Numerical methods. Examples with solution

Numerical methods Examples with solution

CONTENTS Contents. Nonlinear Equations 3 The bisection method............................ 4 Newton s method.............................. 8. Linear Systems LU-factorization.............................. Jacobi iterative method.......................... 3 Gauss-Seidel iterative method....................... 5 3. Interpolation and approximation of functions 7 Polynom interpolation........................... 8 The least square method.......................... 0 4. Numerical integration Trapezoid rule................................ 3 Composite trapezoid rule.......................... 5 Simpson s rule................................ 6 Composite Simpson s rule......................... 8 5. Initial value problems for ODEs 3 Euler s method............................... 3 Runge-Kutta method............................ 34

. NONLINEAR EQUATIONS. Nonlinear Equations 3

. NONLINEAR EQUATIONS Example.. Find the roots of equation by bisection method: x 3 = ln(0 x) with the precision ε = 0 3. Solution: 70 60 50 40 30 0 0 0 0 x 3 ln(0 x) 0 0 3 4 x f(x) -.97. -0.855. -0.4468.3 0.0337.4 0.59.5.349.6.9678.7.7967.8 3.779.9 4.767 5.906 It seen the root (x-intercept) lies in the interval, (from the graph,). The function values are tabulated on this interval f(x) = x 3 ln(0 x) and is found the sign of functional values are changing between. and.3. The function is continuous on a given interval. The equation has one root in the interval.,.3. 4

. NONLINEAR EQUATIONS The root are computed in the interval.,.3, thus a 0 =., b 0 =.3. Compute the first approximation x : x = b0 + a 0 =.3 +. =.5. Compute the functional values of the function f(x) = x 3 ln(0 x) in points a 0, x, b 0 : f(a 0 ) = 0.446, f(x ) = 0.59, f(b 0 ) = 0.0337 and find the interval a, b : f(a 0 )f(x ) 0 a = x 0 =.5, b = b 0 =.3. Find the approximation error b0 a 0 = 0.05 ε = 0 3. The computation continues. Compute the next approximation x = a + b =.5 +.3 =.75 and find the interval a, b : f(a ) = 0.59, f(x ) = 0.0935, f(b ) = 0.0337, f(a )f(x ) 0 a = x =.75 b = b =.3. Find the approximation error b a = 0.05 ε = 0 3. The computation continues. Compute the next approximation x 3 = a + b =.75 +.3 =.875 and find the interval a 3, b 3 : f(a ) = 0.0935, f(x 3 ) = 0.0305, f(b ) = 0.0337, f(a ) f(x 3 ) 0 a 3 = x 3 =.875, b 3 = b =.3. 5

. NONLINEAR EQUATIONS Find the approximation error b a = 0.05 ε = 0 3. The computation continues. Compute the next approximation x 4 = a3 + b 3 =.875 +.3 =.938 and find the interval a 4, b 4 : f(a 3 ) = 0.0305, f(x 3 ) = 0.004, f(b 3 ) = 0.0337, f(a 3 )f(x 3 ) < 0 a 4 = a 3 =.875, b 4 = x 3 =.938 Find the approximation error. b 3 a 3 = 0.006 ε = 0 3. The computation continues. Compute the next approximation x 4 : x 4 = a4 + b 4 =.875 +.938 =.906 and find the interval a 5, b 5 : f(a 4 ) = 0.0305 f(x 4 ) = 0.046 f(b 4 ) = 0.004 f(a 4 ) f(x 4 ) 0 a 5 = x 4 =.906, b 5 = b 4 =.938 Find the approximation error. b 4 a 4 = 0.003 ε = 0 3. The computation continues. Compute the next approximation x 5 : x 5 = a5 + b 5 =.906 +.938 =.9 and find the interval a 6, b 6 : f(a 5 ) = 0.046 f(x 5 ) = 0.0066 f(b 5 ) = 0.004 f(a 5 ) f(x 5 ) > 0 a 6 = x 5 =.906, b 6 = b 5 =.938 Find the approximation error b4 a 4 = 0.006 ε = 0 3. The computation continues. 6

