Subjects
Applied Mathematics for Electrical Engineering - 3130908
Complex Variables and Partial Differential Equations - 3130005
Engineering Graphics and Design - 3110013
Basic Electronics - 3110016
Mathematics-II - 3110015
Basic Civil Engineering - 3110004
Physics Group - II - 3110018
Basic Electrical Engineering - 3110005
Basic Mechanical Engineering - 3110006
Programming for Problem Solving - 3110003
Physics Group - I - 3110011
Mathematics-I - 3110014
English - 3110002
Environmental Science - 3110007
Software Engineering - 2160701
Data Structure - 2130702
Database Management Systems - 2130703
Operating System - 2140702
Advanced Java - 2160707
Compiler Design - 2170701
Data Mining And Business Intelligence - 2170715
Information And Network Security - 2170709
Mobile Computing And Wireless Communication - 2170710
Theory Of Computation - 2160704
Semester
Semester - 1
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Semester - 3
Semester - 4
Semester - 5
Semester - 6
Semester - 7
Semester - 8
Applied Mathematics for Electrical Engineering
(3130908)
AMEE-3130908
Winter-2019
Question-3c ( i )
BE | Semester-
3
Winter-2019
|
26-11-2019
Q3) (c ( i ))
3 Marks
Using Euler’s method find
y
(
0
.
2
)
, given that
dy
dx
=
y
-
2
x
y
;
y
(
0
)
=
1
taking
h
=
0
.
1
.
dy
dx
=
y
-
2
x
y
=
f
x
,
y
Now,
y
(
0
)
=
1
⇒
y
0
=
1
,
x
0
=
0
By Euler’s method :
y
n
+
1
=
y
n
+
h
[
f
(
x
n
,
y
n
)
]
Where,
x
n
=
x
0
+
n
h
,
n
=
0
,
1
,
2
,
3
,
…
For,
n
=
0
For,
x
1
=
x
0
+
h
=
0
+
0
.
1
=
0
.
1
y
1
=
y
0
+
h
f
(
x
0
,
y
0
)
y
1
=
1
+
(
0
.
1
)
f
(
0
,
1
)
y
1
=
1
+
(
0
.
1
)
1
-
2
(
0
)
1
y
1
=
1
.
1
For,
n
=
1
Now,
x
2
=
x
0
+
2
h
=
0
+
0
.
2
=
0
.
2
y
2
=
y
1
+
h
f
(
x
1
,
y
1
)
y
1
=
1
.
1
+
(
0
.
1
)
f
(
0
.
1
,
1
.
1
)
y
1
=
1
+
(
0
.
1
)
1
.
1
-
2
(
0
.
1
)
1
.
1
y
1
=
1
.
9182
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