2
b
g
(Dr = 0)
z2e2z
2
2e z
b D − 2g
2
(D → 1)
2
(D → 0)
2e z
b− 1g
2
P.I.2 = 2ez
P.I.3 =
=
e0 z
b D − 2g
e0z 1
=
4
4
1
z 2 2z
e + 2e z +
2
4
The general solution of Eqn. (2) is
y = C.F. + P.I.
P.I. =
y =
bC + C z g e
y =
bC + C log xg x
1
z 2 e2 z
+ 2e z +
2
4
ez = x or z = log x, we get
Substituting
1
1
2z
2
2
+
2
+
b g
x 2 log x
2
2
+ 2x +
1
4
is the general solution of the equation (1).
d2y
dy
+ 2x
− 12y = x2 log x.
2
dx
dx
Solution. The given Eqn. is
2
3. Solve x
d2y
dy
+ 2x
− 12 y = x2 log x
2
dx
dx
or
Substituting
x = ez
x2
x
dy
= Dy,
dx
...(1)
z = log x, so that
2
and x
d2y
= D (D – 1) y
dx 2
Then Eqn. (1) reduces to
D (D – 1) y + 2 Dy – 12y = e2zz
i.e.,
(D2 + D – 12) y = ze2z
which is the Linear differential equation with constant coefficients.
A.E. is
m2 + m – 12 = 0
...(2)
297
DIFFERENTIAL EQUATIONS–II
i.e.,
(m + 4) (m – 3) = 0
∴
m = – 4, 3
C.F. = C1e–4z + C2e3z
P.I. =
1
ze 2 z
D + D – 12
2
2z
= e
= e2z
z
b D + 2g + b D + 2g − 12
LM z OP
N D + 5D − 6 Q
(D → D + 2)
2
2
1
5
− z−
6
36
– 6 + 5D + D2
z
z–
5
6
5
6
5
6
0
LM
N
2z
P.I. = e −
OP
Q
LM
N
5
– e2z
z 5
−
z+
=
6 36
6
6
∴ General solution of Eqn. (2) is
y = C.F. + P.I.
FG
H
5
e2 z
z+
6
6
z
e = x or z = log x, we get
− 4z
+ C2 e 3z –
y = C1e
Substituting
–4
3
y = C1x + C2 x –
FG
H
IJ
K
5
x2
log x +
6
6
FG
H
5
C1
x2
log x +
+ C2 x 3 –
4
6
6
x
which is the general solution of Eqn. (1).
y =
OP
Q
IJ
K
IJ
K
d2y
dy
–x
+ y = 2 log x.
2
dx
dx
Solution. The given equation is
4. Solve x 2
x2
d2y
dy
–x
+y
2
dx
dx
= 2 log x
...(1)
298
ENGINEERING MATHEMATICS—II
x = ez
Substitute
x
Then
or
z = log x, in Eqn. (1)
2
dy
= Dy,
dx
and
x2
d y
= D (D – 1) y
dx 2
Eqn. (1) reduces to,
D (D – 1) y – Dy + y = 2z
(D2 – 2D + 1) y = 2z
i.e.,
...(2)
which is an equation with constant coefficients
m2 – 2m + 1 = 0
A.E. is
(m – 1)2 = 0
i.e.,
∴
m = 1, 1
C.F. = (C1 + C2z) ez
P.I. =
1
⋅ 2z
D − 2D + 1
2
2z + 4
1 – 2D + D2
2z
2z – 4
4
4
0
P.I. = 2z + 4
∴ The general solution of Eqn. (2) is
y = (C1 + C2z) ez + 2z + 4
Hence the general solution of Eqn. (1) is
y = (C1 + C2 log x) · x + 2 log x + 4.
d2y
dy
– 4x + 6y = cos (2 log x).
2
dx
dx
Solution. The given equation is
5. Solve x 2
d2 y
dy
– 4x + 6y = cos (2 log x)
2
dx
dx
Substituting
x = ez or z = log x, we have
x2
x
...(1)
2
dy
2 d y
= Dy, and x
= D (D – 1) y
dx
dx 2
∴ Eqn. (1) reduces to
D(D – 1)y – 4Dy + 6y = cos 2z
i.e.,
A.E. is
(D2 – 5D + 6) y = cos 2z
2
m – 5m + 6 = 0
...(2)
299
DIFFERENTIAL EQUATIONS–II
i.e.,
(m – 2) (m – 3) = 0
∴
m = 2, 3
C.F. = C1e2z + C2e3z
P.I. =
1
cos 2 z
D − 5D + 6
D2 → – 22
2
=
1
cos 2 z
– 2 − 5D + 6
=
1
cos 2 z
– 5D + 2
=
2 + 5D
1
cos 2 z ×
2 – 5D
2 + 5D
=
2 cos 2 z − 10 sin 2 z
4 − 25D2
=
2 cos 2 z − 10 sin 2 z
104
2
b
1
cos 2z – 5 sin 2 z
52
The general solution of Eqn. (2) is
y = C.F. + P.I.
P.I. =
2z
3z
= C1 e + C 2 e +
D2 → – 22
g
b
1
cos 2 z − 5 sin 2 z
52
g
Hence the general solution is
y = C1 x 2 + C2 x 3 +
b
b
g
b
g
1
cos 2 log x – 5 sin 2 log x .
52
g
d2y
dy
+x
+ 2y = x cos log x .
