Iteration

Second : Iteration
First : ByAns.
Iteration
Third: So So (U)= Q1. Note
question
So
f(1.375)
So-1.25-1
=(1.25)°
f(.25)
root
f(i.5)>0 the
f(2) Bisection
root
root f(1.5) = f(x) Find (Very This Note:|Hours]
TThree
ime:
Year-
2020
Dec.
=1.953-1.25-1
=0.297<0
lies lies 2-2-1
roots 1'-| Attempt : solution
lies =x'-x- tcarries
he
(1.5) =
=0.2246>0 l+1.5 1.5-1=0.875
=3.375- Attempt
between (.375) = +1.5
1.25 between 2 betweenl lies
-|=-1Method
f(x) real Section -A is Short
=1.25 between
=8-3=5
-1.375-1 -1.375 1.25 (positive) -1.5-1 l=0
x-x-1 =root Answer
3
all provided all
Sections
25 and (negative) : marks. Five
and
(negative)
and
(positive) 1.5 and
1
of Each by
1.375 1.5 (positive) questions,
the Questions) Dr. as
2 =0 [3
per
equation: Sangeeta
instructions.
x
5=
Numerical
Methods
15]
Gupta
decimal
places. Iteration
Eighth : Iteration
Seventh : Iteration
Sixth : Iteration
Fifth :
So So
f(1.321) So so
Iteration:
Fourth
root root =0.0183
f(1.329)>0
root
f(1.344) f(1.313)
So
root root
X7=: X6= X5
= X4
=
of lies
=-0.0158< lies lies lies
f(x)
1.321+1.329 between 1.313+1.329 between 1.313+1.344
=0.0837
between =-0.0494
1.313+1.375 between +1.3751.25
= 2
x' 2 2 2
-x 1.321 >
0 1.313 1.313 0 1.313 < IMaximum
Marks:
751
0
-
=1.325 (negative) =1.321
(positive) =1.329 (positive)-=1.344 =1.313
| and and and (negative)
and BCA-S04
=
0 1.329 1.329 1.344 1.375
is
1.32
upto
2
Givenp3. -x belation
put n have
We h=0.1,
Ans.
orA+1=E A=E-1 sf(x) where perator V:
So t
is f or botiesrator A:
where Af(a)difference wher hi valueisan
=0, Euler's given dx
Yn+1=yn +hYn+1dyy--f(«) =(E-1)f(x) -f(x)
Ef(x) = =
between f(x fla Vf(a cal ed
+h
h),backward
first
fa c=al ed
+ f(x) the
y
any
OjeratorDefinE:
inisEf(x)crement e
2,
1, 1,x= =
yy+X dx
+h)-f(x) +h) f(a)h), = operators
=
3,
=yn
method E
f(a)are
f(a first interval. f(x to
... y+x =f(a +h)
forward + the
(Yo x0
+ +hf(x, and are
h) BCA
tXYn
n is 0.1h=
0, the
+h)-f(a) the f(a) - E,
with A: Vth
Yn) values difference values
of andSemester
A
y
= of of
inx
f(x) of f(x) thV.Also
e
for 1 a of function /
x atfunction at function Numerical
x x
= = = obtain
0. a, a, f(x). f(x)
a f(x). a
Find + +
relation
Methods
h. h.
y(0.3)
2020/ /
between
by
113
Euler's
and E
method A.

taking
 Methods / 2020 / 114 BCA Vth Semester /Numerical Methods / 2020/ 115
Semester/Numerical/
8CAVA
Section-B Ilird approximation :
(Short Answer Questions)
ote:.AttemptanyTwo questions. Each
uestioncarries 7.5 marks.
(7.5 x 2 = 15]
/ll-0.:) Picard's method to approximate
=l.l+01 b6.Use
hen x = 0.2, given that y =1 when x 4

0and dy = X-y.
--x - 3 24
dx
(0.2)' (0.2)
yr0.2)=1+(0.2 -(0.2) 3 24
Ans. =x-y=f(x.y) =0.8374
1R-0.27 dx
.IS3-0.2 x =0. y(0)= 1 IVth approximation :
icard's Method :
approximation :
3 4
x

K4-03
=1N4-4.1
2 3 12 120
=13158
2
[at x=02] y (02)-08I74 ARs.

