Numerical Methods for Engineers
Numerical Methods for Engineers
7th Edition
ISBN: 9780073397924
Author: Steven C. Chapra Dr., Raymond P. Canale
Publisher: McGraw-Hill Education
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Chapter 26, Problem 6P

Solve the following initial-value problem from t=4 to 5:

dydt=2yt

Use a step size of 0.5 and initial values of y(2.5)=0.48,y(3)=0.333333, y(3.5)=0.244898, and y(4)=0.1875. Obtain your solutions using the following techniques: (a) the non-self-starting Heun method (εS=1%), and (b) the fourth-order Adams method (εS=0.01%). [Note: The exact answers obtained analytically are y(4.5)=0.148148 and y(5)=0.12.] Compute the true percent relative errors εt for your results.

(a)

Expert Solution
Check Mark
To determine

To calculate: The solution of differential equation, dydt=2yt from t=4tot=5 using the non-self-starting Heun method with step size of 0.5 and initial values y(2.5)=0.48,y(3)=0.333333,y(3.5)=0.244898andy(4)=0.1875. Also, εs=1%. Compute the true percent relative as well.

Answer to Problem 6P

Solution:

First step:

The true error is εt=0.3028%.

Corrector yields:

jyjj+1εa10.14726832.63%20.14769940.292%

Second step:

The true error is εt=0.39%.

Corrector yields:

jyjj+1εa10.119531930.236%

Explanation of Solution

Given Information:

Differential equation, dydt=2yt, t=4tot=5 with step size of 0.5 and initial values y(2.5)=0.48,y(3)=0.333333,y(3.5)=0.244898andy(4)=0.1875. Also, εs=1%.

Formula used:

(1) Predictor for non-self-starting Heun method is given by,

y0i+1=ymi1+f(ti,ymi)2h

(2) Corrector for non-self-starting Heun method is given by,

yji+1=ymi+f(ti,ymi)+f(ti+1,yj1i+1)2h

(3) Predictor modifier is given by:

y0i+1=y0i+1,u+45(ymi,uy0i,u)

where, the subscript u designates that the variable is unmodified.

Calculation:

Consider the given differential equation, dydt=2yt. Here, f(t,y)=2yt.

Predictor for non-self-starting Heun method is given by,

y0i+1=ymi1+f(ti,ymi)2h.. .. .. (1)

For the given differential equation, it can be written as

y0i+1=ymi1+(2yimti)2h

Substitute, i=0,m=0andh=0.5  in above equation,

y01=y01+(2y00t0)2(0.5)=y01+(2y00t0)

The provided initials conditions are y(2.5)=0.48,y(3)=0.333333,y(3.5)=0.244898andy(4)=0.1875.

Now, substitute y01=0.244898 , y00=0.1875 and t0=4.

Thus,

y01=0.244898+(2×0.18754)=0.151148

Corrector for non-self-starting Heun method is given by,

yji+1=ymi+f(ti,ymi)+f(ti+1,yj1i+1)2h

For the given differential equation, it can be written as,

yji+1=ymi+[(2yimti)+(2yi+1j1ti+1)2]h

Substitute i=0,m=0 and  j=1 in above equation,

y11=y00+[(2y00t0)+(2y10t1)2]h

Put y00=0.1875,y01=0.151148,t0=4,t1=4.5 and h=0.5

y11=0.1875+[(2×0.18754)+(2×0.1511484.5)2]0.5=0.147268277

Similarly, for i=0 and j=2 corrector gives y21=0.1476994.

Error is calculated as,

εa=|y21y11y21|×100=|0.14769940.14726830.1476994|×100=0.2918766

Also, the true error can be computed as,

εt=|0.1481480.14769940.148148|×100=0.3028

Hence, the true error is εt=0.3028%.

jyjj+1εa10.14726832.63%20.14769940.292%

Second step:

Predictor

Substitute i=1 and m=0 in equation (1),

y02=y00+(2y10t1)2h

Substitute y00=4.762673,y01=3.913253andt1=2.5 in above equation,

y02=0.1875+(2×0.1511484.5)1=0.12032311

Predictor modifier is given by:

y0i+1=y0i+1,u+45(ymi,uy0i,u)

where, the subscript u designates that the variable is unmodified.