2
dx
dx
Solution. The given equation is
6. Solve x 2
d2y
dy
+x
+ 2 y = x cos (log x)
2
dx
dx
Substitute
x = ez or z = log x,
x2
x
Then we have
...(1)
dy
d2y
= Dy and x 2
= D (D – 1) y
dx
dx 2
∴ Eqn. (1) reduces to
D (D – 1) y + Dy + 2y = ez cos z
i.e.,
A.E. is
i.e.,
(D2 + 2) y = ez cos z
2
m + 2 = 0
m 2 = – 2;
...(2)
300
ENGINEERING MATHEMATICS—II
⇒
m 2 = 2i2
i.e.,
m = ± 2i
C.F. = C1 cos 2 z + C2 sin 2 z
P.I. =
1
e z cos z
D +2
z
= e
= ez
1
b D + 1g
2
+2
cos z
1
cos z
D + 2D + 3
(D2 → –12)
2
LM cos z OP
N –1 + 2 D + 3Q
L cos z OP
e M
N 2D + 2 Q
e L cos z D − 1O
M × P
2 N D + 1 D − 1Q
e L – sin z − cos z O
M D − 1 PQ
2 N
e L – sin z − cos z O
M – 2 PQ
2 N
e
bsin z + cos zg
4
= ez
=
(D → D + 1)
2
2
z
z
=
z
=
(D2 → –12)
2
z
=
z
=
∴ General solution of Eqn. (2) is
y = C.F. + P.I.
= C1 cos 2 z + C2 sin 2 z +
b
ez
sin z + cos z
4
g
∴ The general solution of Eqn. (1), as
y = C1 cos 2 log x + C2 sin 2 log x +
b
g
b g
b
g
x
sin log x + cos log x .
4
sin log x
d2y
dy y
= 5–
·
–
+3
2
x2
dx x
dx
Solution. Multiplying throughout the equation by x, we get
7. Solve 2x
b
g
sin log x
d2y
dy
+ 3x
− y = 5x –
2
dx
x
dx
z
Substitute
x = e or z = log x, in Eqn. (1)
2x2
...(1)
301
DIFFERENTIAL EQUATIONS–II
Then we obtain,
{2D (D – 1) + 3D – 1} y = 5ez – e–z sin z
i.e.,
A.E. is
i.e.,
∴
(2D2 + D – 1) y = 5ez – e–z sin z
...(2)
2
2m + m – 1 = 0
(m + 1) (2m – 1) = 0
1
2
m = – 1,
FG 1 IJ z
C.F. = C1e −2 + C2 e H 2 K
P.I. =
=
d
1
5e z − e – z sin z
2D2 + D − 1
i
1
1
5e z –
e – z sin z
2
2D + D − 1
2D + D – 1
2
= P.I.1 – P.I.2
5e z
P.I.1 =
2 D2 + D – 1
=
P.I.2 =
(D → 1)
5e z
2
1
e – z sin z
2D + D − 1
−z
= e
= e−z
= e−z
= e−z
(D → D – 1)
2
1
b
g b g
LM sin z OP
N 2 D − 3D Q
LM sin z OP
MN 2 d–1 i − 3D PQ
2
2 D −1 + D −1 −1
2
sin z
(D2 → – 12)
2
sin z
– 2 − 3D
−z
= −e
= – e−z
−z
= –e
LM sin z OP
N 3D + 2 Q
LM sin z × 3D − 2 OP
N 3D + 2 3D − 2 Q
LM 3 cos z − 2 sin z OP
N 9D − 4 Q
2
2
(D2 → –12)
302
ENGINEERING MATHEMATICS—II
LM 3 cos z − 2 sin z OP
N – 13 Q
e
b3cos z − 2 sin zg
13
= – e−z
−z
=
b
5 z 1 −z
e + e 3 cos z − 2 sin z
2
13
Complete solution of Eqn. (2) is
P.I. =
g
y = C.F. + P.I.
FG 1 IJ z
y = C1e − z + C2 e H 2 K +
b
g
b g
b g
5 z 1 –z
e +
e 3 cos z − 2 sin z
2
13
∴ The general solution of Eqn. (1) is
y =
3
8. Solve x
C1
+ C2
x
x+
FG
H
5
1
3 cos log x – 2 sin log x .
x+
2
13x
IJ
K
d3y
d2y
1
+ 2x 2
+ 2y = 10 x +
·
3
2
x
dx
dx
Solution. The given equation
FG
H
IJ
K
d 3y
d2y
1
+ 2x2
+ 2 y = 10 x +
2
3
x
dx
dx
Substitute
x = ez or z = log x
x3
x
Hence,
dy
= Dy,
dx
x2
...(1)
d2y
= D (D – 1) y
dx 2
d 3y
= D (D – 1)(D – 2) y
dx 3
Eqn. (1) reduces to linear differential equation as
x3
FG
H
z
[D (D – 1)(D – 2) + 2D (D –1) + 2] y = 10 e +
i.e.,
d
z
−z
(D3 – D2 + 2) y = 10 e + e
i
1
ez
IJ
K
which is linear differential equation with constant coefficients
A.E. is
m3 – m2 + 2 = 0
i.e., (m + 1) (m2 – 2m + 2) = 0
Hence
Therefore,
m = – 1, 1 ± i
C.F. = C1e–z + ez (C2 cos z + C3 sin z)
P.I. =
d
1
10 e z + e – z
D3 – D2 + 2
i
303
DIFFERENTIAL EQUATIONS–II
= 10
RS
TD
1
1
ez + 3
⋅ e– z
2
2
D − D +2
− D +2
3
= 10 [P.I.1 + P.I.2]
P.I.1 =
=
P.I.2 =
=
=
=
P.I.2 =
1
ez
D3 − D2 + 2
UV
W
D→1
ez
2
1
e– z
2
D −D +2
D → –1
3
1
b –1g − b– 1g
3
2
+2
e−z
Dr = 0
1
ze − z
3D − 2 D
D→– 1
2
ze – z
b g
2
b g
3 –1 − 2 –1
ze – z
5
RS e + ze UV
T2 5 W
z
P.I. = 10
–z
Hence the complete solution is
y = C.F. + P.I.
b
g
Substituting
ez = x or z = log x
We get
y =
m
b g
LM e + ze OP
N2 5 Q
z
= C1e – z + e z C2 cos z + C3 sin z + 10
−z
b gr
2 log x
C1
+ x C2 cos log x + C3 sin log x + 5x +
⋅
x
x
b
g
2
d3y
dy
3 d y
2x
– x2
+
+ xy = sin log x .