. NONLINEAR EQUATIONS Compute the next approximation x 4 : x 6 = a6 + b 6 =.9 +.938 =.93 Find the approximation error b6 a 6 = 0.0008 < ε = 0 3. The desired precision is satisfied. All information are written to the table: i a i f(a i ) x i f(x i ) b i f(b i ) b i a i / 0..5.3 + 0.05.5.75.3 + 0.05.75.875.3 + 0.05 3.875.938 +.3 + 0.006 4.875.906.938 + 0.003 5.906.9.938 + 0.006 6.9.930.938 + 0.0008 The root of the equation is: ξ =.93 ± 0.00 Calculation in Matlab: a=., b=.3 f=inline( x^3-log(0-x) ) x=(a+b)/ while abs(b-a)/ > e-3 if f(a)*f(x)<0;b=x;else a=x;end; x=(a+b)/ end given interval definition of a function f(x) initial approximation approximations in the calculation of cycle 7

. NONLINEAR EQUATIONS Example.. Find the roots of equation by Newton s method: x 3 = ln(0 x) with the pecision ε = 0 6. Solution: The equation has one root in the interval.,.3 (see the Example..). Compute the the first and second derivative: f(x) = x 3 ln(0 x), f (x) = 3 x + 0 x, f (x) = 6 x + (0 x), Verify assumptions of method: f(a) f (a) = 0.4468 4.4336 = 0.008 0. = b a and it is seen that the condition is not satisfied. Reduce the interval in which the search root is. 0. 0. 0 0. 0. 0.3 0.4 x 3 ln(0 x) It is seen that the root lies in the interval.5,.3 (from the graph). 0.5..5.5.75.3 And again, verify the method conditions, this time in the interval.5,.3 : f(a) f (a) = 0.59 = 0.0450 < 0.05 = b a 4.808 f(b) f (b) = 0.0337 = 0.0065 < 0.05 = b a 5.849 8

. NONLINEAR EQUATIONS x i f (x i ) f (x i ).5 4.808 7.53.6 4.877 7.573.7 4.953 7.633.8 5.099 7.693.9 5.07 7.753.3 5.849 7.83 It is seen from the first and second derivative values in the interval.5,.3 : Find the initial approximation x 0 = b =.3. Compute the next approximation x : f (x) > 0 na.5,.3 f (x) > 0 na.5,.3. The Newton method conditions are satisfied. x = x 0 f(x 0) 0.03367697433946 =.3 f (x 0 ) 5.84945873563 =.9350485098864 and the approximation error x x 0 = 0.00649549036 ε = 0 6. The computation continues. Compute the next approximation x : x = x f(x ) f (x ) =.9350485098864 0.0006453367995 5.3437883453 =.93478053989 and the approximation error x x = 0.0000304584875 ε = 0 6. The computation continues. Compute the next approximation x 3 : x 3 = x f(x ) f (x ) =.93478053989 0.0000000039978 5.340705040059 =.93478043638 and the approximation error x x 0 = 0.0000000007775 = 7.775 0 0 < ε = 0 6. The desired precision is satisfied. All information are written to the table: i x i x i x i 0.3.9350485098864 0.00649549036.93478053989 0.0000304584875 3.93478043638 0.0000000007775 9

. NONLINEAR EQUATIONS The root of the equation is: ξ =.93473 ± 0 6 Calculation in Matlab: f=inline( x^3-log(0-x) ) input the function f(x) fd=inline( 3*x^+./(0-x) ) input the first derivative f (x) xs=.3 initial approximation x=xs-f(xs)/fd(xs) the next approximation computation while abs(xs-x)>e-6 approximations in the cycle computation xs=x; x=xs-f(xs)/fd(xs) end 0

. LINEAR SYSTEMS. Linear Systems

. LINEAR SYSTEMS Example.. The matrix is given. 3 A = 6 9 0 3 9 5. Compute the LU-decomposition A = LU by using Gaussian elimination method with direct choice of main element. Using the LU-decomposition,compute the solution of a system of linear equations Ax = b,where b = ( 9, 59, 8), inverse matrix A and determinant deta. Solution: Compute by using Gaussian elimination method: 3 6, 3 3 6 9 0 0 3 9 5 0 0 0 3 0 0 0 The matrix U is the result of the Gaussian elimination method, the matrix L we prepare of multipliers, which are written next to matrices : 0 0 3 L = 6 0, U = 0. 3 0 0 0 In solving Ax = b first compute y from Ly = b and then from Ux = y compute x. y = 9 9 Weget : 6y +y = 59 y = 5, 3y +y 3 = 8 x 3x +x 3 = 9 x x 3 = 5 x = 3. x 3 =

