methods of solving ode
DESCRIPTION
This presentation gives the basic idea about the methods of solving ODEs The methods like variation of parameters, undetermined coefficient method, 1/f(D) method, Particular integral and complimentary functions of an ODETRANSCRIPT
Linear ODE
HOMOGENEOUS
CONSTANT COEFFICIENT
VARIABLE COEFFICIENT
SEULER-CAUCHY
NON-HOMOGENEO
US
CONSTANT COEFFICIENT
VARIABLE COEFFICIENT
SEULER-CAUCHY
NON-HOMOGENEOUS O.D.E. WITH CONSTANT COEFFICIENTS
Solution of Non-Homogeneous ODE
Solution of Corresponding Homogeneous ODE(C.F.)
Particular Integral Of Given Non-Homogeneous ODE(P.I)
General Solution =C.F.+P.I.
Method of Finding Particular
Integral
Method of Undetermined Coefficients
Method of Variation of Parameters
1
F(D)
Method of Undetermined Coefficients
The Following Table shows the choice of
The Choice for Yp is made using the following three rules on the basis of the table as well.
r(x)
Basic Rule If r(x) is one of the functions given in the first column of
the table 2.1, choose Yp in the
same line.
Modification Rule If a term in your choice of Yp happens to be a basic solution of the Homogeneous ODE then multiply your
choice by x
Sum Rule If r(x) is sum of
functions in the first column of the Table 2.1, choose for Yp the sum of the functions in the corresponding lines of the second column.
r(x) Basic Rule
The Application of Basic Rule
Solve2y''+ y = x
2
2
0x
h
h
(D +1)y = 0
AuxiliaryEquation :D +1= 0
D = ±i
C.F. y = e (A cosx+Bsinx)
y (C.F.) = A cosx+Bsinx
=
Step-1: General Solution of the homogeneous ODE
Step-2: Solution Yp of the Non-homogeneous ODE
We look for choice Yp from the table for the given 2 nr(x) = x (Of theformx ,n )= 2
From the following table
2p 0 1 2
''p
2
2 22 0 1 2
Let y = K + K x+ K x .
Theny = 2K .
Bysubstituting this iny''+ y = x weget,
2K + K + K x+ K x = x x x= + + 20 0 1
2 0
1
2 02
p
By comparing the coefficients,
(2K + K ) = 0
K = 0
K =1 K = -2
So,y (P.I.) = -2+ x is the required P .I .
Þ
GENERAL SOLUTION(G.S.=C.F.+P.I.)
y(x) A cosx Bsinx x .= + - + 22
h py(x) y (x) y (x)= +
r(x) Modification Rule
The Application of Modification Rule
xSolve: y'' y' y e-+ + = 23 2 30
2
2
x xh
x x
(D + D )y = 0
AuxiliaryEquation :D + D = 0
(D )(D )
D ,
C.F. y = c e c e
y e and y e
- -
- -
+
+
Þ + + =
Þ = - -
= +
= =
21 2
21 2
3 2
3 2
1 2 0
1 2
Step-1: General Solution of the homogeneous ODE
Solution of homogeneous equation : y'' y' y+ + =3 2 0
Step-2: Solution Yp of the Non-homogeneous ODE
We look for choice Yp from the table for the given 2x xr(x) = e (Of theformKe )-30
From the following table
xy e-= 22
' x xp
'' x xp
'' ' xp p p
x x x x x x
x x x x x x
Theny Ke Kxe
&y Ke Kxe
Substitutingthesevaluesininthegiven diff.eqn.
y y y e
Ke Kxe (Ke Kxe ) Kxe e
Ke Kxe Ke Kxe Kxe e
- -
- -
-
- - - - - -
- - - - - -
= -
= - +
+ + =
Þ - + + - + =
Þ - + + - + =
2 2
2 2
2
2 2 2 2 2 2
2 2 2 2 2 2
2
4 4
3 2 30
4 4 3 2 2 30
4 4 3 6 2 30x x
xp
Ke e
By comparing the coefficient
K
y (P.I.) = xe is the required P .I .