3
2
dx
dx
dx
Solution. Dividing throughout the equation by ‘x’, we get
9. Solve x 4
b g
2
sin log x
d 3y
dy
2 d y
2
–x
x
+
+y =
2
3
x
dx
dx
dx
Now substitute
x = ez and z = log x
x3
x
2
dy
2 d y
= Dy, x
= D (D – 1) y,
dx
dx 2
...(1)
304
ENGINEERING MATHEMATICS—II
x3
d 3y
= D (D – 1)(D – 2) y
dx 3
Eqn. (1) reduces to
sin z
ez
[D (D – 1)(D – 2) + 2D (D –1) – D + 1] y =
i.e.,
(D3 – D2 – D + 1) y = e–z sin z
m3 – m2 – m + 1 = 0
A.E. is
⇒ m2 (m – 1) – 1 (m – 1) = 0
⇒
(m2 – 1) (m – 1) = 0
⇒
m2 – 1 = 0,
⇒
m–1=0
m = ± 1, 1
⇒
m = – 1, + 1, + 1
C.F. = (C1 + C2 Z.) ez + C3 e–z and
P.I. =
–z
Taking e
1
b D − 1g b D + 1g
2
e – z sin z
(D → D – 1)
outside and replacing
1
–z
= e
b D − 2g D
–z
= e
dD
–z
= e
2
1
2
sin z
i
+ 4 – 4D D
sin z
sin z
D + 4D – 4D2
(D2 → –12)
3
sin z
–z
= e − D + 4D + 4
–z
= e
4D – 3
sin z
×
4D + 3 4D − 3
–z
= e
LM 4 cos z – 3sin z OP
N 16D − 9 Q
–z
= e
4 cos z – 3 sin z
– 25
=
(D2 → –12)
2
b
– 1 –z
e 4 cos z − 3 sin z
25
g
∴ The complete solution is
y = C.F. + P.I.
=
bC + C z g e
1
2
z
+ C3e – z +
b
1 –z
e 4 sin z − 3 cos z
25
g
305
DIFFERENTIAL EQUATIONS–II
x = ez
z = log x
bC
=
b
g
1
b
g
+ C2 log x ⋅ x +
b
g
b
g
1
C3
4 sin log x – 3 cos log x .
+
25x
x
g
dy
d2y
2
+ 3 3x + 2
− 36y = 3x + 4x + 1.
2
dx
dx
Solution. The given equation is
10. Solve 3x + 2
b3x + 2g
Substitute
2
⋅
2
⋅
b
g
dy
d2y
+ 3 3x + 2
− 36 y = 3x2 + 4x + 1
2
dx
dx
3x + 2 = ez or z = log (3x + 2)
(3 x + 2)
So that
b3x + 2g
2
...(1)
dy
= 3Dy
dx
d2y
= 32 D (D – 1) y
dx 2
ez – 2
3
Substituting these values in Eqn. (1), we get
Also,
x =
F e – 2I
3G
H 3 JK
z
3 D (D – 1) y + 3.3 Dy – 36y =
2
9 (D2 – 4) y =
2
F e – 2I + 1
GH 3 JK
z
+4
e2z − 1
3
d
i
1 2z
e −1
27
Which is a linear differential equation with constant coefficients
i.e.,
Now the A.E.
(D2 – 4) y =
...(2)
m2 – 4 = 0
Whose roots are m2 = 4
m = ±2
C.F. = C1e2z + C2e–
2z
and
d
i
1
1 2z
P.I. = D 2 − 4 27 e − 1
LM e
O
1
−
e P
MN b D − 2gbD + 2g D − 4 PQ
2z
=
1
27
=
1
P.I.1 − P.I.2
27
P.I.1 =
b
oz
2
e2z
D−2 D+2
gb
g
(D → 2)
(Dr = 0)
306
ENGINEERING MATHEMATICS—II
=
ze 2 z
2D
=
ze 2 z
4
(D → 2)
ze oz
P.I.2 =
D2 − 4
=
P.I. =
P.I. =
(D → 0)
1
−4
LM 1 ze + 1 OP
4Q
N4
1
d ze + 1i
108
1
27
2z
2z
General solution is
y = C.F. + P.I.
–2 z
2z
= C1e + C2 e +
b
y = C1 3x + 2
g
2
d
i
1
ze 2 z + 1
108
b
+ C2 3 x + 2
g
–2
+
b
1
3x + 2
108
g
2
b
g
log 3x + 2 + 1 .
EXERCISE 6.2
Solve the following equations:
2
1. x
LMAns. = L cos F 3 log I + sin F 3 log I O + OP
MN y x MMNC GH 2 xJK C GH 2 xJK PPQ x PQ
LMAns. y = x bC + C log xg + x OP
d y
dy
+ 5x
+ 4y = x .