Now, substitute i=1andm=2 in above equation,

y02=y02,u+45(y21,uy01,u)

Substitute, y02,u=0.12032311,y21,u=0.1476994 and y01,u=0.151148 in above equation,

y02=0.12032311+45(0.14769940.151148)=0.11756423

Corrector is given by:

yji+1=ymi+(2yimti)+(2yi+1j1ti+1)2h

Substitute i=1  j=1,m=2andh=0.5, m= 2 in above formula as shown below,

y12=y21+[(2y12t1)+(2y20t2)2]h

Now, substitute y21=0.1476994, y02=0.11756423, t1=4.5 and t2=5 as shown below,

y12=0.1476994+[(2×0.14769944.5)+(2×0.117564235)2]0.5=0.11953193

Error can be computed as,

εa=|y12y02y02|×100=|0.119531930.117564230.11953193|×100=1.646%

The true error can be computed as,

εt=|0.120.119531930.12|×100=0.39%

Hence, the true error is εt=0.39%.

Corrector yields:

jyjj+1εa10.119531930.236%

(b)

Expert Solution
Check Mark
To determine

To calculate: The solution of differential equation, dydt=2yt from t=4tot=5 using the fourth-order Adams method step size of 0.5 and initial values y(2.5)=0.48,y(3)=0.333333,y(3.5)=0.244898andy(4)=0.1875. Also, the exact answers obtained analytically are y(4.5)=0.148148andy(5)=0.12 and εs=0.01%. Compute the true percent relative as well.

Answer to Problem 6P

Solution:

The solution obtained for the given differential equation are shown below in tabular form:

Predictor= 0.1527937

Corrector iteration:

tyεaεt4.50.147605453.5149%0.3322%4.50.148037810.292%0.074%4.50.1480017730.024350.099%

Predictor = 0.12165973

Corrector iteration:

tyεaεt50.11969011.6456%0.258%50.11983780.123%0.133%50.11982670.0092%0.142%

Explanation of Solution

Given Information:

Differential equation, dydt=2yt, t=4tot=5 with step size of 0.5 and initial values y(2.5)=0.48,y(3)=0.333333,y(3.5)=0.244898andy(4)=0.1875. Values obtained analytically y(4.5)=0.148148andy(5)=0.12. Also, εs=0.01%.

Formula used:

(1) The fourth order Adams-Bashforth formula as predictor is given by,

y0i+1=ymi+h24(55fmi59fmi1+37fmi29fmi3)

(2) The fourth order Adams-Bashforth formula as corrector is given by,

yji+1=ymi+h24(9fj1i+1+19fmi5fmi1+fmi2)

(3) Error is given as,

εa=|yji+1yj1i+1yji+1|×100

Calculation:

Consider the given differential equation, dydt=2yt.

Here, f(t)=2yt.. .. .. (1)

Starting values of AdamsBash fourth method are,

At t=2.5, y=0.48. So, y3=0.48. Substitute this value in equation (1)

Thus,

f3=2×0.482.5=0.384

At t=3, y(3)=0.333333. So, y2=0.333333. Substitute this value in equation (1)

Thus,

f2=2×0.3333333=0.222222

At t=3.5, y=0.244898. So, y1=0.244898. Substitute this value in equation (1)

Thus,

f1=2×0.2448983.5=0.1399417

At t=4.0, y=0.1875. So, y0=0.1875. Substitute this value in equation (1)

Thus,

f0=2×0.18754=0.09375

The fourth order Adams-Bashforth formula as predictor is given by,

y0i+1=ymi+h24(55fmi59fmi1+37fmi29fmi3).. .. .. (2)

The fourth order Adams formula as the corrector is,

yji+1=ymi+h24(9fj1i+1+19fmi5fmi1+fmi2).. .. .. (3)

Use predictor to compute a value at t=4.5, substitute i=0 and m=0 in equation (2),

y01=y00+h24(55f059f1+37f29f3)

Substitute values f0=0.09375, f1=0.1399417, f2=0.222222, f3=0.384 and h=0.5.

y01=0.1875+0.524×(55(0.09375)59(0.1399417)+37(0.222222)9(0.384))=0.1527937

Hence, at t=4.5,y01=0.1527937

f1=2×0.15279374.5=0.0679083

Thus, f1=0.0679083.