. LINEAR SYSTEMS Inverse matrix is calculated by solving the system Aa i = e i, i =,, 3, where e i are the columns of the unit matrix I = (e,e,e 3 ). The results a i are the columns of inverse matrix, i.e. A = (a,a,a 3 ). The solution of systems is again decomposed into two steps. y =, 0, 0 0 0 6y +y = 0,, 0 Y = 6 0, 3y +y 3 = 0, 0, 3 0 x 3x +x 3 =, 0, 0 5 3 8 x x 3 = 6,, 0 A = 0. x 3 =, 0, 3 0 Determinant det A be calculated as the product of the diagonal elements of the matrix L and U: deta = detldetu = ( ) (( ) ( ) ) =. Example.. Solve the system of linear equations x +x 3x 3 = 4, x +5x +x 3 = 5, 4x x +x 3 =, using Jacobi iterative method with the pecision ǫ = 0. Solution: To ensure convergence, modify the system in form with strong diagonal dominant matrix. Replace the first and third equation 4x x +x 3 =, x +5x +x 3 = 5, x +x 3x 3 = 4. 3

. LINEAR SYSTEMS The modified system is rewrite to iterative form x = 4 ( + x x 3 ), x = 5 (5 x x 3 ), x 3 = 3 ( 4 x x ) The recurrent formulas for Jacobi method are obtained by iteration indices. x (k+) = ( + x(k) 4 x (k) 3 ), x (k+) = (5 x(k) 5 x (k) 3 ), x (k+) 3 = ( 4 x(k) 3 x (k) ). The computation converges for any initial approximation x (0) = (x (0), x (0), x (0) 3 ), so the following initial approximation is choosen x (0) = ( 0, 0, 0 ). Substitute x (0) into the right side of the recurrent formulas. Compute continue equally x () = ( 3,, 4 3 ). x () = ( 4, 9 5, 3 ), x (3) = (.8500,.333, 0.8389 ), atd. Find that the true is x (k) x (k ) R 0 3, compute thus x () x (0) R = max{ 3 0, 0, 4 0 } = 3, 3 x () x () R = max{ 4 + 3, 9, 5 4 4 } =, 3 3 5 x (3) x () R = 0.5667, etc. The results are written to the table: Stop after the eleventh iteration, because x () x (0) R = 0.0077 0 3, and the result is x = 3.00 ± 0, x =.00 ± 0, x 3 =.00 ± 0. 4

. LINEAR SYSTEMS k x (k) x (k) x (k) 3 x (k) x (k ) R 0 0 0 0 3.0000.0000.3333 3.0000 3.467.9333 0.6667 0.9333 3.8500.333 0.8389 0.5667 4.86.97.78 0.889 5 3.0708.989.0370 0.097 6 3.0388.009 0.9494 0.00 7.9694.056 0.9940 0.0694 8.9906.9890.087 0.0367 9 3.0.995 0.9995 0.05 0 3.0.995 0.9995 0.05.9955.009.00 0.0077 Example.3. Solve the system of linear equations x +x 3x 3 = 4, x +5x +x 3 = 5, 4x x +x 3 =, using the Gauss-Seidel iterative methods with the pecision ǫ = 0. Solution: To ensure convergence, modify the system in form with strong diagonal dominant matrix. Replace the first and third equation 4x x +x 3 =, x +5x +x 3 = 5, x +x 3x 3 = 4. The modified system is rewrite to iterative form x = ( + x 4 x 3 ), x = (5 x 5 x 3 ), x 3 = ( 4 x 3 x ) 5