Qe Defne Sinpson's three-eight ule for Numerical integration. -0.2

Q7. Find the value of y when x = 10 for
the following table:
ss SmTSUt'Y 38 Rde: 2
=0.82 5 6
12 13 14 16
lnd approximation :
z=in *z imcam wtich 2ssumessthe ialues p } , corresponding to the valas af Ans. x =5, x,=6, x=9, x,=|1
Y, = 12. y, = 13, y, = 14, y;= i6
by Lagrange's formulay x-10
(00- 6X10-9%10-11),12
(5-6X5-9X5-11)
Wra io you uderstanó by Gausss (10-5X10-9X10-11)
elimánations method ? (6-5X6-9X6-11)
-x13

00-5X10-6X10-1) x<4
(9-SX9- 6X9 -11)
=i+: (10-5X10-6X10-9),16
{Fur Mre leformatisn Pleas Refer 92 (-5X11-6x1] -9)
Unit-IV Page-88| )02) =1 +(02)* -(02) -
(02)3
6
=083867
 / 11n
Numorcal Mothods / 2020 acA VIh Sementor /
BCA VVn Semostor / Numericol Meothods / 2020 / 117
5x 4x() Section - C
4×l-I) 5Ix)13+ 4x 3x(2) (Long Answer Questions)
433+ Il.67+ 5.33
Attempt any Three questions. Each Now 3) 20 17- y2) +22)}
15 marks.
14.7 Ans. estion carries i
[15 x 3 = 45] 17--(-1.0275) +20.0109))
20
following data : Apply Gauss-Seidal iteration method
Q8. UsingBessel's formula, find y(25) from the olve theequations. 10025
20 24
32
28
35
32
40 20x + y 2z 17 y3) -[-18-
20 3x(3) +(2)1
24
3x 420y - Z -18
Ans. Bessel's Formula :
2x -3y + 20z = 25 -18- 3(1.0025)+1.0109]
u(u GauSS-Seidal 0.9998
s.
Iteration Method :
2 3
t....
20x + y- 2z = 17 z 2 5 - 2x(3) +3y(3)|
Difference table is 3x+ 20y -z=-18
2% -3y + 20z = 25 25-2(1.0025) + 3(-0.9998)]
Ayu hese equation can be written as =0.9998
20 x(4) 17-y5) +22(3))
24
24
-5
x=(07-y+22) ..(1)
32-4
28 35
5
y= -18-3x + z) ..(2) l7-(-0.9998) +2(0.9998))
32 40 =0.9997
1
Z= (25-2x +3y) ...(3) y(4) =-18-3x(4) +z1
et start with x) = 0, y) = 0, z(l) = 0 in eq. (1)
u-X-225-240.2s
h 4 A
(2)
20 -18- 3(0.9997)+0.9998]
-7-y) +22)--0.85
20
=-0.9999

>aago.9-(-y-s.l). 3!
ut x(2) = 0.85, z(1) =0 in equation (2)

y2) =-18-3x(2) +z01

(4) 25-2x(4) - 3y(4)1
[25- 2(0.9997) + 3(-0.9999)]
1 =1.0000

=33.5 -0.75 +0.1406+0.054687 -18-3x 0.85 +0] Hence, x= 1.0, y = (-1.0), z = L.0 Ans.

=-1.0275
=32.945287 Ans. Q10. Use Runge-Kutta method of fourth
From equation (3) order, to find y (0.2) for the equation.
dy y y(o= 1, take h =0.2
z(2) =[25
20
-2x(2) +3y(2) dx y+x
Ans.
[25-2(0.85) + 3(-1.0275)]
dy_yX
dx y + x
-f(x,y)
=1.0109
Yo1 at xo =0, h 0.2
 ocA Vh Semester /Numerical Methods
BCA VO Senesteri Numerca
Methods / 2020 /e 118 2020 / 119
Reege-ketta Method :
using
hod,nearer to
Newton-Raphson's
findthe root of x4- - 10
x =: 2 correct to three
0 Pron (l ).(2). (3)
ofdecimal
2-0.2 02 LAlso obtain the rate of
enceo f Newton-Raphson's

=02
+0.1-0.1 #x)= x*-x-10

2 Using Tayiorstheorerm
Raphson's Method :
=92 (108335 -0.1 fxn) flay e, fia
.98385-0.1 =0.1662
4

(Yo*ks-9021662+ 02) =0.14144 L1662 - 0.27

43 -1
4

=16727 A.
4%? -1 aE f"ia) -
-9, 1,2,3, ... in equation (1)
by using
(aj Trapezitel re put xy =2
(b) Simpsort's 1/3 nde.

3u2+10-=171
2 2 4 4/2-1

441 871)' -1
31+19 341256, 19-= 1254 As. 2?2

413-1 441256)' -!
14,-141e Rate of
Converyesee :
L t differert from the rot of fx) -9 i a

Hesce the Rzte of Cunveszese o c

hy bievon's Mettod: Kaçhoon method is Qadratic Secoss
-1741-171911 Am.
2,
(-f1-ý-4,
(4-54-Sf-4
-24
t13-2 6512

24

24 -11%+34s-24j (-192A1)
15 24
1
-55522 -170x +120-8-82 -192x -51'-35*-30r -2:-10x -8x]
129
-(-20% +189%-400% +120}
249

3)=+9,2 -20x +6] Ans.

