UPSC Exam > UPSC Notes > Mathematics Optional Notes for UPSC > Ordinary Differential Equation
Ordinary Differential Equation | Mathematics Optional Notes for UPSC PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
Edurev123
5. Ordinary Differential Equation
5.1 Use Euler's method with step size ???
=?? .???? to compute the approximate value
of ?? (?? .?? ) correct upto few decimal places from the initial value problem
?? ?=?? (?? +?? )-?? ?? (?? )?=??
(2013 : 15 Marks)
Solution:
We use Euler's predictor equation
?? ?? =?? ?? -1
+h(?? 0
+(?? -1)h,?? ?? -1
)
to predict value at a point and then the equation
?? ?? ?? =?? ?? -1
+
h
2
[?? (?? ?? -1
,?? ?? -1
)+?? (?? ?? ,?? ?? ?? -1
)]
to refine it upto 5 decimal places
?? 0
=0,h=0.15,?? 0
=0
?? ?? '
=?? (?? +?? )-??
Mean Slope
(M.S.)
?? (?? )
=?? ?? -1
+h× M.S.
0 -2 - -0.30000
0.25 -2.02250 -2.01125 -0.30169
0.15 -2.02275 -2.01138 -0.30171
0.15 -2.02276 -2.01138 -0.30171
???? (0.15)=-0.30171
0.15 -2.02276 - -0.60512
0.30 -2.09154 -2.05715 -0.61028
0.30 -2.09308 -2.05792 -0.61039
Page 2
Edurev123
5. Ordinary Differential Equation
5.1 Use Euler's method with step size ???
=?? .???? to compute the approximate value
of ?? (?? .?? ) correct upto few decimal places from the initial value problem
?? ?=?? (?? +?? )-?? ?? (?? )?=??
(2013 : 15 Marks)
Solution:
We use Euler's predictor equation
?? ?? =?? ?? -1
+h(?? 0
+(?? -1)h,?? ?? -1
)
to predict value at a point and then the equation
?? ?? ?? =?? ?? -1
+
h
2
[?? (?? ?? -1
,?? ?? -1
)+?? (?? ?? ,?? ?? ?? -1
)]
to refine it upto 5 decimal places
?? 0
=0,h=0.15,?? 0
=0
?? ?? '
=?? (?? +?? )-??
Mean Slope
(M.S.)
?? (?? )
=?? ?? -1
+h× M.S.
0 -2 - -0.30000
0.25 -2.02250 -2.01125 -0.30169
0.15 -2.02275 -2.01138 -0.30171
0.15 -2.02276 -2.01138 -0.30171
???? (0.15)=-0.30171
0.15 -2.02276 - -0.60512
0.30 -2.09154 -2.05715 -0.61028
0.30 -2.09308 -2.05792 -0.61039
0.3 -2.09312 -2.05794 -0.61040
0.3 -2.09312 -2.05794 -0.61040
???? (0.3)=-0.61040
0.3 -2.09312 - -0.92437
0.45 -2.21345 -2.15329 -0.93339
0.45 -2.21753 -2.15532 -0.93370
0.45 -2.21766 -2.15539 -0.93371
0.45 -2.21767 -2.15539 -0.93371
???? (0.45)=-0.93371
0.45 -2.21767 - -1.26636
0.60 -2.39981 -2.30874 -1.28002
0.60 -2.40801 -2.31283 -1.28063
0.60 -2.40838 -2.31302 -1.28066
0.60 -2.40840 -2.31303 -1.28066
???? (0.6)=-1.28066
5.2 Use Runge-Kutta formula of fourth order to find the value of ?? at ?? =?? .?? ,
where
????
????
=v?? +?? , ?? (?? .?? )=?? .???? . Take the step length ?? =?? .?? .
(2014 : 20 Marks)
Solution:
Given that
????
????
=v?? +??
