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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 )
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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.
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