hls is required Cubic Polynomial.
 aCAVth Semester /
Nmerical Metbods / 2021 / 122
Numerical Methods the,
Find
form ofthe function from folowing given data:
Year - Dec. 2021 3
f(x): 3 6

IMaxiBCA-
mum Ma504rky:
11 1%
TimeThre Hngrs
ate Aemgt aSertis per istrcts f(x)
This seigrine is pred br Dr, Unvashi Chaudhary sf(x) s'(s)'f) s
6
Section - A
(Very Short Answer Questions)
ote : Attempt any Two questions. Each 18
question carries 7.5 9
Q1 Find a root of the equation f(x) =x- 4x marks. [2x7 27

met7.5-h1odi5)n
four itestions. -9 =0 using the
bisection -=X

1-9= -13
ton Forward Interpolation formula :
y=f(ko)+ Af
2

":*5-=2-3)=25
The
25=(25 -425)-9=3375 ie.negative
s
=3+3x +(x-x)
tew 25 3.
ierzxs2: A=25, 4=3 Ans.

6
.-21=(25-3) =275 (3. Evaluate
dx
by Trapezoidat Rule.
2=05 -4275,-9=9.7687>0 6l+?
ie.positive
iterztun 3:L,-25 £ x, 2.75 |ABs. Divide the Range of integration (0,6) into six
equal parts each width 6-0
-=1, hence h-1
6

(2625) =2625)
(25-275)= 2.625
The -42.625)
ros s ttween
2625 arnd 2.75. -9=3.5878>0 i.e.positive X0 =0y =lil+)= 1
lteratison4:Lan
2625 ard , =2.75 1/2=0.5
X)=2|1I/5 =0.2

f(26815)=
:The "3*2)=a425+275)-9==)2..363911<0815 ie.negative
rooA ies(2.6275) - 42.6075)
tetween 2.625 arnd 2.6875.
X3 =3
X4=4
WI0=0.10
/I7=0.0588235
T/26 =0.384615
1/37 =0.0270270
 Matrix Given
Ans. Q5. B range 0 Q4.
Ans. <xs05
1ly-z=33 8x-3y
4x+ 12 Y6=Y5
yS)4-0I*;y) )4=Y3
10-0liai*)=0+0.x0' +o»)=0 x Tratezndal
Rue
+y+4z= 2x Solve =0.
y2*))-01(a)=0+0.10.1'
0140025+
0.+(0.001))=0.05
X(04)
+08;)=0001+0.x(0.2)° +o'3*y2)=0.001 n+l=Yn Euier's
Given Use
form 8x-3y
4x+ 0+0.1 (sx<0.5
fx.y)=x*y
8 |2 2x Euler's
-3 1 of 2z= + +-0.1(1)=0.005
by + Formula dx
the 0.x2+y)=0.030022+0.1(0.5)*
11y
Gauss-elimination
23 +
y+
hf(x + 0.1, = =+y,
2 4 given - 42
2z=Z Method
: x,
=|23
=B
AX
Xsystem =
33 2312
= yn
) 0.2,
= Jn,nSD827=1.2569 An.
h- with 2/0
BCA
VIh
of..3) .2) .(1) =0.3, x, 0.1, 502+0|
h
+06588 Sesester
equation
method. 0.X(0.3) + x, =0,
y, 0.1
=

given
is 0.4, = x, find
to
Numeric/
/
+(0.005))=0.0140025 x=0.5 =0
by: +(0.030022)' ) the +0.3846)
(00140025)2y + solution Methods
+0).0270271 2021 /
of
0.551I1 =
)=0.030022 /e
dx dy
134

+y,
y(l}=0inthe

Pt Now matNoWrxform
from marks.
0.
By
Note
carries15
I=3,y=2,
=1z 3
x=3.35
Put
value
y=1.857x2
the
value
elimínate y 412707 [2
I equation(1
aiste
the Attempt :
means
2x=6.715 2x+ 7y=13.002 7y+ 14z=
of
7y+Z==0.857
25 x1 7 +4y1iz 42 from x
21 18 147y
2 25
following 1.857+ and y 14 of 21z= 18 142= + y 14
0.857
x inz from |
(l)
of any inz equation 25
equation
by and
Newton's 4(0.857) =25 2
table. One
equation
25
= nd
question (4) (5)
f(x): | Ans. =(1)
12
dividedQuestions)
Answer
(Short with .5)
48 4 the
out
Section-B help
100| difference
of of
294 the equaztion
following
900 | formula
10
11
13
(4)
1210 |
find Three
2028
tquestions.
he
value
of
f(8) [1 Eachx
and question
15
f(15) 15] =
 Ant. Q7.
From f15) I8)- fix,
=3150 =48- n
the 48=fíkgi-
á8
445
2.0 |4 10
2.7183 -
123.3201 dy given 52(15- 52/8 (x
7.38916.04964.95304.0352 -4X52} (x
Jy12 table. - -
4} 4)
- g)4firg) 2028 13
1210i1 BCA
(15 (8
*00.735|
3.318) 07351| 0.412 Find - (x 04
V
L0751 02 G.8978
33931.0966 y
- 4N8
4X1s-SX45) 4*x-SNI5) -
Semester
dy dy - -ir
2.0 18 16 1.2 10 5x45)-(8 409 310
- /Numerkcal
-0.08135 0.198%
0,2429 0.1627 0.1333
1.2.x=
at
okK
+(15- - (r
7.38916.04964.93304.05523.32012.7183 4X8-x)4f(xg)+
-4Xx
+ 4X15-5X15-7) Methods
0.01203- 0.04410.03610.0294 5\8 - -SXx
o361-(0.09) -7) 2021
/