Now, use corrector,

Put i=0 ,  j=1  in equation (3),

y11=y00+h24(9f1+19f05f1+f2)

Substitute values f0=0.09375, f1=0.1399417, f2=0.222222, f1=0.0679083 and h=0.5.

y11=0.1875+0.524(9(0.0679083)+19(0.09375)5(0.1399417)+(0.222222))=0.14760545

At t=4.5andy11=0.14760545

f1=2×0.147605454.5=0.0656024

Thus, f1=0.0656024.

This result can be substituted back into equation (3) to iteratively correct the estimate,

y21=0.1875+0.524(9(0.0656024)+19(0.09375)5(0.1399417)+(0.222222))=0.14803781

Again,

At t=4.5andy21=0.14803781

f1=2×0.148037814.5=0.0657946

Thus, f1=0.0657946.

This result can be substituted back into equation (3) to iteratively correct the estimate,

y31=0.1875+0.524(9(0.0657946)+19(0.09375)5(0.1399417)+(0.222222))=0.148001773

At t=4.5andy31=0.148001773

f1=2×0.1480017734.5=0.0657786

Thus, f1=0.0657786.

Now, use the predictor to find the value at t=5.

Substitute i=1 in equation (2),

y02=y1+h24(55f159f0+37f19f2)

Substitute values, f1=0.0657786 f0=0.09375, f1=0.1399417 and f2=0.222222 in above equation, to get

y02=0.148001773+0.524×(55(0.0657786)59(0.09375)+37(0.1399417)9(0.222222))=0.12165973

At t=5andy02=0.12165973.

f2=2×0.121659735=0.0486639

Thus, f2=0.0486639

Now error is given as,

εa=|yji+1yj1i+1yji+1|×100

Calculate εa as shown below,

εa=|0.147605450.15279370.14760545|×100=3.5149%

The true percent relative errors can be calculated as,

εt=|0.1481480.147605450.148148|×100=0.36622%

Corrector formula is given as,

yji+1=ymi+h24(9fj1i+1+19fmi5fmi1+fmi2)

Substitute i=1andj=1 in above formula,

y12=y1+h24(9f02+19f15f0+f1)

Substitute values, f1=0.0657786 f0=0.09375, f1=0.1399417 and f2=0.0486639 in above equation, to get

y12=0.148001773+0.524(9(0.0486639)+19(0.0657786)5(0.09375)+(0.1399417))=0.1196901

At t=5andy12=0.1196901

f2=2×0.11969015=0.0478760

Thus, f2=0.0478760.

Hence,

y22=0.148001773+0.524(9(0.0478760)+19(0.0657786)5(0.09375)+(0.1399417))=0.11983783

At t=5andy22=0.11983783

f2=2×0.119837835=0.0479351

Thus, f2=0.0479351.

Hence,

y32=0.148001773+0.524(9(0.0479351)+19(0.0657786)5(0.09375)+(0.1399417))=0.11982675

Now error is given as,

εa=|yji+1yj1i+1yji+1|×100

Calculate εa as shown below,

εa=|0.11969010.121659730.1196901|×100=1.6456%

The true percent relative errors can be calculated as,

εt=|0.120.11969010.12|×100=0.25825%

Results are shown below in tabular form:

Predictor= 0.1527937

tyεaεt4.50.147605453.5149%0.3322%4.50.148037810.292%0.074%4.50.1480017730.024350.099%

Corrector iteration:

Predictor = 0.12165973

Corrector iteration:

tyεaεt50.11969011.6456%0.258%50.11983780.123%0.133%50.11982670.0092%0.142%

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