To find ?? (0.6; :
Page 3
Edurev123
5. Ordinary Differential Equation
5.1 Use Euler's method with step size ???
=?? .???? to compute the approximate value
of ?? (?? .?? ) correct upto few decimal places from the initial value problem
?? ?=?? (?? +?? )-?? ?? (?? )?=??
(2013 : 15 Marks)
Solution:
We use Euler's predictor equation
?? ?? =?? ?? -1
+h(?? 0
+(?? -1)h,?? ?? -1
)
to predict value at a point and then the equation
?? ?? ?? =?? ?? -1
+
h
2
[?? (?? ?? -1
,?? ?? -1
)+?? (?? ?? ,?? ?? ?? -1
)]
to refine it upto 5 decimal places
?? 0
=0,h=0.15,?? 0
=0
?? ?? '
=?? (?? +?? )-??
Mean Slope
(M.S.)
?? (?? )
=?? ?? -1
+h× M.S.
0 -2 - -0.30000
0.25 -2.02250 -2.01125 -0.30169
0.15 -2.02275 -2.01138 -0.30171
0.15 -2.02276 -2.01138 -0.30171
???? (0.15)=-0.30171
0.15 -2.02276 - -0.60512
0.30 -2.09154 -2.05715 -0.61028
0.30 -2.09308 -2.05792 -0.61039
0.3 -2.09312 -2.05794 -0.61040
0.3 -2.09312 -2.05794 -0.61040
???? (0.3)=-0.61040
0.3 -2.09312 - -0.92437
0.45 -2.21345 -2.15329 -0.93339
0.45 -2.21753 -2.15532 -0.93370
0.45 -2.21766 -2.15539 -0.93371
0.45 -2.21767 -2.15539 -0.93371
???? (0.45)=-0.93371
0.45 -2.21767 - -1.26636
0.60 -2.39981 -2.30874 -1.28002
0.60 -2.40801 -2.31283 -1.28063
0.60 -2.40838 -2.31302 -1.28066
0.60 -2.40840 -2.31303 -1.28066
???? (0.6)=-1.28066
5.2 Use Runge-Kutta formula of fourth order to find the value of ?? at ?? =?? .?? ,
where
????
????
=v?? +?? , ?? (?? .?? )=?? .???? . Take the step length ?? =?? .?? .
(2014 : 20 Marks)
Solution:
Given that
????
????
=v?? +??
To find ?? (0.6; :
Here
?? 0
=0.4,?? 0
=0.41,h=0.2
?? (?? 0
,?? 0
)=v0.81
???? 1
=h?? (?? 0
,?? 0
)=(0.2)v0.81=0.18
?? 2
=h?? (?? 0
+
h
2
,?? 0
+
?? 1
2
)=(0.2)?? (0.5,0.51)=0.2
?? 3
=h?? (?? 0
+
h
2
,?? 0
+
?? 2
2
)=(0.2)?? (0.5,0.51)=0.20099
?? 4
=h?? (?? 0
+h,?? 0
+?? 3
)=(0.2)?? (0.6,0.61099)=0.220099
?? =
1
6
(?? 1
+2?? 2
+2?? 3
+?? 4
)
=
1
6
[0.18+2(0.2)+2(0.20099)+0.220089]=
1
6
(1.202069982)
=0.2003449
?? 1
=?? (0.6)=?? 0
+?? =0.41+0.2003449=0.6103449
To find ?? (0.8) :
Here,
?? 1
?=0.6,?? 1
=0.6103449,h=0.2
?? 1
?=h?? (?? 1
,?? 1
)=(0.2)?? (0.6,0.6103449)=0.220031
?? 2
?=h?? (?? 1
+
h
2
,?? 1
+
?? 1
2
)=(0.2)?? (0.7,0.72036)=0.23836
?? 3
?=h?? (?? 1
+
h
2
,?? 1
+
?? 2
2
)=(0.2)?? (0.7,0.72952)=0.23913
?? 4
?=h?? (?? 1
+h,?? 1
+?? 3
)=(0.2)?? (0.8,0.85257)=0.257105
?? ?=
1
6
(?? 1
+2?? 2
+2?? 3
+?? 4
)=0.238686
?? 2
?=?? (0.8)=?? 1
+?? =0.6103449+0.238686
?=0.8490309
5.3 Solve the initial value problem
????
