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UNIT –IV MA2264 NUMERICAL METHODS
INITIAL VALUE PROBLEMS FOR ORDINARY DIFFERENTIAL EQUATIONS
PART-A
1. Explain the Taylor series algorithm for the first order differential equation.
2.
What is the disadvantage in using the Taylor series method
3. Using Taylor’s series find y(0.1) correct to 4 decimal places if y(x) satisfies y'=x+y , y(0) =1
Ans : 1.1103
4. Write down Euler and modified Euler algorithm for solving a first order differential equation
5. Using Modified Euler’s method, compute y(0.1) from
dy
2x
, y(0) = 1
 y
dx
y
Ans: 1.09548
6. Given y    y and y (0) =1, determine the values of y at x= 0.01 by Euler method Ans : 0.99
7. Write the Runge-kutta formulae of Fourth order for solving a first order differential equation
8. Write the Runge-kutta formulae of Fourth order for solving a second order differential equation
9. Write the Milne’s and Adam’s Predictor-Corrector formulae.
10. Compare Runge-kutta and Predictor-Corrector methods for solving an initial value problem.
PART-B
1. Using Taylor method , Compute y(0.2) and y(0.4) correct to 4 decimal places given
dy
 1  2 xy and y(0) = 0.
Ans: y(0.2)=0.19475,y(0.4) =0.35988
dx
dy
 e x  y 2 ,y(0) =1,find y(0.1) Ans:1.005
2. Using Taylor series for (i)
dx
dy
 2 y  3e x ,y(0)=0,find y at x=0.1 , 0.2
(ii)
Ans: 0.349 , 0.811
dx
3.
Using Euler’s method find y(0.3) of y(x)satisfies the initial value problem
dy 1 2
 ( x  1) y 2 ,y(0.2)=1.1114.
Ans : y(0.3)= 1.1756
dx 2
dy
 x  y  xy , y(0)=1 compute y at x=0.1,by taking h=0.05.
4.
Using Euler’s method solve
dx
Ans=1.0527
dy
 ( y  x 2 )3 ,y(1)=0,find y(1.2) Ans :0.44
5.
Using Modified Euler method (i)
dx
(ii) y'=1-y,y(0)=0,find x=(0.1)(0.1)(0.2) Ans: 0.095, 0.1809, 0.2587
dy x  y
(iii)
,y(2)=1 find y(1) by taking h=-0.2 0.73207

dx x  y
6.
Using Runge-kutta method of fourth order compute. (i)y(0.2), y(0.4)and y(0.6) given
Ans:2.073,2.452,3.023
y   x 3  y ,y(0)=2,
dy
xy

(ii) y(0.1) given
, y(0)=1
Ans:1.00492
dx 1  x 2
dy y 2  x 2
(iii)y at x=0.2,0.4
,y(0) =1 Ans: 1.1959, 1.3751

dx y 2  x 2
dy
1
y ( 0)  1
(iv) y at 0.1 and 0.2 for
Ans: 1.0914,1.1696

dx x  y
7.
Using the fourth order Runge-kutta method solve
d2y
i)
Ans:1.0013, 1.0107
 xy 2 , y (0)  1, y (0)  0 at x=02 and 0.4
dx 2
d2y
dy
 2  2 y  e 2 x sin x with y(0)=-0.4, y'(0)=-0.6
dx 2
dx
2
(v)
y"-0.1(1-y )y'+y=0 with y(0), y'(0)=1
(vi)
Given y  xy  y =0,y(0),y'(0)=0,find the value of y(0.1) by R-K method of 4 order
.
Ans :0.9950
1
8.
Using Milne’s Predictor and Corrector methods (i) Find y(2) if dy/dx= (x+y) ,
2
y(0)=2,y(0.5)=2.636,y(1)= 3.595 and y(1.5)=4.968
Ans y4,p=6.8710 , y4,c=6.8732
y(0.2) for
(ii) Find y(0.8) and y(1) if y'=x-y2 , y(0)=0 , y(0.2) = 0.02, y(0.4)= 0.0795, y(0.6)= 0.1762
Ans:yp =0.3049 , 0.4553 yc= 0.3046 , 0.4515
9.
Given y'=1-y , and y(0)=0 ,find (i) y(0.1) by Euler method (ii) y(0.2) by modified Euler method
(iii) y(0.3) = 0.2629 (iv) y(0.4) by Milne’s method. Ans (i)0.1 (ii) 0.1855(iv) yp 0.3280,yc 0.3333
10.
Given y'=x3+y,y(0)=2 the values of y(0.2)=2.073,y(0.4)=0.452 and y(0.6)=3.023 are got by R-K
method of 4th order ,Find y(0.8) by Milne’s.(h=0.2).
Ans yp=4.1664, yc= 3.7956
1
11.
Find y(2) ,given y'= (x+y), y(0)=2, y(0.5)=2.636, y(1)=3.595, y(1.5)=4.968 by Adam2
Bashforth Method
Ans :yp=6.8708, yc=6.8731.
12.
Given y'=x2(1+y) ,y(1)=1 , y(1.1)= 1.233 ,y(1.2)= 1.548 , y(1.3)= 1.979 , evaluate y(1.4) by
Adam’s method.
Ans: yp=2.5722, yc= 2.5749
13.
Find y(0.1),y(0.2),y(0.3) from y'=xy+y2 , y(0)=1 by using R-K 4th order and hence obtain y(0.4)
by Adam’s method. Ans :y(0.1)=1.1169,y(0.2)=1.2774,y(0.3)=1.5041 yp=1.8341, yc=1.8389
14.
Using Adam Bashforth method find y(0.4) given
xy
y '  ,y(0)=1,y(0.1)=1.01,y(0.2)=1.022,y(0.3)=1.023
Ans yp=1.0408, yc= 1.0410
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