36 Q
dx
dx
N
LMAns. y = C x + C x – L x + x + 1 OOP
d y
dy
– 20 y = (x + 1) .
+ 2x
MN 14 9 20 PQPQ
dx
dx
MN
d2y
+ y = 3x2.
dx 2
2
1
2
2
2. x
2
1
2
2
4. x
4
–5
2
1
2
2
d2y
dy
–x
+ 2 y = x sin (log x).
2
dx
dx
LMAns. y = x C cos blog xg + C sin blog xg – x log x cos blog xg OP
2
N
Q
LMAns. y = x C + C log x + x + x blog xg OP
d y
dy
1
+ 3x
+ y = x+ .
dx
x
dx
4
2
MN
PQ
1
2
2
2
5. x
2
2
2
2
3. x
4
−2
4
2
−1
–1
1
2
2
307
DIFFERENTIAL EQUATIONS–II
2
6. x
dy
d2y
+ 4x
+ 2 y = ex.
2
dx
dx
Ans. y = C1 x –1 + C2 x −2 e x
dy
d2y
–x
+ 4 y = cos (log x) + x sin (log x).
2
dx
dx
1
x
3 cos log x − 2 sin log x + sin log x
Ans. y = x C1 cos 3 log x + C2 sin 3 log x +
13
2
2
7. x
LM
N
e
2
8. x
j
e
b g
j
dy
d2y
– 3x
+ 5 y = x2 sin (log x).
2
dx
dx
LMAns. y = x
N
2
b g
b
2
11. x
d2y
dy
+ 5x
+ 4 y = x log x.
2
dx
dx
3
12. x
2
d3y
dy
2 d y
3
x
+
+x
+ 8 y = 65 cos (log x).
2
3
dx
dx
dx
–2
1
15.
–1
1
b
g
+ x C2 cos 3 log x + C2 sin
b2 x + 1g
b g
2
e
b2 x – 1g
2
b gOQP
j
b g
b g
x2
log x cos log x
2
3 log x + 8 cos log x – sin log x
b g
C1 cos log x + C2 sin log x –
b gOP
Q
d2y
dy
+ 1+ x
+ y = 4 cos log (1 + x).
2
dx
dx
Ans. y = C1 cos log 1 + x + C2 sin log 1 + x + 2 log 1 + x sin log 1 + x
2
⋅
b g
b
g
b g
3
b g
d2y
dy
– 2 2x + 1
– 12 y = 16x.
dx
dx 2
LMAns. y = C b2 x + 1g
N
3
1
16.
–1
3
1
LMAns. y = x
N
2
2
–2
d2y
dy
– 3x
+ 5 y = x2 sin (log x).
2
dx
dx
b1 + xg
g
LMAns. y = x bC + C log xg + C x + 1 x log x OP
4
N
Q
LMAns. y = x bC + C log xg + x FG log x – 2 IJ OP
9H
3K Q
N
2
d 3y
dy
3 d y
2
– x2
x
+
+ xy = 1.
2
3
dx
dx
dx
14.
b gOP
Q
x2
log x cos log x
2
Ans. y = C1 + C2 log x x + log x + 2
4
10. x
2
13. x
b g
C1 cos log x + C2 sin log x –
9. x2y″ – xy′ + y = log x.
LMAns. y = C x
N
b gOPQ
b g
b
g
b
g
+ C2 2 x − 1
–1
−
b g
b
g OPQ
b
g
3
1
2x + 1 +
16
4
d 3y
dy
+ 2x – 1
+ 2 y = 0.
3
dx
dx
LMAns. y = 2 x − 1 LC + C 2 x − 1
b g MM
b g
MN
N
1
2
3
2
+ C2 2 x − 1
– 3
2
OPOP
PQPQ
308
ENGINEERING MATHEMATICS—II
6.3 SOLUTION OF INITIAL AND BOUNDARY VALUE PROBLEMS
The differential equation in which the conditions are specified at a single value of the independent
variable say x = x0 is called an Initial Value Problem (IVP).
If y = y(x), the initial conditions usually will be of the form.
y(x0) = x0, y′(x0) = y1, ... y(n –1) (x0) = yn–1
The differential equation in which the conditions are specified for a given set of n values of the
independent variables is called a Boundary Value Problem (BVP).
If
y = y(x) the n boundary conditions will be
y (x1) = y1 ,
y(x2) = y2 ,
y(x3) = y3 ...
y(xn) = yn .
We can also have problems involving a system of d.e. (simultaneous de.s) with these type of
conditions.
WORKED OUT EXAMPLES
1. Solve the initial value problem
d 2x
dx
+5
+ 6x = 0, given that
2
dt
dt
bg
dx
0 = 15.
dt
Solution. We have (D2 + 5D + 6) y = 0
x (0) = 0,
A.E..
m2 + 5m + 6 = 0
(m + 2) (m + 3) = 0
⇒
m = – 2, – 3
Therefore general solution is
x = x (t) = C1e–2t + C2e–3t
...(1)
This is the general solution of the given equation
Now, consider,
x (0) = 0
Eqn. (1), becomes
x (0) = C1(1) + C2(1)
C1 + C2 = 0
i.e.,
...(2)
Also we have from Eqn. (1),
dx
= – 2C1e–2t– 3C2e–3t
dt
Applying the conditions,
bg
dx
0 = 15
dt
We obtain
– 2C1 – 3C2 = 15
Solving equations (2) and (3), we get C1 = 15, C2 = – 15
thus
x (t) = 15 (e–2t – e–3t).
...(3)
309
DIFFERENTIAL EQUATIONS–II
d2y
dy
+4
+ 3y = e–x subject to the conditions y(0) = y′(0).