0.002) 0.0080.0067 x! -7) /

|x 12A
1x
0.0013

Approximation
second :
First f:(x, Given approximationAh
Approximation:
Third
+Ji(.ygdds
Y4=yo
Approximation
Fourth :
Approximation PimetcUsaringdhod's
oroximsoatliountion
y3 =1+x+
-1-,1-xydx y) iy
-lt**** ==1+x+xX
Yo X*y.y= =
=l+*+**+ y obtain
upto a
+} 2
x. pCA
f(x.y2) y0) of Vh
3
6 at the
12012
3
x =I of
Semester
= equation.
24 4 dx 0 successive
24 4
Numerical /
\+x
dx

the
Approximation
Third : Second : Approximation
Approximation First : Regulaequation
given
Since Ans. 09.question Note:
We x-2= Methods
Required Sinilarly
L.695.
f(x)=(1.652)
-2=-0.2207 -(1.5))*2)-x2f(x)
-(1.652) ,2)K2)-x>iix;)f(x=(1.5j
-2-0.875 have
Given
Find folfowing
Questions)
Answer
(Long
Attempt
x}=|1.6956 f(x,)
Falsi i)<9. f(2)-2ffl)-2-x-2
f(0) fx)=x
0 a
carries 2021 /
xit) L.652(2) f(x)
root, -2(-0.2207)
x=15.x,2 fxz)-
f(x) x'-x- by real Section C
-
x 20-0.875)
1.52) - 2-(-2) Method lies
2-(-02207)= =-0.2207, = = 2 False /
correct |.6953 L.652, 2--0.875) -0.875, between 2) root
22.5 Five any 127
2 Two
to x, formula Position= of
f{x,) 0, marks.
0 tquestions.
thce i(x,) 2 bence
1 he questions
and [2
-2 =1.652 =2 : 2.real equation
decinai
6865 |
Method. 22.5 x
root
pluces of =Each out
the x45] of
is
 GiAnsven, Q11.Find Gaus' Ans. thefollowing
table. Q10,
Forward Interpolate
X 66 the =32.398
-012)-.S.S-S(-1.5|-25).
dx value interpolation
by BCA
the of
93 PopuBatiYeona()
r
09729O91 Ans mean
integral 2 Formula 10
2 ofSemesteVtrh
Tear Gauss's
20
1951 1941 1931 1921 1911 1901
: (1)
6 dy
13
formula /Numerical
Popl(in ation
thousand)
the Methods
20 1
080.923
0,5902|05 6 sing (y) population
Simpson's 2021 /
/
6 4 for 120

6
and
hia. 3

lterationIist Given s e04,04. he

et9.0)=020-02 Using
Runge-Kuta =07833979
rule, we
get
=0.783958 18 aCA
Vth
Semester
method
Numerical /
fourth of
Mthods
order
solve 2024 I
11
L.2056+0 1) [1.2-005601 |121+001)L21-01
=01967 =1aty(wi0th)
0)-=0.1%7
 equation
Rewrite
the Ans. Q13. Iteration:
Secnsd
y0.4)=y1 k4 kË
=hf(x(
y04-7x
-134)B) +4z]A)
x*5-1ly 292=71
+8y+ 3x+11y-
+13z=104
7x+52y4z=9583x 3x83x+ Find
7x+52y+ =1.37515
=1.1959 hf(*j =
29 + the -7;)
&y+ 1ly *k + 34209
solution h,y1
+10.1891 =0.2
292 13z 4z- +2k2 +kg)
=
La.1959)
(1.1959)-(0.2)(139018
71 of =02
104 =95the + 2k3 +
3) 1) 2(0.1795)a.5752-(04)|
La37s2
+(o4) La28565
system
ka) + +(0.3 | L0.29045)
(0.3) + (0.2)2 +
=0.1959
by 2(0.1793)
+ =0+0.2-0.2
}=4 +h
Gauss-Seidel (147018
+0.1688]
|=0.2:
f1.562895
I.742895 4)=09456(0.2)
Method.
=0.16879
=0.1688 +0.04
43209
=0.1793
=0.17949
x0.1795 =0.1891