. LINEAR SYSTEMS The recurrent formulas for Jacobi method are obtained by iteration indices. x (k+) = ( + x(k) 4 x (k) 3 ), x (k+) = (5 x(k+) 5 x (k) 3 ), x (k+) 3 = ( 4 x(k+) 3 x (k+) ). The computation converges for any initial approximation x (0) = (x (0), x (0), x (0) 3 ), so the following initial approximation is choosen x (0) = ( 0, 0, 0 ). Substitute x (0) into the right side of the recurrent formulas. Compute x () = ( 3.0000,.000,.0667 ). continue equally x () = (.9833,.9800, 0.9989 ), atd. Find that the true is x (k) x (k ) R 0, compute thus x () x (0) R = max{ 3 0,. 0,.0667 0 } = 3, x () x () R = max{.9833 + 3,.98., 0.9989.0667 } = 0.00, atd. The results are written to the table: k x (k) x (k) x (k) 3 x (k) x (k ) R 0 0 0 0 3.0000.000.0667 3.0000.9833.9800 0.9989 0.00 3 3.0044.000 0.999 0.00 4.999.9998.000 0.0054 Stop after the four iteration, because x (4) x (3) R = 0.0054 0, and the result is x = 3.00 ± 0, x =.00 ± 0, x 3 =.00 ± 0. 6

3. INTERPOLATION AND APPROXIMATION OF FUNCTIONS 3. Interpolation and approximation of functions 7

3. INTERPOLATION AND APPROXIMATION OF FUNCTIONS Example 3.. The nodes x i and function values f i are given below. i 0 x i 3 f i Find the interpolation polynom for theese datas: a) in basic form, b) in Lagrang form, c) in Newton form. Solution: a)interpolating polynomial is finding in form p(x) = a 0 + a x + a x. Replace the interpolation conditions p(x i ) = f i, i = 0,, p( 3) = = a 0 3a + 9a =, p( ) = = a 0 a + a =, p() = = a 0 + a + a =. These equalities represent system of linear equations, which can be written in the form 3 9 a 0 a = a. The results are a 0 =, a 8 = 3, a = 5, the interpolation polynomial has the 8 form p(x) = 8 + 3 x + 5 8 x. b) Interpolation polynomial in Lagrange form is prescribed p(x) = f 0 ϕ 0 (x) + f ϕ (x) + f ϕ (x), where ϕ 0 (x), ϕ (x), ϕ (x) are Lagrange basis polynomials: ϕ 0 (x) = (x + )(x ) ( 3 + )( 3 ) = (x + )(x ), 8 8

3. INTERPOLATION AND APPROXIMATION OF FUNCTIONS ϕ (x) = ϕ (x) = (x + 3)(x ) ( + 3)( ) = (x + 3)(x ), 4 (x + 3)(x + ) = (x + 3)(x + ). ( + 3)( + ) 8 It is obtained p(x) = 8 (x + )(x ) + 4 (x + 3)(x ) + (x + 3)(x + ). 4 c) Interpolation polynomial in Newton form is prescribed p(x) = f 0 + f[x, x 0 ](x x 0 ) + f[x, x, x 0 ](x x 0 )(x x ), where f[x, x 0 ] a f[x, x, x 0 ] are the relative difference st and nd order. The results are written to the table: i x i f i st order nd order 0 3 5 8 3 Using the computed values in the first row of the table, find interpolation polynomial: p(x) = (x + 3) + 5 (x + 3)(x + ). 8 9

3. INTERPOLATION AND APPROXIMATION OF FUNCTIONS Example 3.. Aproximate the following data x i.5 3 5 7 y i 0 6.5 6 5 by the line ϕ(x) = c x + c by the least squares method. Solution: Find the system of linear equations for the unknown coefficients c, c. 5 5 c x i + c x i = i= i= 5 5 c x i + c = i= i= 5 y i x i i= 5 y i i= Compute individual totals: 5 x i = +.5 + 3 + 5 + 7 = 90.5 i= 5 x i = +.5 + 3 + 5 + 7 = 8.5 i= 5 y i x i = 0 +.5 + 6.5 3 + 6 5 + 5 7 = 57 i= 5 y i = 0 + + 6.5 + 6 + 5 = 8.5 i= Replace the system of linear equations: 90.5c + 8.5c = 57 8.5c + 5c = 8.5 Solve the system and the coefficients c =.3647, c = 3.0493 are obtained. The line ϕ(x) =.3647x 3.0493 is the result. 0

3. INTERPOLATION AND APPROXIMATION OF FUNCTIONS 6 4 zadane body nalezena funkce 0 8 6 4 0 3 4 5 6 7 Calculation in Matlab: x=[,.5,3,5,7] y=[0,,6.5,6,5] n=length(x) imput x i imput y i number of points A=[sum(x.^) sum(x);sum(x) n] matrix system b=[sum(y.*x);sum(y)] c=a\b f=c()*x+c() plot(x,y, o,x,f) vector of right-hand sides calculation of the coefficients line values in points x i graph