2
dx
dx
Solution. We have (D2 + 4D + 3) y = e–x
2. Solve
A.E.
or
m2 + 4m + 3 = 0
(m + 1) (m + 3) = 0
m = – 1, – 3
C.F. = C1e–x + C2e–3x
P.I. =
=
=
e–x
D2 + 4 D + 3
D→– 1
e– x
b –1g + 4b –1g + 3
Dr = 0
2
x e– x
2D + 4
D→– 1
x e– x
2
y = C.F. + P.I.
=
y = C1 e – x + C2 e –3 x +
x e–x
2
...(1)
d
dy
1
= – C1 e – x – 3 C2 e –3 x +
– x e– x + e− x
dx
2
Consider the conditions y(0) = 1 and y′(0) = 1
y′ =
i
...(2)
Eqn. (1) and (2) become,
1 = C1 + C2 and 1 = – C1 – 3 C2 +
1
.
2
By solving these equations we get,
C1 =
Thus
y =
7
–3
and C2 =
4
4
7 – x 3 –3 x x e − x
e − e +
is the particular solution.
4
4
2
d2x
dx
+6
+ 25x = 0. If the particle
2
dt
dt
is started at x = 0 with an initial velocity of 12 ft/sec to the left, determine x in terms of t.
3. A particle moves along the x-axis according to the law
Solution. We have (D2 + 6D + 25) x = 0
From the given data, the initial conditions x = 0 when t = 0 and
dx
= – 12 when t = 0
dt
310
ENGINEERING MATHEMATICS—II
A.E.
m2 + 6m + 25 = 0
m = – 3 ± 4i
∴
x = x(t) = e–3t (C1 cos 4t + C2 sin 4t)
–3t
Now
x′(t) = + 3e
(– C1 sin 4t · 4 + 4 C2 cos 4t)
–3t
= – 12 e
Consider
...(1)
(C1 cos 4t – C2 sin 4t)
...(2)
x (0) = 0 and x′(0) = – 12
Eqns. (1) and (2) become,
0 = C1 and – 12 = 4C2 – 3C1
∴
C = 0 and C2 = – 3
x (t) = – 3e–3t sin 4t.
EXERCISE 6.3
Solve the following initial value problems:
1.
d2y
dy
2
+6
+ 9 y = 12e–3x = y (0) =0, = y ′ (0)
2
dx
dx
Ans. y = e –3x ⋅ x 4
LMAns. y = 1 bcos h x + cos x gOP
2
N
Q
LMAns. y = 1 d– 5 + e + 4 cos t – 2 sin t iOP
10
N
Q
d4y
– y = 0; y (0) = 1 and y′(0) = 0 = y″(0) = y ′′′ ( 0).
2.
dx 4
bg
3. y ′′′( t ) + y ′ t = e2t, y(0) = 0 = y′(0) = y ′′′ ( 0 ).
2t
ADDITIONAL PROBLEMS (From Previous Years VTU Exams.)
1. Using the method of variation of parameters solve:
Solution. We have (D2 + 1) y =
d2y
1
+y =
2
1 + sin x
dx
1
1 + sin x
m2 + 1 = 0 and hence m = ± i
C.F. = C1 cos x + C2 sin x
y = A(x) cos x + B(x) sin x
be the complete solution of the given d.e. where A(x) and B(x) are to be found.
y2 = sin x
We have
y 1 = cos x
A.E. is
∴
y1′ = – sin x
W = y1 y2′ − y2 y1′ = 1
y2′ = cos x
Also
φ(x) =
1
1 + sin x
...(1)
311
DIFFERENTIAL EQUATIONS–II
bg
Now,
A′ =
– y2 φ x
W
i.e.,
A′ =
– sin x
1 + sin x
consider
A′ =
A =
b
and
and
g = –1 +
– 1 + sin x − 1
z LMN
1 + sin x
–1 +
= – x+
= – x+
B′ =
OP
Q
B′ =
bg
y1 φ x
W
cos x
1 + sin x
1
sin x
1
dx + k1
1 + sin x
z
zd
1 − sin x
dx + k1
cos2 x
i
sec 2 x − sec x tan x dx + k1
A = – x + tan x – sec x + k1
Also,
B′ =
B =
=
b
...(2)
g
cos x 1 + sin x
cos x
1 − sin x
=
=
2
1 + sin x
cos x
cos x
z
zb
1 − sin x
dx + k 2
cos x
g
sec x − tan x dx + k 2
= log (sec x + tan x) + log (cos x) + k2
= log
FG 1 + sin x IJ + log bcos x g + k
H cos x K
2
= log (1 + sin x) – log (cos x) + log (cos x) + k2
B = log (1 + sin x) + k2
...(3)
Using Equations (2) and (3) in (1), we have
y = [– x + tan x – sec x + k1] cos x + [log (1 + sin x) + k2] sin x
i.e.,
y = k1 cos x + k2 sin x – x cos x + sin x – 1 + sin x log (1 + sin x)
The term sin x can be neglected in view of them k2 sin x present in the solution
Thus,
y = k1 cos x + k2 sin x – (x cos x + 1) + sin x log (1 + sin x).
2. Solve by the method of variation of parameters
d2y
+ 4y = 4 tan 2x.
dx 2
Solution. Refer page no. 282. Example 2.
3. Solve
d2y
dy
−2
+ 2y = ex tan x using the method of variation of parameters
2
dx
dx
Solution. Refer page no. 287, Example 7.
312
ENGINEERING MATHEMATICS—II
d2y
+ a 2 y = tan ax.