Approximation
Fourth : Approximation
Third : Approximation
ucoNd : Putx
Since
at
x4x3*95-I1(141187)
Z4 Y4=71-3(1.05)
X=1.05, Third Zy 271-3X0.9877)
-8(1.4187)]
=1.9566
Y3 =104-7(0.9877) 83
Putx-|1446y,=1.L.34354599 1.,LADp1446,.1446z=0ro-x).iZm=0ationin:
Z=18207
1
=71-3(1.057) 29 1
1 83 1 and
[104-7(1.057)
y=
and 29 1
95 95-110.8459) equation
(A).we
Fourth
-110.369) 04-71.05) y
= in
ACA
1.366, equation
Approximation, Vh
z=1.96 - in
8(1.369)] SemasterMsmercaf
-
&1.367]-13(1.962)41.962)] +
-13(1.9566)] -13(L.3207)]41.3207)] + equation(C B).
+4(1.9566)] we
zet

= zet
the =1.96 1.962
values =1.367 =1.057 =1 =
369 l.05 = 1.41137 = 0.9877 zet
we
of
x, ethorts
y,
are z
same, 2021A F

correct 131
to
second

decimal

place
 -1375-1=0246
.f1373)=(379
leration :3 leratlon2: Now Iteration .The fa)-x-1*0 Ans us0ng Q1.0btalnquestion Note (Very
Funcion,decimal
places.
solutlThionaisNote: TiHoThrnue:ne Year-
Now . Now . 2022
Dec.
().29)=(125-26-|=-9296815
root
f03)=(0.5-1.5)-\=0315 12)-2-2-1*S
1 ro¡t bisection Attempt : Alfempt
y*|lroatie3s +13-1373 root lies Lta1,b2 : lics f(x)=x-1a0 real a carries Short
Section A
- pronided
ies betwoen betwecn
method al
bebween between root marks. 3Answerall Sectloms
and 1 of by
125 1.25S 1.S.
and I
2 correct the questions.
Questjons) Dr. per as
and and equation
(3x5=15) Urvashi
instructions.
1.375 1.5 Numeri
Methodscat
threeto Each Chaudhary
Iteration :6 lleration
:A0.5:3125)=(0.roThte 3125
-\3125-\=N515 Iteration4:
called isiference
Afinite Dference
ForwardAs, palrs:forward Q2.
ordered f(1303)=
Hence-.187634 -1.3281-|
:fO323)=(03231)}
=-0014S
lteratlon7:
=00ZAI1Nothew,
-134315-\A13479=134979
Define the rot lies
forvard dlifference roct
Aijg)=ix+h)-f) forward of 13125+13281 lies between
Gifference. he
2
betweæn 3-133-13125 |MMaa:"
almunm
table equation 13125
S+1313,=\343T5 1.3125 BCA-S04
differenca,
forthefollowing -=|3203 and
l3203 is and 1.375
1.3435
Make
(20,7MO.
9.0250)(22,7183),
3891),
MIst
we
|Ans,
+92=16 2X+y+2= 10
Ihe X+4y 3x+2y Solve
M. rex
=22, Findt X4
4 X3
3
y+Jz=6 elimx+4y+z
inate l6=2xty+z= 10
+2y+3z= 3x
I8 given
= the =9.0128
+3z18 02
18| 4.95304.16
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27183
X
value
system following (1L2,3.3201), of BCA
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from 159+-(03964)+-(0)--0094-0019--00
linearof atx=22for Semester
the system L.6359 13395 1.0966 0.3978 0.7351 06018 Vy
second equations (1.4,4.052),(1.6, Numerical /
and .) .) .4) equations
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. 06. question the Note Methods
Hence, f(x)=(2.058823)
-2(2.0S8823)-s find f[x) questions, Find following :
f(2.08126366)
root 3)=3-2x3-s=16 1-2x|-5=6 f)=f[0) =x'false Attempt(Short
=2.089639211
the 2.08126366(6)-3(-0.147204057) =2.08126366
=-0.390799917
root2x(16)-3(-)_35af(b)-b(f(a) A2)=2-2
liex2-5=
between s method
- of a 2022 /
root in
lie-3(-0.390799917)
s 2.058823(16) lies l6-(-1) -f(a)
f(b) =( =x-
f(x) real carries Section
this between -2 2x = Questions)
root Answer /
=2.094460846of =2.092739575
=2.09388371 X6 X4 between - Three any 134
=2.094305452
the X6 way, = x =)$ 7.5
16.147204057 2.08126366
-0.147204057 16.390799917 and 2 0-5=2x-50 ofquestiion
outot Two
we equation: the tmarks.
)=x'-2x obtain
2.058823 17
so:
3, position. -
=2.058823 (7.5: B
a=2&b G2=15)
&3. &3
-5 Hence Each
is
2.0945. we
jas mpson's Evaluate 8. missing 4|81 2|9
I= teX==31
rm
=-124124-4x
4x =0 sy=0 3 the
12 4 6 12