4. NUMERICAL INTEGRATION 4. Numerical integration

4. NUMERICAL INTEGRATION Example 4.. Compute the integral by composite trapezoid tule for n = 8: 3 e x dx Solution: Compute the step h in the interval a, b, h = b a n = 3 ( ) 8 = 0.5. Write to the table division points x 0,...,x 8 and the values of integrated functions in these points. i x i y i = e x i 0 -.783-0.5.84 0 3 0.5.84 4.783 5.5 9.4877 6 54.598 7.5 58.08 8 3 803.0839 Finally, compute the approximate value of the integral: I 0.5 ( ) ( ) n y0 + y n y0 + y 8 7 = h + y i = 0.5 + y i = i= i= (.783 + 803.0839 = 0.5 +.84 + +.84 +.783 + 9.4877+ ( ) 805.80 +54.598 + 58.08) = 0.5 + 588.385 = 30.64305 The result is: I 0.5 = 30.64305 3

4. NUMERICAL INTEGRATION Calculation in Matlab: a=-,b=3 interval bounds a, b n=8 number of segments on dividing n h=(b-a)/n x=a:h:b computation of step h computation of x i y=exp(x.^) computation of y i I=h*((y()+y(n+))/+sum(y(:n))) computation of I = h ( y 0 +y n + ) n i= y i In Matlab is indexed from, and we index points from 0, so the indexes in the formula in Matlab are shifted by. 4

4. NUMERICAL INTEGRATION Example 4.. Compute the integral by composite trapezoid rule with the precision ε = 0 3 : 3 e x dx Solution: Compute the integral composite trapezoid formula for n =, 4, 8, 6,..., until the error is less than the precision, i.e. I h I h < ε. n h I h I h I h ε = 0 3 = 0.00 8.388-4 4.76 4000.0 8 30.643 790.5745 6 4 688.983 63.748 3 8 508.640 80.88 64 6 46.798 46.8474 8 3 449.96.8307 56 64 446.9969.965 5 8 446.55 0.748 04 56 446.0696 0.855 048 5 446.03 0.0464 4096 04 446.06 0.06 89 048 446.0087 0.009 6384 4096 446.0080 0.0007 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 < 0.00 The result is: 3 e x = 446.008 ± 0.00 5

4. NUMERICAL INTEGRATION Example 4.3. Compute the integral by composite Simpson rule for n = 6: 0 arctg (x) + x + dx Solution: Compute the step h in the interval a, b, h = b a n = 0 = 0.5. 6 Write to the table division points x 0,...,x 6 and the value of integrated functions in these points. i x i y i = arctg (x i ) + x i + 0 0 0.50.405 0.500.300 3 0.3750.5037 4 0.5000.750 5 0.650.9370 6 0.7500.64 7 0.8750.397 8.669 i x i y i = arctg (x i ) + x i + 9.50.8376 0.500 3.059.3750 3.64.5000 3.4659 3.650 3.6636 4.7500 3.8560 5.8750 4.043 6 4.58 Finally, compute the approximate value of the integral: I 0.5 = h 3 (y 0 + y n + 4 S L + S S ) = 0.5 3 (y 0 + y 6 + 4 (y + y 3 + y 5 + +y 7 + y 9 + y + y 3 + y 5 ) + (y + y 4 + y 6 + y 8 + y 0 + y + y 4 ) = = 0.047 (5.58 + 4 0.7797 + 8.808) = 5.00 6

4. NUMERICAL INTEGRATION The result is: I 0.5 = 5.00 Calculation in Matlab: a=0,b= interval bounds a, b n=6 number of segments on dividing n h=(b-a)/n computation of steph x=a:h:b computation of x i y=atan(x).^+x+ computation of y i SL=sum(y(::n)) computation of S L SS=sum(y(3::n-)) computation of S S I=h/3*(y()+y(n+)+4*SL+*SS) computation of I In Matlab is indexed from, and we index points from 0, so the indexes in the formula in Matlab are shifted by. 7