4. Solve
2
dx
Solution. We have (D2 + a2) y = tan ax
A.E. is m2 + a2
C.F. :
be the complete solution
We have
y1
= 0 ⇒ m = ± ia
= C1 cos ax + C2 sin ax
of the d.e. where A and B are functions of x to be found.
= cos ax
y2 = sin ax
y1′ = – a sin ax
y2′ = a cos ax
W = y1 y2′ − y2 y1′ = a
φ(x) = tan ax
Also
bg
– y2 φ x
W
B′ =
y1 φ x
W
A′ =
– 1 sin 2 ax
a cos ax
B′ =
cos ax tan ax
a
A′ =
2
– 1 1 − cos ax
a
cos ax
B′ =
sin ax
a
d
i
zb
A =
y =
Thus
RS 1
Ta
2
z
g
1
B =
cos ax − sec ax dx + k1 ,
a
1
⇒
A = 2 sin ax − log sec ax + tan ax + k1
a
cosax
+ k2
B = –
a2
Substituting these values in Eqn. (1), we get
⇒
bg
A′ =
b
g
sin ax
dx + k 2
a
UV
W
RS
T
sin ax − log sec ax + tan ax + k1 cos ax + –
y = k1 cos ax + k 2 sin ax −
b
g
UV
W
cos ax
+ k 2 sin ax
a2
1
log sec ax + tan ax cos ax .
a2
5. Using the method of variation of parameters find the solution of
ex
x
A.E. is m2 – 2m + 1 = 0 ⇒ (m – 1)2 = 0
m = 1, 1 are the roots of A.E.
∴
C.F. = (C1 + C2 x) ex
y = (A + Bx) ex
where A = A(x), B = B(x)
be the complete solution of the d.e. and we shall find A, B, we have
y2 = x ex
y1 = ex,
d2y
dy
ex
−2
+y=
·
2
dx
x
dx
Solution. We have (D2 + 2D + 1) y =
...(1)
313
DIFFERENTIAL EQUATIONS–II
y1′ = ex
∴
y2′ = (x + 1) ex
2x
W = y1 y2′ − y2 y1′ = e
Also
Further, we have
A′ =
bg
– y2 φ x
W
– xe x ⋅
A′ =
and
⇒
A =
e
x
bg
y1 φ x
W
ex ⋅
B′ =
e2 x
z
B′ =
ex
x
x
ex
x
e2 x
1
B′ = x
A′ = – 1
i.e.,
φ(x) =
z
1
dx + k 2
x
B = log x + k2.
–1⋅ dx + k1
B =
i.e.,
A = – x + k1
Using these values in Eqn. (1), we have
y = (– x + k1) ex + (log x + k2) x ex
i.e.,
y = (k1 + k2 x) ex + (log x – 1) x ex
x
The term – xe can be neglected in view of the term k2 xex present in the solution.
Thus
y = (k1 + k2 x) ex + x log x ex.
3
6. Solve x
d3y
d2y
dy
+ 3x 2
+x
+ 8 y = 65 cos (log x).
2
3
dx
dx
dx
Solution. Put
Thus, we have
t = log x or x = et
xy′ = Dy, n 2 y ′′ = D (D – 1) y
x 3 y ′′′ = D (D – 1) (D – 2) y
where D =
Hence, the given d.e. becomes
[D (D – 1) (D – 2) + 3D (D – 1) + D + 8] y = 65 cos t
i.e.,
= (D3 – 3D2 + 2D + 3D2 – 3D + D + 8) y = 65 cos t
or
(D3 + 8) y = 65 cos t
⇒
m3 – 23 = 0
A.E. :
m3 + 8 = 0
(m + 2) (m2 – 2m + 4) = 0
m = – 2 and
m2 – 2m + 4 = 0
By solving m2 – 2m + 4 = 0, we have
m =
2 ± 4 − 16 2 ± 2i 3
=
= 1± i 3
2
2
o
t
–2 t
C.F. = C1e + e C2 cos 3t + C3 sin 3t
t
d
dt
314
ENGINEERING MATHEMATICS—II
Also
P.I. =
65 cos t
D3 + 8
P.I. =
65 cos t 65 8 + D cos t ,
=
–D + 8
64 − D2
=
D2 → –12 = –1
b
g
b
65 8 cos t − sin t
g
65
P.I. = 8 cos t – sin t
Complete solution : y = C.F. + P.I.
with
=
{
C1
+ x C2 cos
x
D2 → –1
e
t = log x,
et = x
j
3 log x + C3 sin
e
3 log x
j}
+ 8 cos (log x) – sin (log x).
2
7. Solve x
d3y
d 2 y dy
3x
+
+
= x2 log x.
dx 3
dx 2 dx
Solution. Multiplying the equation by x, we have
x3
put
then
d 3y
d2y
dy
3
x
+
+x
= x3 log x
3
2
dx
dx
dx
t = log x or
...(1)
x = et
x 2 y ′′ = D (D – 1) y
xy′ = Dy,
x 3 y ′′′ = D (D – 1) (D – 2) y
Hence Eqn. (1) becomes
[D (D – 1) (D – 2) + 3D (D – 1) + D] e3t . t
i.e.,
D3 y = 0
A.E. :
m 3 = 0 and hence m = 0, 0, 0
∴
C.F. = (C1 + C2 t + C3 t2) eot
C.F. = C1 + C2 t + C3 t2
P.I. =
e 3t t
t
,
= e 3t
3
2
D
D−3
b
3t
= e
t
D 3 + 9 D 2 + 27 D + 27
P.I.3 is found by division
27 + 27D + 9D2 + D3
g
t
1
−
27 27
t
t +1
–1
–1
0
D→D+3
315
DIFFERENTIAL EQUATIONS–II
3t
P.I. = e ⋅
bt − 1g
27
The complete solution : y = C.F. + P.I.
with
t = log x and
Thus
{C + C log x + C blog xg } + 27x blog x − 1g .
b
g
2
b
1
2
3
g
dy
d2y
− 2x + 3
− 12y = 6x.