0.93060485 0.84089640.707106780.5087426
0

y=Vsinx
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3 is
31
4 124 -x
81x-9
|90-2x X-15
6 Ay
mis terming
0.982815 0 dx, x 4 ay pcA
with 81 3
in Vth
-3x105 X-19 Semester
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12 x
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s'y /Numerical
using

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of Simpson
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Find Ans. the =0 real question=3.60523926]=1.1872836 12x3
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the f(x) correct
root
Questions)
Answer
(Long
root followingAttempt 2022 /
= =-0.349373236
f(x)=2.55901
Xn+|=Xn 5826 f(x) rootNewton-Raphson
= +1] 0.8408964 +2(y2
+0.982815)
-0485)
0+4(0.5087426+
-0.00001274,=0.36880-4=1.X3 f(x2) f(x)=-0.056587 =0.2+. x
=-0.f(x2)002715
5830287 f(x)=1.75330
=0.33653+ X)
between 4e* lies carries
to of Section-C rule /
between
sin 3the +y4 135
decimal
x 15 any
0 -| equation marks. Five + +
it (0.349373236) f(a,) f(x)
2.559015826 and =0 Three 2(0.70710678+ y6
foll1.ows58302870.002775 0056587
1.75330 0 t..tYn-2)+yn]
0.5 places, and questions.
that method questions
the=0.37055
0.36880 = 0.5. 4e (15
=0.33653 given sin x3=45]
required to 0.9306
that -1fixnd Each out
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table.the
Ans.
Q11. Q10.
the
Lagrange's
Use values Using
fox)
Simpson's Lagr
(x(X-(XDX-3XX
=(x
(0- = = in
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-4)12)+-0X(24)
-1)[x-7x (x-0Xx
(3-0)(3-1)(3 -I)Xx
4) the
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-1)[x- 3Xx I)(0 formula
n= - - 0X2 N0)(x BCA
1)(x-4) 3\0
rule Sx - formulainterpolatlon Vth
+ + 4) - -)(x
to
12] 4),-12)+(X-0)(X-
12- XD(x2 Semes
-
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2x - ter
X3)-x3) 12

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-x)2 Numerlcnl
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(4-0)(4-1)(4
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find
the
1 6 3 8x (1-
+ -3)(10)(1
y=
J01+x 2x-6x] -4) (x|
0.54545 0.667 0.8571 (x3 (x Methods
0.5 0.6 0.75 + (x-
1+X 2x(x 3)(X - -
- X0(X| X0(x
X0X0)(x
4) /
with 1)(x (X3 functionform 2022
(o) - - 24
- - X2(x
X2
n=
-3) X|(X|)(x-
X3 /
(x| the of 136
6
-X3) -
-x2) X)
x2)3
1

y(x)
from

Iteration
Second :
Iteration
First : Ans.
-2X+z=9
10X.+
w2.Solve
nethodbiven
2x2y
Kimpson
Iteration
Third :
Put Put Put Put Put Rewrite
+ +
48 0.5] +48 +2[y3
+2(0.667) 3h rul2th8e
Put Put
x X= y X= 20y below
y Y=-2.38 x =0.9= 0,y=io-22-
z=
x11.09165=0.69322
3y 1+3(0.8571+
x =ky x Z=1.667
=y
xy=yin =
x,Z=0 the +
=
Xg= Zy1.809
1.407
Xy1.2094
yY2=-2.108 = z=0 [22+ 9-2y1-] 1the + - by Y6+3(y1
X 2z system
y2,Z=Z,in equation 10z +
,y=y2 ZZin , Z=z in Y9+)
=-44 Gauss's + BCA
in
equation = y2
in equation
2x 2x 22 0.75
in equation equation of + + Vh
equation + + yn] y4
equation equation 3y] 22] linear
Siedel + Semester
(1) 0.6 +
y5
(2) +0.54545)
(2) (1) (3) equations +
(1) (3) Iterative yt..] /
Numerical
we
...(3) "..(2) ...(1)
get
Iteration
Fourth :
Fifth PutMethods
Put
where methodQ13. Put
Step Ans. four x= y = x
hf(x0.Yo) kË
= 1 1,y=-2,z=2 Iteration =Y3 yY3-2.1196
Z,= Y3, = X3
=0.2[1
=0.2
: (x)
dxdydecimal
y(O)
X=1.1310
Using Ys=-2.064
1.807 Z,
Y4=-2.0599
1.809Z4
=
xy=1.1394) X , 2022
/
=0.202
-ozfo-o-o-o)
0.0102]
=0.2[1
+ =0.202
x0=0, -=|+y,
=
find
:
z= 1.845
X3
Z=,
/
+ 1 = y(0.2)
fourth-order in in 137
(0. Yo + places,
0 equation
equation
1)-] y?
0,= y(0)
h and
-0.2 =0, given (2)
y(0.4) (3)
h=0.2
Ranga-Kutta
dx
correct
=1
+
y² to
 +02=04227
+k=0.
04)=yj
2027 2:h-0+02-0.2 Nep
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=02027
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0404-044 BCA
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2160)=0.2027
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02062 )2/ /
2
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138