4. NUMERICAL INTEGRATION Example 4.4. For the function f(x) = sin x + arctanx + x 4 compute approximately the first derivative on the interval 0, by the formula f (x) f(x + h) f(x h) h with a step h = 0.5 a h = 0.. Compare the results with values of exact calculated derivative. Solution: The exact derivative is found by the formula: f (x) = x cosx + +x 4 +x 4x 3 arctan x ( + x 4 ). Find the nodes x i = ih, i = 0,..., 8 and compute the functional values in them f i = f(x i ) to compute approximate derivative with step h = 0.5. There are the first three columns in the table: i x i f i f[x i+,x i ] f (x i ) e i 0 0.00 0.0000 -.0000-0.5 0.3065.3676.43 0.0538 0.50 0.6838.433.565 0.085 3 0.75.0.008.84 0.076 4.00.34 0.4764 0.545 0.0688 5.5.603 0.5880 0.4570 0.30 6.50 0.940.600.948 0.0347 7.75 0.803 3.637 3.6746 0.409 8.00 0.697 -.754 - The formula for calculation of approximate derivative is written in form f (x i ) f i+ f i h = f[x i+, x i ], 8

4. NUMERICAL INTEGRATION. Compute the approximate derivative values in all internal nodes, eg f (0.75).34 0.6838 0.5 =.008;. This is the fourth column of the table. The fifth column contains the exact values of derivatives that are in the last column compared with the approximate values: e i = f[x i+, x i ] f (x i ). For h = 0. the procedure is analogous, we obtain more accurate values, as shown in the following images: h=0.5 0... derivace aproximace derivace 3 4 0 0. 0.4 0.6 0.8..4.6.8 h=0. 0... derivace aproximace derivace 3 4 0 0. 0.4 0.6 0.8..4.6.8 9

4. NUMERICAL INTEGRATION Calculation in Matlab: a=0,b= interval bounds a, b h=0.5 step h x=[a:h:b] computation of nodes x i n=length(x) number of nodes n + df= *x.*cos(x.^) +((+x.^4)./(+x.^)-... 4*x.^3.*atan(x))./(+x.^4).^ computation of f (x i ) f=sin(x.^)+atan(x)./(+x.^4) computation of f i = f(x i ) adf=[nan;(f(3:n)-f(:n-))/(*h);nan] approximate derivative tab=[[0:n-],x,f,adf,df,abs(adf-df)] table plot(x,df, :,x,adf, - ) graph 30

5. INITIAL VALUE PROBLEMS FOR ODES 5. Initial value problems for ODEs 3

5. INITIAL VALUE PROBLEMS FOR ODES Example 5.. Solve the differential equation with the initial condition: y = 0. y, y( ) =, x + by the Euler method in the interval, 3 with the step h = 0.5. Solution: First, compute the number of dividing segments in the interval a, b : n = b a h = 3 ( ) 0.5 = 0. The initial condition is x 0 =,, y 0 =. Compute x, y by the Euler metod: x = x 0 + h = + 0.5 =.5 ( ) ( ) y = y 0 + h x 0 + 0. y 0 = + 0.5 0. ( ) = 0.85 ( ) + Compute x, y, x 3, y 3 : x y = x + h =.5 + 0.5 = ( ) ( ) = y + h x + 0. y = 0.85 + 0.5 0. ( 0.85) = (.5) + = 0.6537 x 3 = x + h = + 0.5 = 0.5 ( ) ( ) y 3 = y + h x + 0. y = 0.6537 + 0.5 0. ( 0.6537) = ( ) + = 0.37 Other approximate values of solution x 4, y 4,..., x 0, y 0 compute accordingly. 3

5. INITIAL VALUE PROBLEMS FOR ODES The approximate values of the solution are in the table and graph: i x i y i 0 - - -.5-0.85 - -0.6537 3-0.5-0.37 4 0 0.0476 5 0.5 0.545 6 0.979 7.5. 8.98 9.5.588 0 3.648 y.5 0.5 0 0.5 Priblizne reseni.5 0.5 0 0.5.5.5 3 x Calculation in Matlab: a=-,b=3 h=0.5 n=(b-a)/h x()=-,y()=- for i=:n x(i+)=x(i)+h y(i+)=y(i)+h*(/(x(i)^+)-0.*y(i)) end [x y ] plot(x,y) interval bounds a, b input step h number of dividing segments n input initial condition computation of x i, y i in the cycle extract into the table graph 33