????
=?? (?? -?? ),?? (?? )=?? in the interval [?? ,?? .?? ]
using the Runge-Kutta fourth-order method with step size ?? =?? .?? .
(2015 : 15 Marks)
Solution:
Given :??????????????????????????????????????????
????
????
?=?? (?? -?? )=?? (?? ,?? )
Page 4
Edurev123
5. Ordinary Differential Equation
5.1 Use Euler's method with step size ???
=?? .???? to compute the approximate value
of ?? (?? .?? ) correct upto few decimal places from the initial value problem
?? ?=?? (?? +?? )-?? ?? (?? )?=??
(2013 : 15 Marks)
Solution:
We use Euler's predictor equation
?? ?? =?? ?? -1
+h(?? 0
+(?? -1)h,?? ?? -1
)
to predict value at a point and then the equation
?? ?? ?? =?? ?? -1
+
h
2
[?? (?? ?? -1
,?? ?? -1
)+?? (?? ?? ,?? ?? ?? -1
)]
to refine it upto 5 decimal places
?? 0
=0,h=0.15,?? 0
=0
?? ?? '
=?? (?? +?? )-??
Mean Slope
(M.S.)
?? (?? )
=?? ?? -1
+h× M.S.
0 -2 - -0.30000
0.25 -2.02250 -2.01125 -0.30169
0.15 -2.02275 -2.01138 -0.30171
0.15 -2.02276 -2.01138 -0.30171
???? (0.15)=-0.30171
0.15 -2.02276 - -0.60512
0.30 -2.09154 -2.05715 -0.61028
0.30 -2.09308 -2.05792 -0.61039
0.3 -2.09312 -2.05794 -0.61040
0.3 -2.09312 -2.05794 -0.61040
???? (0.3)=-0.61040
0.3 -2.09312 - -0.92437
0.45 -2.21345 -2.15329 -0.93339
0.45 -2.21753 -2.15532 -0.93370
0.45 -2.21766 -2.15539 -0.93371
0.45 -2.21767 -2.15539 -0.93371
???? (0.45)=-0.93371
0.45 -2.21767 - -1.26636
0.60 -2.39981 -2.30874 -1.28002
0.60 -2.40801 -2.31283 -1.28063
0.60 -2.40838 -2.31302 -1.28066
0.60 -2.40840 -2.31303 -1.28066
???? (0.6)=-1.28066
5.2 Use Runge-Kutta formula of fourth order to find the value of ?? at ?? =?? .?? ,
where
????
????
=v?? +?? , ?? (?? .?? )=?? .???? . Take the step length ?? =?? .?? .
(2014 : 20 Marks)
Solution:
Given that
????
????
=v?? +??
To find ?? (0.6; :
Here
?? 0
=0.4,?? 0
=0.41,h=0.2
?? (?? 0
,?? 0
)=v0.81
???? 1
=h?? (?? 0
,?? 0
)=(0.2)v0.81=0.18
?? 2
=h?? (?? 0
+
h
2
,?? 0
+
?? 1
2
)=(0.2)?? (0.5,0.51)=0.2
?? 3
=h?? (?? 0
+
h
2
,?? 0
+
?? 2
2
)=(0.2)?? (0.5,0.51)=0.20099
?? 4
=h?? (?? 0
+h,?? 0
+?? 3
)=(0.2)?? (0.6,0.61099)=0.220099
?? =
1
6
(?? 1
+2?? 2
+2?? 3
+?? 4
)
=
1
6
[0.18+2(0.2)+2(0.20099)+0.220089]=
1
6
(1.202069982)
=0.2003449
?? 1
=?? (0.6)=?? 0
+?? =0.41+0.2003449=0.6103449
To find ?? (0.8) :
Here,
?? 1
?=0.6,?? 1
=0.6103449,h=0.2
?? 1
?=h?? (?? 1
,?? 1
)=(0.2)?? (0.6,0.6103449)=0.220031
?? 2
?=h?? (?? 1
+
h
2
,?? 1
+
?? 1
2
)=(0.2)?? (0.7,0.72036)=0.23836
?? 3
?=h?? (?? 1
+
h
2
,?? 1
+
?? 2
2
)=(0.2)?? (0.7,0.72952)=0.23913
?? 4
?=h?? (?? 1
+h,?? 1
+?? 3
)=(0.2)?? (0.8,0.85257)=0.257105
?? ?=
1
6
(?? 1
+2?? 2
+2?? 3
+?? 4
)=0.238686
?? 2
?=?? (0.8)=?? 1
+?? =0.6103449+0.238686
?=0.8490309
5.3 Solve the initial value problem
????