Raphson
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toines 3Short This tmtTeHo:hreurs Vear-De2c0.23
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is From
4 |35 2|1 1|0
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=
Vf(x)=f(x)-f(k-h)
5 Find =VE=& EV(1) VE=A =fx+h)-f(x)
8 6 and =f(x+h)-{(x)
f(x given the
=hf(*)
12-02 1 1
(2)
4! 2!
01|5 86 below first
I|23|45 (Maximum
Marks:13)
3 derivative . 2) BCA-504
4 |-6 the at
pointx
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1.2 =the
 |z=2.5 =5
22 X-y+z=1 0y=-6’R3-3R2 Gauss-elimination
method: 4. Ans.
z==2.5 1|x1-10-1
[1 R3 0-1
0y=-6
2]z5 -30 ’R3-2RË R3R2 -52| -32 -
-6y=-6 2x
1-5y 3x Solve =0.23
y=l| ’R2 +2y-
X-y+Z=1
=]-0.9-1.08 0.95
1Tx]1 -1. 4(0.2)
+ the
3R1 + 4z=53z=-6 -3r+ following 2(0.2)
2x-5y -1S(0.2)
X-y+z=1 24
- 1
X3
2y-
+ + SCA
4z equations
=5=-6 3z 22(0.2) 3(0.2)- Vth

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by
using /Numerical

iteration II: iteration :I xty²Ans.y'= of 05.

y y) method
the X-1+2.5=1
x=0.5 Methods/
=1+ =yo initialPerform
=1+x+1+x*+: =yo =l+|(x+)
=l+|(+I) dx dx f(x,
dx
to 2023
JO
Jx0 +
[+
2
y) y'=x+y;y(=10.) find
value two
+ =x 7
2-1x,|d f(x,y)
X
f(,yo) +
y? approXimate
problem
Picard's
an
iterations
of
140
y(0) 1=
dx
4
+ 2
dx
2x solution
dx

root
Iapproximation1=2-4=2
: f)=1-4
f(2) +Ans.
o6. estion Not
apositivesrequired hee:
(Short approximatiokonl is
x(2);
Now, Here Let Using following =1+
Here =-1.0369
f(1.6667) Attempt
526 lies =
8-8+1 carries Answer I+x+
=142-1||-2|
x) =1 f(x) method
root 2
=1.6667
xl2) + Y1=-2,y2 x1=l, x0) between =
not Section -B 23x2 +
X
+
y1=-1.0369,=1.6667,X1
=1.8364 =14x2|-2|+|1|=X+M2-X|x2ly| x* any =1 -
exceeding
of three
4x-4x the of marks.
+1=0.
7.5
2x3 2x3 BCA
3 and
1 twTYpe
o 3
=2 + False
equation
Vh
=1.6667
Iy1|+ly2l 2,.
1 questions. questions 4
x
20 X
Semester
2.03690.3333
y2 =2
X2 Questions)
position, [7.5 200Short 20 x
4 4
x1.0369 =! words.x
answer
2 Numerical /
= Each out
find 15] of
x3):Methods
Here
x(5); Here =-0.0172
f(1.8581)
Q7.Using
formula,
takes ’ Here =-0.00195
f(1.8605)
put knowthatWe Ans.
real =1.8605
x)+ =1.8364x(5) + /
f(a+
a =1.8608
=1.8581+x(4)
=1.8581
X=1.8364,
y1=-0.1526, 2023
=nh) on root y1=-0,0172,
=l
y2 X1
=1.8581,
0, 4 3 2 =1.860s,
y1=-0.00195,y2 XË
=| =1.8605 /
n the find
=x,f(a)+ = 123 5013 Newton's 1.8608
is
141
h=| following
0
the
73 37 13 A 1.001950.1395 1.15260.1636
"Cjafa) 1 cubic
forward
1.01720.1419
=2
X2
y2 =2
=1 X2
13
<0.00195 x0.1526
values
polynomial -x0.0172
- 50 3
"C,ari(a) 123
interpolation
4

which
+..
 2023 / 14a
Semester/Numerical Methods /
BCA Vh BCAVth Semester/ Numerical Methods / 2023 / 143