5. INITIAL VALUE PROBLEMS FOR ODES Example 5.. Solve the differential equation with the initial condition: y = sin(x) y x 3 + y, y() = 0, by the 4th order Runge Kutta method in the interval, with the step h = 0.. Solution: First, compute the number of dividing segments in the interval a, b : n = b a h The initial condition is x 0 =,, y 0 = 0. = 0. = 5. Compute x, y by the 4th order Runge Kutta method: k = h f (x 0, y 0 ) = h (sin (x 0 ) y 0 x 3 0 + y 0) = 0. ( sin () 0 3 + 0 ) = k k 3 = 0. ( = h f x 0 + h, y 0 + k ) = ( = h sin x 0 + h ) ( y 0 + k ) ( x 0 + h ) 3 + y 0 + k = ( ( = 0. sin + 0. ) ( 0 + 0. ) ( + 0. ) 3 + 0 + 0. ) = 0.304 ( = h f x 0 + h, y 0 + k ) = ( = h sin x 0 + h ) ( y 0 + k ) ( x 0 + h ) 3 + y 0 + k = ( ( = 0. sin + 0. ) ( 0 + 0.304 ) ( + 0. ) 3 + 0 + 0.304 ) = = 0.337 k 4 = h f (x 0 + h, y 0 + k 3 ) = = h (sin ) (x 0 + h) (y 0 + k 3 ) (x 0 + h) 3 + y 0 + k 3 = = 0. (sin ( + 0.) (0 0.337) ( + 0.) 3 + 0 0.337 ) = 0.4707 x = x 0 + h = + 0. =. 34

5. INITIAL VALUE PROBLEMS FOR ODES y = y 0 + 6 (k + k + k 3 + k 4 ) = = 0 + ( 0. + ( 0.304) + ( 0.337) 0.4707) = 0.3 6 Compute x, y : k = h f (x, y ) = h (sin (x ) y x 3 + y ) = = 0. (sin (.) ( 0.3) 3 0.3 ) = = 0.4696 ( k = h f x + h, y + k ) = ( = h sin x + h ) ( y + k ) ( x + h ) 3 + y + k = ( ( = 0. sin. + 0. ) ( 0.3 + 0.4696 ) (. + 0. ) 3 0.3+ + 0.4696 ) = 0.6577 ( k 3 = h f x + h, y + k ) = ( = h sin x + h ) ( y + k ) ( x + h ) 3 + y + k = ( ( = 0. sin. + 0. ) ( 0.3 + 0.6577 ) ( + 0. ) 3 + 0+ + 0.6577 ) = 0.6946 k 4 = h f (x + h, y + k 3 ) = h (sin ) (x + h) (y + k 3 ) (x + h) 3 + y + k 3 = = 0. (sin ( + 0.) (0 0.337) ( + 0.) 3 + 0 0.337 ) = 0.95 x = x + h =. + 0. =.4 y = y + 6 (k + k + k 3 + k 4 ) = = 0.3 + ( 0.4696 + ( 0.6577) + ( 0.6946) 0.95) =.0087 6 Other approximate values of solution x 3, y 3,..., x 5, y 5 compute accordingly. 35

5. INITIAL VALUE PROBLEMS FOR ODES The approximate values of the solution are in the table and graph: 0 Priblizne reseni i x i y i 0 0. -0.30.4 -.0087 3.6 -.357 4.8-4.6586 5-8.535 y 3 4 5 6 7 8 9...3.4.5.6.7.8.9 x Calculation in Matlab: f=inline( sin(x)*y-x^3+y ) right sidef(x, y) a=,b= h=0. n=(b-a)/h x()=,y()=0 for i=:n k=h*f(x(i),y(i)) k=h*f(x(i)+h/,y(i)+k/) k3=h*f(x(i)+h/,y(i)+k/) k4=h*f(x(i)+h,y(i)+k3) x(i+)=x(i)+h y(i+)=y(i)+/6*(k+*k+*k3+k4) end [x y ] plot(x,y) interval bounds a, b input step h number of dividing segments n input initial condition computation of x i, y i in the cycle extract into the table graph 36