????
=?? (?? -?? ),?? (?? )=?? in the interval [?? ,?? .?? ]
using the Runge-Kutta fourth-order method with step size ?? =?? .?? .
(2015 : 15 Marks)
Solution:
Given :??????????????????????????????????????????
????
????
?=?? (?? -?? )=?? (?? ,?? )
Step size??????????????????????????????????????????h=0.2
?? (2)?=3
Taking ?? 0
=?? (?? ),?? 0
=??
?? 1
?=h?? (?? 0
,?? 0
)=0.2?? (2,3)
?=0.4
?? 2
?=h?? (?? 0
+
h
2
,?? 0
+
?? 1
2
)=0.2?? (2.1,3.2)
?=0.462
?? 3
?=h?? (?? 0
+
h
2
,?? 0
+
?? 2
2
)=0.2?? (2.1.,3.231)
?=0.475
?? 4
?=h?? (?? 0
+h
1
?? 0
+?? 3
)=0.2?? (2.2,3.475)=0.561
?? ?=
1
6
(?? 1
+2?? 2
+2?? 3
+?? 4
)
?=
1
6
(0.4+0.462×2+0.475×2+0.561)
?=0.4725
?? 1
?=?? 0
+?? =?? (2.2)=?? (2)+?? =3+0.4725
?=3.4725( approx. )
Taking ?? 1
=?? (2.2)=3.4725,?? 1
=2.2
?? 1
=h?? (?? 1
,?? 1
)=0.2?? (2.2,3.4725)=0.5599
?? 2
=h?? (?? 1
+
h
2
,?? 1
+
?? 2
2
)=0.2?? (2.3,3.75245)=0.6681
?? 3
?=h?? (?? 1
+
h
2
,?? 1
+
?? 2
2
)=0.2?? (2.4,4.1655)=0.84744
?? ?=
1
6
(?? 1
+2?? 2
+2?? 3
+?? 4
)=0.6883
?? 2
?=?? (2.4)=?? 1
+?? =3.4725+0.6883=4.1608 (approx.)
???? (2.2)?=3.4725 (approx.)
?? (2.4)?=4.1608 (approx.)
5.4 Using Runge-Kutta method of fourth order, solve
????
????
=
?? ?? -?? ?? ?? ?? +?? ?? with ?? (?? )=?? at ?? =
?? .?? . Use four decimal places for calculation and step length 0.2 .
(2019 : 10 Marks)
Solution:
Page 5
Edurev123
5. Ordinary Differential Equation
5.1 Use Euler's method with step size ???
=?? .???? to compute the approximate value
of ?? (?? .?? ) correct upto few decimal places from the initial value problem
?? ?=?? (?? +?? )-?? ?? (?? )?=??
(2013 : 15 Marks)
Solution:
We use Euler's predictor equation
?? ?? =?? ?? -1
+h(?? 0
+(?? -1)h,?? ?? -1
)
to predict value at a point and then the equation
?? ?? ?? =?? ?? -1
+
h
2
[?? (?? ?? -1
,?? ?? -1
)+?? (?? ?? ,?? ?? ?? -1
)]
to refine it upto 5 decimal places
?? 0
=0,h=0.15,?? 0
=0
?? ?? '
=?? (?? +?? )-??