2
dx
dx
Solution. Put
t = log (2x + 3)
Hence
x =
Also, we have
and
3
2
y =
8. Solve 2x + 3
x = et
b2 x + 3g y′
b2 x + 3g y ′′
2
d
or et = 2x + 3
i
1 t
e −3
2
= 2Dy
= 22 D (D – 1) y
Hence, the given d.e. becomes
d
i
1 t
e −3
2
i.e., 2 (2D2 – 2D – D – 6) y = 3 (et – 3)
[4D (D – 1) – 2D – 12] y = 6 ⋅
i.e.,
A.E. is
d
i
3 t
e −3
2
2m2 – 3m – 6 = 0
(2D2 – 3D – 6) y =
m =
∴
∴
Also
3 ± 9 + 48 3 ± 57
=
4
4
C.F. = C1 e
C.F. =
P.I. =
3t
e4
d
e3+ 57 j t
4
LM
MNC e
1
+ C2 e
57
t
4
e3+ 57 j t
+ C2 e
3e t
2 2 D 2 − 3D − 6
4
−
57
t
4
OP
PQ
−
i 2 d2 D
3e t
9e ot
−
=
2 2 − 3− 6 2 0 − 0− 6
b
3e t 3
+
14 4
y = C.F. + P.I.
t = log (2x + 3), et = 2x + 3
P.I. = −
Complete solution
with
g b
9e ot
2
g
i
− 3D − 6
316
ENGINEERING MATHEMATICS—II
3t
R|
S|
T
y = e 4 C1 e
Thus,
57
t
4
+ C2 e
– 57
t
4
where t = log (2x + 3) and et = 2x + 3.
9. Solve the initial value problem:
U| 3
V| − 14 e
W
t
+
3
4
dy
d2y
dy
= 1 at x = 0 .
+4
+ 5y + 2cos hx = 0, given y = 0,
2
dx
dx
dx
Solution. We have (D2 + 4D + 5) y = – 2 cos hx
i.e., (D2 + 4D + 5) y = – (ex + e–x)
A.E. is m2 + 4m + 5 = 0
– 4 ± 16 − 20
= –2±i
2
C.F. = e–2x (C1 cos x + C2 sin x)
m =
∴
P.I. =
– ex
e– x
− 2
D + 4D + 5 D + 4D + 5
2
=
– ex
e–x
−
1+ 4 + 5 1+ 4 + 5
=
– e x e– x
−
10
2
LM e + e OP
N10 2 Q
x
P.I. = –
–x
Complete solution: y = C.F. + P.I.
g FGH
b
e x e– x
–2 x
+
C1 cos x + C2 sin x −
y = e
10
2
Now, we apply the given initial conditions, y = 0,
I
JK
...(i)
dy
= 1 at x = 0
dx
From Eqn. (i), we get
b
g
b
dy
= e –2 x – C1 sin x + C2 cos x – 2e –2 x C1 cos x + C2 sin x
dx
Using
y = 0 at x = 0, Eqn. (i) becomes
0 = C1 –
Using
FG 1 + 1 IJ
H 10 2 K
or
C1 =
3
5
dy
= 1 at x = 0, Eqn. (ii) becomes
dx
1 = C2 − 2C1 −
1 1
+
10 2
or
C2 − 2 C1 =
3
5
g
−
e x e– x
+
10
2
...(ii)
317
DIFFERENTIAL EQUATIONS–II
3
9
, we get C2 =
5
5
Thus, the required particular solution from Eqn. (i) is given by
Using
C1 =
y =
g FGH
b
I
JK
3 –2 x
e x e– x
·
e
cos x + 3 sin x −
+
5
10
2
dy
d2y
dy
= 2, y = 1 at x = 0.
–4
+ 5y = 0 . Subject to the conditions
2
dx
dx
dx
Solution. We have (D2 – 4D + 5) y = 0
A.E. : m2 – 4m + 5 = 0
10. Solve
m =
4 ± 16 − 20 4 ± 2i
=
= 2 ±i
2
2
∴
∴
C.F. = e2x (C1 cos x + C2 sin x)
y = e2x (C1 cos x + C2 sin x)
Also
dy
dx
Consider
= e2x (– C1 sin x + C2 cos x) + 2e2x (C1 cos x + C2 sin x)
y = 1
1 = 1 (C1 + 0)
Also by the condition
Using
Thus
dy
dx
...(1)
...(2)
at x = 0, Eqn. (1) becomes
∴ C1 = 0
= 2 at x = 0, Eqn. (2) becomes
2 = C2 + 2C1
C1 = 1, we get C2 = 0
y = e2x (cos x) is the particular solution.
11. Solve the initial value problem
bg
d2y
+ y = sin (x + a) satisfying the condition y = (0) = 0,
dx 2
y ′ 0 = 0.