- -
-
Þ - =
= -
-
2 2
2
30
30
30
But if we look at the C.F. it has the solution
which is same as choice of x
py Ke-= 2
So, by modification rule we have to multiply the above choice of Yp
by x, i.e., we have the modified choice xp
y Kxe-= 2
GENERAL SOLUTION(G.S.=C.F.+P.I.)
x x xy(x) c e c e xe .- - -= + -2 21 2
30
h py(x) y (x) y (x)= +
r(x) Sum Rule
The Application of Sum Rule
Solve: y'' y' y x sin x+ + = +24 5 25 13 2
x
hx
h
(D D )y = 0
AuxiliaryEquation :
D D = 0
D i,
C.F. y = e (A cosx+Bsinx)
y (C.F.) = e (A cosx+Bsinx)
-
-
+ +
+ +
- + -= = - ±
=
2
2
2
2
4 5
4 5
4 16 202
2
Step-1: General Solution of the homogeneous ODE
Solution of homogeneous equation : y'' y' y+ + =4 5 0
Step-2: Solution Yp of the Non-homogeneous ODE
We look for choice Yp from the table for the given r(x) = x sin x r (x) r (x)+ = +2
1 225 13 2
From the following table
n(Here r (x)is of theformx ,n
& r (x) is of the formksin x, ) =
=1
2
2
2
2p 0 1 2
'p''p
Let y = (K + K x+ K x ) (K sin x K cos x).
Theny = K K x K cos x K sin x.
y = K K sin x K cos x.
+ +
+ + -
- -
3 4
1 2 3 4
2 3 4
2 2
2 2 2 2 2
2 4 2 4 22
20 1 2
Bysubstituting this iny''+ y' y = x sin x,weget,
( K K sin x K cos x)
(K K x K cos x K sin x)
(K + K x+ K x K sin x K cos x)
x x sin x cos x
+ +
- -
+ + + -
+ + +
= + + + +
2 3 4
1 2 3 4
3 42
4 5 25 13 2
2 4 2 4 2
4 2 2 2 2 2
5 2 2
0 0 25 13 2 0 2
( K K K ) ( K K )x K x
( K K K )sin x ( K K K )cos x
x sin x
Þ + + + - - +
+ - - + + - + +
= +
22 1 0 2 1 2
3 4 3 4 3 42
2 4 5 8 5 5
4 8 5 2 4 8 5 2
25 13 2
( K K K ) ( K K )x K x
(K K )sin x (K K )cos x x sin x
Þ + + + - - +
+ - + + = +
22 1 0 2 1 2
23 4 4 3
2 4 5 8 5 5
8 2 8 2 25 13 2
( K K K ) ( K K )x K x
(K K )sin x (K K )cos x x sin x
Þ + + + - - +
+ - + + = +
22 1 0 2 1 2
23 4 4 3
2 4 5 8 5 5
8 2 8 2 25 13 2
By comparing the coefficients we get,
K K K , K K , K ,
K K , K K
+ + = - - = =
- = + =2 1 0 2 1 2
3 4 4 3
2 4 5 0 8 5 0 5
8 13 8 0
KK K &K K .
& K K , K K ( i.e.,K K )
Þ - - = = Þ = - = -
- = + = = -
22 1 2 1
3 4 4 3 4 3
88 5 0 5 8
58 13 8 0 8
K ( K ) gives K K
K K &K K .
using these in K K K ,
we have K K
Þ - - = + =
Þ = Þ = = - = -
+ + =
- + = Þ =
3 3 3 3
3 3 4 3
2 1 0
0 0
8 8 13 64 13
1 865 13 8
5 52 4 5 0
2210 32 5 0
5
pSo, y (P.I.) x x (sin x cos x)= - + + -222 1
8 5 2 8 25 5
GENERAL SOLUTION(G.S.=C.F.+P.I.)
xy(x) e (A cosx+Bsinx) x x (sin x cos x).
-= + - + + -
2 222 18 5 2 8 2
5 5
h py(x) y (x) y (x)= +
4251 3
0011 0010 1010 1101 0001 0100 1011Method of Finding Particular Integral
Method of Variation of Parameters
4251 3
0011 0010 1010 1101 0001 0100 1011
Method of Variation of Parameters
4251 3
0011 0010 1010 1101 0001 0100 1011
x
h
h
(D )y = 0
AuxiliaryEquation :D = 0 D i,
C.F. y = e (A cosx+Bsinx)
y (C.F.) = A cosx+Bsinx
Next,we find theP.I.
Here we have y cosx and y sinx
+
+ Þ = ±
=
= =
2
2
0
1 2
1
1
Solution of homogeneous equation : y'' y+ = 0
' 'So,W W(y ,y ) y y y y
cosx(cosx) sinx( sinx) cos x sin x
= = -
= - - = +
=
1 2 1 2 2 12 2
1
4251 3
0011 0010 1010 1101 0001 0100 1011
Next, we have the formula,
General solution is
Method of variation of parameres for higher order ODE