Mean Slope
(M.S.)
?? (?? )
=?? ?? -1
+h× M.S.
0 -2 - -0.30000
0.25 -2.02250 -2.01125 -0.30169
0.15 -2.02275 -2.01138 -0.30171
0.15 -2.02276 -2.01138 -0.30171
???? (0.15)=-0.30171
0.15 -2.02276 - -0.60512
0.30 -2.09154 -2.05715 -0.61028
0.30 -2.09308 -2.05792 -0.61039
0.3 -2.09312 -2.05794 -0.61040
0.3 -2.09312 -2.05794 -0.61040
???? (0.3)=-0.61040
0.3 -2.09312 - -0.92437
0.45 -2.21345 -2.15329 -0.93339
0.45 -2.21753 -2.15532 -0.93370
0.45 -2.21766 -2.15539 -0.93371
0.45 -2.21767 -2.15539 -0.93371
???? (0.45)=-0.93371
0.45 -2.21767 - -1.26636
0.60 -2.39981 -2.30874 -1.28002
0.60 -2.40801 -2.31283 -1.28063
0.60 -2.40838 -2.31302 -1.28066
0.60 -2.40840 -2.31303 -1.28066
???? (0.6)=-1.28066
5.2 Use Runge-Kutta formula of fourth order to find the value of ?? at ?? =?? .?? ,
where
????
????
=v?? +?? , ?? (?? .?? )=?? .???? . Take the step length ?? =?? .?? .
(2014 : 20 Marks)
Solution:
Given that
????
????
=v?? +??
To find ?? (0.6; :
Here
?? 0
=0.4,?? 0
=0.41,h=0.2
?? (?? 0
,?? 0
)=v0.81
???? 1
=h?? (?? 0
,?? 0
)=(0.2)v0.81=0.18
?? 2
=h?? (?? 0
+
h
2
,?? 0
+
?? 1
2
)=(0.2)?? (0.5,0.51)=0.2
?? 3
=h?? (?? 0
+
h
2
,?? 0
+
?? 2
2
)=(0.2)?? (0.5,0.51)=0.20099
?? 4
=h?? (?? 0
+h,?? 0
+?? 3
)=(0.2)?? (0.6,0.61099)=0.220099
?? =
1
6
(?? 1
+2?? 2
+2?? 3
+?? 4
)
=
1
6
[0.18+2(0.2)+2(0.20099)+0.220089]=
1
6
(1.202069982)
=0.2003449
?? 1
=?? (0.6)=?? 0
+?? =0.41+0.2003449=0.6103449
To find ?? (0.8) :
Here,
?? 1
?=0.6,?? 1
=0.6103449,h=0.2
?? 1
?=h?? (?? 1
,?? 1
)=(0.2)?? (0.6,0.6103449)=0.220031
?? 2
?=h?? (?? 1
+
h
2
,?? 1
+
?? 1
2
)=(0.2)?? (0.7,0.72036)=0.23836
?? 3
?=h?? (?? 1
+
h
2
,?? 1
+
?? 2
2
)=(0.2)?? (0.7,0.72952)=0.23913
?? 4
?=h?? (?? 1
+h,?? 1
+?? 3
)=(0.2)?? (0.8,0.85257)=0.257105
?? ?=
1
6
(?? 1
+2?? 2
+2?? 3
+?? 4
)=0.238686
?? 2
?=?? (0.8)=?? 1
+?? =0.6103449+0.238686
?=0.8490309
5.3 Solve the initial value problem
????
????
=?? (?? -?? ),?? (?? )=?? in the interval [?? ,?? .?? ]
using the Runge-Kutta fourth-order method with step size ?? =?? .?? .
(2015 : 15 Marks)
Solution:
Given :??????????????????????????????????????????
????
????
?=?? (?? -?? )=?? (?? ,?? )
Step size??????????????????????????????????????????h=0.2
?? (2)?=3
Taking ?? 0
=?? (?? ),?? 0
=??