Solution. We have (D2 + 1) y = sin (x + a)
A.E. :
m2 + 1 = 0 and hence m = ± i
∴
C.F. = C1 cos x + C2 sin x
P.I. =
b
sin x + a
D2 + 1
The denominator becomes zero
P.I. = x
=
P.I. =
g
D2 → – 12 = – 1
b
g
b
g
sin x + a
D
×
2D
D
x cos x + a
2
2 D2
b
– x cos x + a
2
D2 → – 12 = – 1
g
318
ENGINEERING MATHEMATICS—II
∴ The complete solution is y = C.F. + P.I.
b
– x cos x + a
2
y = C1 cos x + C2 sin x
b
Using
g
...(1)
b
cos x + a
x sin x + a
−
2
2
y ′ = – C1 sin x + C2 cos x
Now,
g
g
...(2)
y (0) = 0, y ′ (0) = 0 in Eqns. (1) and (2) respectively, we have
cos a
2
Thus by using these values in Eqn. (1), we get the particular solution,
C1 = 0 and C2 =
y =
b
x cos x + a
cos a
sin x –
2
2
b
g
1
cos a sin x − x cos x + a
2
12. Solve the initial value problem
=
g
bg
dx
d2x
dx
0 = 15 .
+4
+ 29x = 0, given x(0) = 0,
2
dt
dt
dt
Solution. We have (D2 + 4D + 29) y = 0
A.E. : m2 + 4m + 29 = 0
– 4 ± 16 − 116 – 4 ± 10i
=
= – 2 ± 5i
2
2
x(t) = e–2t (C1 cos 5t + C2 sin 5t)
m =
∴
Now,
...(1)
dx
= x′(t) = e–2t (– 5C1 sin 5t + 5C2 cos 5t) – 2e–2t (C1 cos 5t + C2 sin 5t) ...(2)
dt
us consider x(0) = 0 and x′(0) = 15
Let
Equations (1) and (2) respectively becomes
0 = C1 and 15 = 5C2 – 2C1
∴
C1 = 0
and C2 = 3
Thus,
x(t) = 3e–2t sin 5t is the required particular solution.
OBJECTIVE QUESTIONS
1. Match the following and find the correct alternative
I. Cauchy’s equation
II. Bernoulli’s equation
(i)
b x + 2g
2
(ii) x ⋅
2
b
g
d2y
dy
+ x+2
+ y=5
2
dx
dx
d 3y
d2y
– x 2 = ex
3
dx
dx
319
DIFFERENTIAL EQUATIONS–II
III. Method of variation of parameters
(iii)
dy
+ xy = x 2
dx
dy
+ xy = x 2 y 2
dx
(v) y dx2 – x dy2 = 0
(vi) (D2 + a) y = tan x
(iv)
(vii)
dy y − x
=
dx y + x
(a) I (i), II (iii), III (vii)
(b) I (ii), II (iii), III (v)
(c) I (i), II (iv), III (vi)
(d) I (ii), II (iv), III (vi)
Ans. d
2. The homogeneous linear differential equation whose auxillary equation has roots 1, 1 and
– 2 is
(b) (D3 + 3D – 2) y = 0
(a) (D3 + D2 + 2D + 2) y = 0
(c) (D3 + 3D + 2) y = 0
3. The general solution of
(a) y = C1 + C2 ex
(x2
(d) (D + 1)2 (D – 2) y = 0.
D2
(c) y = C1 + C2 x2
Ans. c
– xD), y = 0 is
(b) y = C1 + C2 x
(d) y = C1 x + C2 x2.
Ans. c
4. Every solution of y ′′ + ay ′ + by = 0, where a and b are constants approaches to zero as
x → α provided.
(a) a > 0, b > 0
(b) a > 0, b < 0
(c) c < 0, b < 0
(d) a < 0, b > 0.
Ans. a
5. By the method of variation of parameters y ′′ + a 2 y = sec ax, the value of A is
(a)
b
– log sec ax
a
2
g+k
1
(b)
– log sec ax
+ k1
a
log sec ax
+ k1
(d) None.
a3
6. By the method of variation of parameters, the value of W is called
(a) The Demorgan’s function
(b) Euler’s function
(c)
(c) Wronskian of the function
(d) Robert’s function.
Ans. a
Ans. c
7. The method of variation of parameters, the formular for A′ is
bg
(a)
y1φ x
W
(c)
– y2 φ x
W
bg
(b)
bg
y2 φ x
W
(d) None.
Ans. c
320
ENGINEERING MATHEMATICS—II
b
g
8. The equation a0 (ax + b)2 y ′′ + a1 ax + b y ′ + a2 y = φ(x) is called
(a) Legendre’s linear equation
(b) Method of undetermined coefficients
(c) Cauchy’s linear equation
(d) Simultaneous equation
Ans. a
b)2
9. The D.E. (ax +
y″ is
3
(a) a D (D – 1) (D – 2) y
(c) D · Dy
(b) a2 D (D – 1) y
(d) None.
Ans. b
10. The equation a0
y″ + a1 xy′ + a2y = φ(x) is called
(a) Legendre’s linear equation
(b) Cauchy’s linear equation
(c) Simultaneous equation
(d) Method of undetermined coefficients.
x2
Ans. b
11. If t = log x, the value of x is
(a) et
(c) xet
12. The initial value problem,
(a) C1 – C2 = 0
(c) C1 = 0
13.
(b) ex
(d) ext.
Ans. a
d 2x
dx
+5
+ 6 x = 0 x(0) = 0 is
2
dt
dt
(b) C1 + C2 = 0
(d) C2 = 0
Ans. b
d2y
dy
+4
+ 3 y = e – x. Subject to the condition y (0) = 1 is
dx
dx 2
(b) C1 + C2 = 0
(a) C1 + C2 = 1
(c) C1 = 0
(d) C2 = – 1
Ans. a
GGG
© Copyright 2024 ExpyDoc