CAf0)+ Caf(0) +-,..... 3- COS X

f(x)=f(0 fx)-2x -
letK0)=4
=-0.99
2! 3! AI) -1.999, f(1.5)
f(2) =0.0006
=-l+x+ N(X -I)x + x(x -3x + 2) x2 rootlies between .5 and 2, since f(1.5).f(2) <0
=-l+x+6x-óx +2(0x-3x+ 2x)
1.5+2 =1.75, f(1.75) =-0.499
=-}-Zx +6x2+ -6r+4x 2
=2x-x -1 lies between 1.75 and 2
Ans. Soroot
1.75+2
=1.875, f(0.875) =-0.249
Q8. Using Simpson's rule, find the f(x)dx given that: 2
So root lies between 1.875 and 2
1 2 3 4 5
f(x) 10 50 70 80 100 (0)=1.9375, f(1.9375) =-0.12
root lies between 1.9375 and2
Ans.
(4) =1.96875, f(1.96875) = -0.061
root lies between 1.96875 and 2
By Simpson's 3 Rule
x) =1.9844, fU.9844) =-0.03
(x0 +nh
y(áx)=o+Ya)+4(y| +y3 t...) +2(y2 +y4 +..) root lies between 1.9844 and 2

f(x) =y (6) =1.9922, f(1.9922) =-0.01

1 10= yo So root lies between 1.9922
2 50 =y1
3
10=y2 )=1.9961
So root is 1.99 Ans.
4
80= y3
5 100 =y4 Q10. Using Lagrange's
x=2fromthefollowinginterpolation formula, find the value of y
=ffydalyo *y4)+391*y)+2y2) table:
X 3
corresponding to
4
-10-100) +3S0 +80)+2× 70] Ans,
6 S0 105

+390+140]
a

|x0
-{640] =213.33 Ans. 6 50 105
f(a) (b) f(c) (d)
Lagrange's Interpolation Formula :
Note: Atempt any three(Long Answer Section -C f(x)=(%-b(x -cXx- d) (x-aXx-Cx-)f(b)+ (x - ajx -b)x -d)
carries 15 marks. Answerquestions
is out ofType Questions)
the following (a -b\a-cXa -d) xf(a) + (b-aXb- cXb-d) (c- ujc - b}c -d) xf{c) +,..
Q9. Find a
ARs. 2x =3 +positive root of
required in detail. five questions. Each question (X-Jx-3Xx -4),c (x-0XX - 3Xx -4),63-0X -j-4),50
cos% the equation 2x =3 +cos x by [15 x3 45]
(0-1X0-3X0- 4) (|-0X1-- 3X1-4) (3-0)X3 - j(3- 4)
Bisection menod (x -0Xx-X%-3), 05
(4-0X4-1E4-3)
 Simpsoa's By
1vauate
0.6923 .9000
=ls-94695
1.6]= -
=-0.5)
rule :
Y6

Simpson'
usingsby
3(0.9737 +
,

12.5695
L6
+0.9+0.6925 )
=0,7856

+0.5903) rule.

Ans.
2x+
0.3

iterationII: lteration
Put | : 12
=0, y
iteration IV
:

yl=1-*+z]=3.555s
z021-+2y]=6944
=1-x2)+1=7.037 ) y
iterationIIl :

2-**291
X-x+3y-
2yty+z|||z3x Solvo byt
y3) +2y]=9.s601
D21-2 =0 +
D-y)-0=3088
0)-21-x)+2y]=9963 ==0.333 1 in 4z X-2y+4721
Gauss-Seidel
-1-+z=7.383 cquation 21 BCA

y(4)=-5.61x(4)
219)=8.858
=11.081 ..(3) ..(2) ...(1) VIIh
(1)
Somoster
method

Numorlcal /

iteration :V
iteration VI
: Methods
Ans. (take whenorderQ13.
Step1: Here
h=0.1
k4 kË dxdy Apply
dx h x to
hf(x0.yo) =
=0.1[0 -= 2023 /
=0.|0.1 = =0.1 f(x, = =0.2, find =10.83z6) =8.0yl--6.313
0) x6)
hf(x0 0.1) x+y»
+ y) Runge-Kutta /
++h, 1]= = an =1.571y5)
=12.3697) x(5) 145
(1.I
yo 0.1 = +y', x and given
approximate -6.313 =
168)*]=0.1347 +k3) 0. x=0,
y=l at
1I[x0 y that
=
1 method
=1.1152
yá] + h=0.1
when
Ans.
x value
= fourth
0 of
y
146
/
2023
/
Methods

Numerlcal =0.1551 =0.1576

0.1576)]=0.1823
=0.I[0.1+0.1+(|.1165+
Semoster/
=0011(Ll65491347|
2k+kal-0.165 ka]=0.1571
l65)²]=0.1347 2
VIh h=0.1
BCA
1L65
+22) +k3) + Ans.

+y 2k3 =1.l165
+0.1571
+ tk- (1.1
Y=1.1165, +h,y1 +
2k, kË
=hf[x1,y)
+ 0.1 2
0.1
0.1+
=0.12 +2kz +k =1.2736
k=k+y
=0.1| x hf(x1 k=-kjy
y(0.1)- =hf =
y(0.2)
Step
2: ky =
k4
So