?? 1
?=h?? (?? 0
,?? 0
)=0.2?? (2,3)
?=0.4
?? 2
?=h?? (?? 0
+
h
2
,?? 0
+
?? 1
2
)=0.2?? (2.1,3.2)
?=0.462
?? 3
?=h?? (?? 0
+
h
2
,?? 0
+
?? 2
2
)=0.2?? (2.1.,3.231)
?=0.475
?? 4
?=h?? (?? 0
+h
1
?? 0
+?? 3
)=0.2?? (2.2,3.475)=0.561
?? ?=
1
6
(?? 1
+2?? 2
+2?? 3
+?? 4
)
?=
1
6
(0.4+0.462×2+0.475×2+0.561)
?=0.4725
?? 1
?=?? 0
+?? =?? (2.2)=?? (2)+?? =3+0.4725
?=3.4725( approx. )
Taking ?? 1
=?? (2.2)=3.4725,?? 1
=2.2
?? 1
=h?? (?? 1
,?? 1
)=0.2?? (2.2,3.4725)=0.5599
?? 2
=h?? (?? 1
+
h
2
,?? 1
+
?? 2
2
)=0.2?? (2.3,3.75245)=0.6681
?? 3
?=h?? (?? 1
+
h
2
,?? 1
+
?? 2
2
)=0.2?? (2.4,4.1655)=0.84744
?? ?=
1
6
(?? 1
+2?? 2
+2?? 3
+?? 4
)=0.6883
?? 2
?=?? (2.4)=?? 1
+?? =3.4725+0.6883=4.1608 (approx.)
???? (2.2)?=3.4725 (approx.)
?? (2.4)?=4.1608 (approx.)
5.4 Using Runge-Kutta method of fourth order, solve
????
????
=
?? ?? -?? ?? ?? ?? +?? ?? with ?? (?? )=?? at ?? =
?? .?? . Use four decimal places for calculation and step length 0.2 .
(2019 : 10 Marks)
Solution:
Given:
?
?? '
=
?? 2
-?? 2
?? 2
+?? 2
at ?? =0,?? (0)=1,h=0.2,?? (0.2)=?
Using Runge-Kutta method for fourth order
?? (0.2)=?? (0)+?? ?? =
1
6
[?? 1
+?? 4
+2(?? 2
+?? 3
)]
?? 1
=h?? (?? 0
,?? 0
)
?? 2
=h?? (?? 0
+
h
2
,?? 0
+
?? 1
2
)
?? 3
=h?? (?? 0
+
h
2
,?? 0
+
?? 2
2
)
?? 4
=h?? (?? 0
+h
1
,?? 0
+?? 3
)
?? 1
=0.2?? (0,1)=0.2×1=0.2
?? 2
=0.2?? (0.1,1.1)=0.2×0.98361=0.19672
?? 3
=0.2?? (0.1,1.09836)=0.2×0.98356=0.19672
?? 4
=0.2?? (0.2,1.19671)=0.2×0.94566=0.18913
??
?? (0.2)=1+
1
6
{(0.2+0.18913+2(0.19672+0.19671)}
?? (0.2)=1+
1
6
(0.38913+0.78686)
?? (0.2)=1+
1
6
(1.17599)
?? (0.2)=1+0.195998
?? (0.2)=1.960( Required solution )
Read More
|
387 videos|203 docs
|
FAQs on Ordinary Differential Equation - Mathematics Optional Notes for UPSC
| 1. How are ordinary differential equations used in real-life applications? |  |
Ans. Ordinary differential equations are used in various real-life applications such as modeling population growth, predicting weather patterns, analyzing chemical reactions, and designing electrical circuits.
| 2. What is the difference between ordinary differential equations and partial differential equations? | |
Ans. Ordinary differential equations involve functions of a single variable, whereas partial differential equations involve functions of multiple variables. Additionally, ordinary differential equations deal with derivatives with respect to one independent variable, while partial differential equations involve derivatives with respect to multiple independent variables.
| 3. Can ordinary differential equations be solved analytically or numerically? | |
Ans. Ordinary differential equations can be solved both analytically and numerically. Analytical methods involve finding exact solutions using techniques like separation of variables, integrating factors, and power series. Numerical methods involve approximating solutions using algorithms such as Euler's method, Runge-Kutta methods, and finite difference methods.
| 4. What are the types of ordinary differential equations? | |
Ans. The types of ordinary differential equations include first-order differential equations, second-order differential equations, higher-order differential equations, linear differential equations, nonlinear differential equations, homogeneous differential equations, and nonhomogeneous differential equations.
| 5. How important are ordinary differential equations in the field of engineering? | |
Ans. Ordinary differential equations play a crucial role in engineering as they are used to model and analyze various physical systems such as mechanical vibrations, electrical circuits, control systems, and fluid dynamics. Engineers rely on solving differential equations to design and optimize systems for different applications.
Related Exams
About this Document
|
Nov 21, 2024 Last updated |
Document Description: Ordinary Differential Equation for UPSC 2024 is part of Mathematics Optional Notes for UPSC preparation.
The notes and questions for Ordinary Differential Equation have been prepared according to the UPSC exam syllabus. Information about Ordinary Differential Equation covers topics
like and Ordinary Differential Equation Example, for UPSC 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Ordinary Differential Equation.
Introduction of Ordinary Differential Equation in English is available as part of our Mathematics Optional Notes for UPSC
for UPSC & Ordinary Differential Equation in Hindi for Mathematics Optional Notes for UPSC course.
Download more important topics related with notes, lectures and mock test series for UPSC
Exam by signing up for free. UPSC: Ordinary Differential Equation | Mathematics Optional Notes for UPSC
Description
Full syllabus notes, lecture & questions for Ordinary Differential Equation | Mathematics Optional Notes for UPSC - UPSC | Plus excerises question with solution to help you revise complete syllabus for Mathematics Optional Notes for UPSC | Best notes, free PDF download
Information about Ordinary Differential Equation
In this doc you can find the meaning of Ordinary Differential Equation defined & explained in the simplest way possible. Besides explaining types of
Ordinary Differential Equation theory, EduRev gives you an ample number of questions to practice Ordinary Differential Equation tests, examples and also practice UPSC
tests
|
Explore Courses for UPSC exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Ordinary Differential Equation | Mathematics Optional Notes for UPSC
,Important questions
,past year papers
,ppt
,study material
,Exam
,Sample Paper
,Viva Questions
,Free
,Objective type Questions
,video lectures
,Semester Notes
,MCQs
,Ordinary Differential Equation | Mathematics Optional Notes for UPSC
,Ordinary Differential Equation | Mathematics Optional Notes for UPSC
,practice quizzes
,mock tests for examination
,Previous Year Questions with Solutions
,Summary
,Extra Questions
,shortcuts and tricks
;
Additional Information about Ordinary Differential Equation for UPSC Preparation
Ordinary Differential Equation Free PDF Download
The Ordinary Differential Equation is an invaluable resource that delves deep into the core of the UPSC exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Ordinary Differential Equation now and kickstart your journey towards success in the UPSC exam.
Importance of Ordinary Differential Equation
The importance of Ordinary Differential Equation cannot be overstated, especially for UPSC aspirants.
This document holds the key to success in the UPSC exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Ordinary Differential Equation Notes
Ordinary Differential Equation Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Ordinary Differential Equation.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Ordinary Differential Equation Notes on EduRev are your ultimate resource for success.
Ordinary Differential Equation UPSC Questions
The "Ordinary Differential Equation UPSC Questions" guide is a valuable resource for all aspiring students preparing for the
UPSC exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Ordinary Differential Equation on the App
Students of UPSC can study Ordinary Differential Equation alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Ordinary Differential Equation,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Ordinary Differential Equation is prepared as per the latest UPSC syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup