Solve the following initial-value problem from t = 4 to 5 : d y d t = − 2 y t 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 ( ε S = 1 % ) , and (b) the fourth-order Adams method ( ε S = 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 ε t for your results.

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Answer 1

Textbook Question

Chapter 26, Problem 6P

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

$d y d t = − 2 y t$

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 $( ε S = 1 % )$ , and (b) the fourth-order Adams method $( ε S = 0.01 % )$ . [Note: The exact answers obtained analytically are $y ( 4.5 ) = 0.148148 and y ( 5 ) = 0.12$ .] Compute the true percent relative errors $ε t$ for your results.

(a)

Expert Solution

To determine

To calculate: The solution of differential equation, $dydt=−2yt$ from $t=4 to t=5$ using the non-self-starting Heun method with step size of 0.5 and initial values $y(2.5)=0.48, y(3)=0.333333, y(3.5)=0.244898 and y(4)=0.1875$ . 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:

$jyj+1jεa10.14726832.63%20.14769940.292%$

Second step:

The true error is $εt=0.39%$ .

Corrector yields:

$jyj+1jεa10.119531930.236%$

Explanation of Solution

Given Information:

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

Formula used:

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

$yi+10=yi−1m+f(ti,yim)2h$

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

$yi+1j=yim+f(ti,yim)+f(ti+1,yi+1j−1)2h$

(3) Predictor modifier is given by:

$yi+10=yi+1,u0+45(yi,um−yi,u0)$

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,

$yi+10=yi−1m+f(ti,yim)2h$ .. .. .. (1)

For the given differential equation, it can be written as

$yi+10=yi−1m+(−2yimti)2h$

Substitute, $i=0, m=0 and h=0.5$ in above equation,

$y10=y−10+(−2y00t0)2(0.5)=y−10+(−2y00t0)$

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

Now, substitute $y−10=0.244898 , y00=0.1875 and t0=4$ .

Thus,

$y10=0.244898+(−2×0.18754)=0.151148$

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

$yi+1j=yim+f(ti,yim)+f(ti+1,yi+1j−1)2h$

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

$yi+1j=yim+[(−2yimti)+(−2yi+1j−1ti+1)2]h$

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

$y11=y00+[(−2y00t0)+(−2y10t1)2]h$

Put $y00=0.1875, y10=0.151148, t0=4, t1=4.5 and h=0.5$

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

Similarly, for $i=0 and j=2$ corrector gives $y12=0.1476994$ .

Error is calculated as,

$εa=|y12−y11y12|×100=|0.1476994−0.14726830.1476994|×100=0.2918766$

Also, the true error can be computed as,

$εt=|0.148148−0.14769940.148148|×100=0.3028$

Hence, the true error is $εt=0.3028%$ .

$jyj+1jεa10.14726832.63%20.14769940.292%$

Second step:

Predictor

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

$y20=y00+(−2y10t1)2h$

Substitute $y00=4.762673, y10=3.913253 and t1=2.5$ in above equation,

$y20=0.1875+(−2×0.1511484.5)1=0.12032311$

Predictor modifier is given by:

$yi+10=yi+1,u0+45(yi,um−yi,u0)$

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

Now, substitute $i=1 and m=2$ in above equation,

$y20=y2,u0+45(y1,u2−y1,u0)$

Substitute, $y2,u0=0.12032311, y1,u2=0.1476994 and y1,u0=0.151148$ in above equation,

$y20=0.12032311+45(0.1476994−0.151148)=0.11756423$

Corrector is given by:

$yi+1j=yim+(−2yimti)+(−2yi+1j−1ti+1)2h$

Substitute $i=1 j=1, m=2 and h=0.5$ , m= 2 in above formula as shown below,

$y21=y12+[(−2y12t1)+(−2y20t2)2]h$

Now, substitute $y12=0.1476994$ , $y20=0.11756423$ , $t1=4.5$ and $t2=5$ as shown below,

$y21=0.1476994+[(−2×0.14769944.5)+(−2×0.117564235)2]0.5=0.11953193$

Error can be computed as,

$εa=|y21−y20y20|×100=|0.11953193−0.117564230.11953193|×100=1.646%$

The true error can be computed as,

$εt=|0.12−0.119531930.12|×100=0.39%$

Hence, the true error is $εt=0.39%$ .

Corrector yields:

$jyj+1jεa10.119531930.236%$

(b)

Expert Solution

To determine

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

Formula used:

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

$yi+10=yim+h24(55fim−59fi−1m+37fi−2m−9fi−3m)$

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

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$

(3) Error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

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, $y−3=0.48$ . Substitute this value in equation (1)

Thus,

$f−3=−2×0.482.5=−0.384$

At $t=3$ , $y(3)=0.333333$ . So, $y−2=0.333333$ . Substitute this value in equation (1)

Thus,

$f−2=−2×0.3333333=−0.222222$

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

Thus,

$f−1=−2×0.2448983.5=−0.1399417$

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,

$yi+10=yim+h24(55fim−59fi−1m+37fi−2m−9fi−3m)$ .. .. .. (2)

The fourth order Adams formula as the corrector is,

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$ .. .. .. (3)

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

$y10=y00+h24(55f0−59f−1+37f−2−9f−3)$

Substitute values $f0=−0.09375$ , $f−1=−0.1399417$ , $f−2=−0.222222$ , $f−3=−0.384$ and $h=0.5$ .

$y10=0.1875+0.524×(55(−0.09375)−59(−0.1399417)+37(−0.222222)−9(−0.384))=0.1527937$

Hence, at $t=4.5, y10=0.1527937$

$f1=−2×0.15279374.5=−0.0679083$

Thus, $f1=−0.0679083$ .

Now, use corrector,

Put $i=0 , j=1$ in equation (3),

$y11=y00+h24(9f1+19f0−5f−1+f−2)$

Substitute values $f0=−0.09375$ , $f−1=−0.1399417$ , $f−2=−0.222222$ , $f1=−0.0679083$ and $h=0.5$ .

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

At $t=4.5 and y11=0.14760545$

$f1=−2×0.147605454.5=−0.0656024$

Thus, $f1=−0.0656024$ .

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

$y12=0.1875+0.524(9(−0.0656024)+19(−0.09375)−5(−0.1399417)+(−0.222222))=0.14803781$

Again,

At $t=4.5 and y12=0.14803781$

$f1=−2×0.148037814.5=−0.0657946$

Thus, $f1=−0.0657946$ .

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

$y13=0.1875+0.524(9(−0.0657946)+19(−0.09375)−5(−0.1399417)+(−0.222222))=0.148001773$

At $t=4.5 and y13=0.148001773$

$f1=−2×0.1480017734.5=−0.0657786$

Thus, $f1=−0.0657786$ .

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

Substitute $i=1$ in equation (2),

$y20=y1+h24(55f1−59f0+37f−1−9f−2)$

Substitute values, $f1=−0.0657786$ $f0=−0.09375$ , $f−1=−0.1399417$ and $f−2=−0.222222$ in above equation, to get

$y20=0.148001773+0.524×(55(−0.0657786)−59(−0.09375)+37(−0.1399417)−9(−0.222222))=0.12165973$

At $t=5 and y20=0.12165973$ .

$f2=−2×0.121659735=−0.0486639$

Thus, $f2=−0.0486639$

Now error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

Calculate $εa$ as shown below,

$εa=|0.14760545−0.15279370.14760545|×100=3.5149%$

The true percent relative errors can be calculated as,

$εt=|0.148148−0.147605450.148148|×100=0.36622%$

Corrector formula is given as,

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$

Substitute $i=1 and j=1$ in above formula,

$y21=y1+h24(9f20+19f1−5f0+f−1)$

Substitute values, $f1=−0.0657786$ $f0=−0.09375$ , $f−1=−0.1399417$ and $f2=−0.0486639$ in above equation, to get

$y21=0.148001773+0.524(9(−0.0486639)+19(−0.0657786)−5(−0.09375)+(−0.1399417))=0.1196901$

At $t=5 and y21=0.1196901$

$f2=−2×0.11969015=−0.0478760$

Thus, $f2=−0.0478760$ .

Hence,

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

At $t=5 and y22=0.11983783$

$f2=−2×0.119837835=−0.0479351$

Thus, $f2=−0.0479351$ .

Hence,

$y23=0.148001773+0.524(9(−0.0479351)+19(−0.0657786)−5(−0.09375)+(−0.1399417))=0.11982675$

Now error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

Calculate $εa$ as shown below,

$εa=|0.1196901−0.121659730.1196901|×100=1.6456%$

The true percent relative errors can be calculated as,

$εt=|0.12−0.11969010.12|×100=0.25825%$

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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Answer 2

Textbook Question

Chapter 26, Problem 6P

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

$d y d t = − 2 y t$

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 $( ε S = 1 % )$ , and (b) the fourth-order Adams method $( ε S = 0.01 % )$ . [Note: The exact answers obtained analytically are $y ( 4.5 ) = 0.148148 and y ( 5 ) = 0.12$ .] Compute the true percent relative errors $ε t$ for your results.

(a)

Expert Solution

To determine

To calculate: The solution of differential equation, $dydt=−2yt$ from $t=4 to t=5$ using the non-self-starting Heun method with step size of 0.5 and initial values $y(2.5)=0.48, y(3)=0.333333, y(3.5)=0.244898 and y(4)=0.1875$ . 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:

$jyj+1jεa10.14726832.63%20.14769940.292%$

Second step:

The true error is $εt=0.39%$ .

Corrector yields:

$jyj+1jεa10.119531930.236%$

Explanation of Solution

Given Information:

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

Formula used:

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

$yi+10=yi−1m+f(ti,yim)2h$

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

$yi+1j=yim+f(ti,yim)+f(ti+1,yi+1j−1)2h$

(3) Predictor modifier is given by:

$yi+10=yi+1,u0+45(yi,um−yi,u0)$

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,

$yi+10=yi−1m+f(ti,yim)2h$ .. .. .. (1)

For the given differential equation, it can be written as

$yi+10=yi−1m+(−2yimti)2h$

Substitute, $i=0, m=0 and h=0.5$ in above equation,

$y10=y−10+(−2y00t0)2(0.5)=y−10+(−2y00t0)$

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

Now, substitute $y−10=0.244898 , y00=0.1875 and t0=4$ .

Thus,

$y10=0.244898+(−2×0.18754)=0.151148$

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

$yi+1j=yim+f(ti,yim)+f(ti+1,yi+1j−1)2h$

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

$yi+1j=yim+[(−2yimti)+(−2yi+1j−1ti+1)2]h$

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

$y11=y00+[(−2y00t0)+(−2y10t1)2]h$

Put $y00=0.1875, y10=0.151148, t0=4, t1=4.5 and h=0.5$

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

Similarly, for $i=0 and j=2$ corrector gives $y12=0.1476994$ .

Error is calculated as,

$εa=|y12−y11y12|×100=|0.1476994−0.14726830.1476994|×100=0.2918766$

Also, the true error can be computed as,

$εt=|0.148148−0.14769940.148148|×100=0.3028$

Hence, the true error is $εt=0.3028%$ .

$jyj+1jεa10.14726832.63%20.14769940.292%$

Second step:

Predictor

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

$y20=y00+(−2y10t1)2h$

Substitute $y00=4.762673, y10=3.913253 and t1=2.5$ in above equation,

$y20=0.1875+(−2×0.1511484.5)1=0.12032311$

Predictor modifier is given by:

$yi+10=yi+1,u0+45(yi,um−yi,u0)$

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

Now, substitute $i=1 and m=2$ in above equation,

$y20=y2,u0+45(y1,u2−y1,u0)$

Substitute, $y2,u0=0.12032311, y1,u2=0.1476994 and y1,u0=0.151148$ in above equation,

$y20=0.12032311+45(0.1476994−0.151148)=0.11756423$

Corrector is given by:

$yi+1j=yim+(−2yimti)+(−2yi+1j−1ti+1)2h$

Substitute $i=1 j=1, m=2 and h=0.5$ , m= 2 in above formula as shown below,

$y21=y12+[(−2y12t1)+(−2y20t2)2]h$

Now, substitute $y12=0.1476994$ , $y20=0.11756423$ , $t1=4.5$ and $t2=5$ as shown below,

$y21=0.1476994+[(−2×0.14769944.5)+(−2×0.117564235)2]0.5=0.11953193$

Error can be computed as,

$εa=|y21−y20y20|×100=|0.11953193−0.117564230.11953193|×100=1.646%$

The true error can be computed as,

$εt=|0.12−0.119531930.12|×100=0.39%$

Hence, the true error is $εt=0.39%$ .

Corrector yields:

$jyj+1jεa10.119531930.236%$

(b)

Expert Solution

To determine

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

Formula used:

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

$yi+10=yim+h24(55fim−59fi−1m+37fi−2m−9fi−3m)$

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

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$

(3) Error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

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, $y−3=0.48$ . Substitute this value in equation (1)

Thus,

$f−3=−2×0.482.5=−0.384$

At $t=3$ , $y(3)=0.333333$ . So, $y−2=0.333333$ . Substitute this value in equation (1)

Thus,

$f−2=−2×0.3333333=−0.222222$

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

Thus,

$f−1=−2×0.2448983.5=−0.1399417$

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,

$yi+10=yim+h24(55fim−59fi−1m+37fi−2m−9fi−3m)$ .. .. .. (2)

The fourth order Adams formula as the corrector is,

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$ .. .. .. (3)

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

$y10=y00+h24(55f0−59f−1+37f−2−9f−3)$

Substitute values $f0=−0.09375$ , $f−1=−0.1399417$ , $f−2=−0.222222$ , $f−3=−0.384$ and $h=0.5$ .

$y10=0.1875+0.524×(55(−0.09375)−59(−0.1399417)+37(−0.222222)−9(−0.384))=0.1527937$

Hence, at $t=4.5, y10=0.1527937$

$f1=−2×0.15279374.5=−0.0679083$

Thus, $f1=−0.0679083$ .

Now, use corrector,

Put $i=0 , j=1$ in equation (3),

$y11=y00+h24(9f1+19f0−5f−1+f−2)$

Substitute values $f0=−0.09375$ , $f−1=−0.1399417$ , $f−2=−0.222222$ , $f1=−0.0679083$ and $h=0.5$ .

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

At $t=4.5 and y11=0.14760545$

$f1=−2×0.147605454.5=−0.0656024$

Thus, $f1=−0.0656024$ .

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

$y12=0.1875+0.524(9(−0.0656024)+19(−0.09375)−5(−0.1399417)+(−0.222222))=0.14803781$

Again,

At $t=4.5 and y12=0.14803781$

$f1=−2×0.148037814.5=−0.0657946$

Thus, $f1=−0.0657946$ .

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

$y13=0.1875+0.524(9(−0.0657946)+19(−0.09375)−5(−0.1399417)+(−0.222222))=0.148001773$

At $t=4.5 and y13=0.148001773$

$f1=−2×0.1480017734.5=−0.0657786$

Thus, $f1=−0.0657786$ .

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

Substitute $i=1$ in equation (2),

$y20=y1+h24(55f1−59f0+37f−1−9f−2)$

Substitute values, $f1=−0.0657786$ $f0=−0.09375$ , $f−1=−0.1399417$ and $f−2=−0.222222$ in above equation, to get

$y20=0.148001773+0.524×(55(−0.0657786)−59(−0.09375)+37(−0.1399417)−9(−0.222222))=0.12165973$

At $t=5 and y20=0.12165973$ .

$f2=−2×0.121659735=−0.0486639$

Thus, $f2=−0.0486639$

Now error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

Calculate $εa$ as shown below,

$εa=|0.14760545−0.15279370.14760545|×100=3.5149%$

The true percent relative errors can be calculated as,

$εt=|0.148148−0.147605450.148148|×100=0.36622%$

Corrector formula is given as,

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$

Substitute $i=1 and j=1$ in above formula,

$y21=y1+h24(9f20+19f1−5f0+f−1)$

Substitute values, $f1=−0.0657786$ $f0=−0.09375$ , $f−1=−0.1399417$ and $f2=−0.0486639$ in above equation, to get

$y21=0.148001773+0.524(9(−0.0486639)+19(−0.0657786)−5(−0.09375)+(−0.1399417))=0.1196901$

At $t=5 and y21=0.1196901$

$f2=−2×0.11969015=−0.0478760$

Thus, $f2=−0.0478760$ .

Hence,

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

At $t=5 and y22=0.11983783$

$f2=−2×0.119837835=−0.0479351$

Thus, $f2=−0.0479351$ .

Hence,

$y23=0.148001773+0.524(9(−0.0479351)+19(−0.0657786)−5(−0.09375)+(−0.1399417))=0.11982675$

Now error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

Calculate $εa$ as shown below,

$εa=|0.1196901−0.121659730.1196901|×100=1.6456%$

The true percent relative errors can be calculated as,

$εt=|0.12−0.11969010.12|×100=0.25825%$

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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Answer 3

Textbook Question

Chapter 26, Problem 6P

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

$d y d t = − 2 y t$

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 $( ε S = 1 % )$ , and (b) the fourth-order Adams method $( ε S = 0.01 % )$ . [Note: The exact answers obtained analytically are $y ( 4.5 ) = 0.148148 and y ( 5 ) = 0.12$ .] Compute the true percent relative errors $ε t$ for your results.

(a)

Expert Solution

To determine

To calculate: The solution of differential equation, $dydt=−2yt$ from $t=4 to t=5$ using the non-self-starting Heun method with step size of 0.5 and initial values $y(2.5)=0.48, y(3)=0.333333, y(3.5)=0.244898 and y(4)=0.1875$ . 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:

$jyj+1jεa10.14726832.63%20.14769940.292%$

Second step:

The true error is $εt=0.39%$ .

Corrector yields:

$jyj+1jεa10.119531930.236%$

Explanation of Solution

Given Information:

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

Formula used:

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

$yi+10=yi−1m+f(ti,yim)2h$

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

$yi+1j=yim+f(ti,yim)+f(ti+1,yi+1j−1)2h$

(3) Predictor modifier is given by:

$yi+10=yi+1,u0+45(yi,um−yi,u0)$

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,

$yi+10=yi−1m+f(ti,yim)2h$ .. .. .. (1)

For the given differential equation, it can be written as

$yi+10=yi−1m+(−2yimti)2h$

Substitute, $i=0, m=0 and h=0.5$ in above equation,

$y10=y−10+(−2y00t0)2(0.5)=y−10+(−2y00t0)$

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

Now, substitute $y−10=0.244898 , y00=0.1875 and t0=4$ .

Thus,

$y10=0.244898+(−2×0.18754)=0.151148$

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

$yi+1j=yim+f(ti,yim)+f(ti+1,yi+1j−1)2h$

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

$yi+1j=yim+[(−2yimti)+(−2yi+1j−1ti+1)2]h$

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

$y11=y00+[(−2y00t0)+(−2y10t1)2]h$

Put $y00=0.1875, y10=0.151148, t0=4, t1=4.5 and h=0.5$

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

Similarly, for $i=0 and j=2$ corrector gives $y12=0.1476994$ .

Error is calculated as,

$εa=|y12−y11y12|×100=|0.1476994−0.14726830.1476994|×100=0.2918766$

Also, the true error can be computed as,

$εt=|0.148148−0.14769940.148148|×100=0.3028$

Hence, the true error is $εt=0.3028%$ .

$jyj+1jεa10.14726832.63%20.14769940.292%$

Second step:

Predictor

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

$y20=y00+(−2y10t1)2h$

Substitute $y00=4.762673, y10=3.913253 and t1=2.5$ in above equation,

$y20=0.1875+(−2×0.1511484.5)1=0.12032311$

Predictor modifier is given by:

$yi+10=yi+1,u0+45(yi,um−yi,u0)$

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

Now, substitute $i=1 and m=2$ in above equation,

$y20=y2,u0+45(y1,u2−y1,u0)$

Substitute, $y2,u0=0.12032311, y1,u2=0.1476994 and y1,u0=0.151148$ in above equation,

$y20=0.12032311+45(0.1476994−0.151148)=0.11756423$

Corrector is given by:

$yi+1j=yim+(−2yimti)+(−2yi+1j−1ti+1)2h$

Substitute $i=1 j=1, m=2 and h=0.5$ , m= 2 in above formula as shown below,

$y21=y12+[(−2y12t1)+(−2y20t2)2]h$

Now, substitute $y12=0.1476994$ , $y20=0.11756423$ , $t1=4.5$ and $t2=5$ as shown below,

$y21=0.1476994+[(−2×0.14769944.5)+(−2×0.117564235)2]0.5=0.11953193$

Error can be computed as,

$εa=|y21−y20y20|×100=|0.11953193−0.117564230.11953193|×100=1.646%$

The true error can be computed as,

$εt=|0.12−0.119531930.12|×100=0.39%$

Hence, the true error is $εt=0.39%$ .

Corrector yields:

$jyj+1jεa10.119531930.236%$

(b)

Expert Solution

To determine

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

Formula used:

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

$yi+10=yim+h24(55fim−59fi−1m+37fi−2m−9fi−3m)$

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

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$

(3) Error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

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, $y−3=0.48$ . Substitute this value in equation (1)

Thus,

$f−3=−2×0.482.5=−0.384$

At $t=3$ , $y(3)=0.333333$ . So, $y−2=0.333333$ . Substitute this value in equation (1)

Thus,

$f−2=−2×0.3333333=−0.222222$

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

Thus,

$f−1=−2×0.2448983.5=−0.1399417$

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,

$yi+10=yim+h24(55fim−59fi−1m+37fi−2m−9fi−3m)$ .. .. .. (2)

The fourth order Adams formula as the corrector is,

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$ .. .. .. (3)

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

$y10=y00+h24(55f0−59f−1+37f−2−9f−3)$

Substitute values $f0=−0.09375$ , $f−1=−0.1399417$ , $f−2=−0.222222$ , $f−3=−0.384$ and $h=0.5$ .

$y10=0.1875+0.524×(55(−0.09375)−59(−0.1399417)+37(−0.222222)−9(−0.384))=0.1527937$

Hence, at $t=4.5, y10=0.1527937$

$f1=−2×0.15279374.5=−0.0679083$

Thus, $f1=−0.0679083$ .

Now, use corrector,

Put $i=0 , j=1$ in equation (3),

$y11=y00+h24(9f1+19f0−5f−1+f−2)$

Substitute values $f0=−0.09375$ , $f−1=−0.1399417$ , $f−2=−0.222222$ , $f1=−0.0679083$ and $h=0.5$ .

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

At $t=4.5 and y11=0.14760545$

$f1=−2×0.147605454.5=−0.0656024$

Thus, $f1=−0.0656024$ .

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

$y12=0.1875+0.524(9(−0.0656024)+19(−0.09375)−5(−0.1399417)+(−0.222222))=0.14803781$

Again,

At $t=4.5 and y12=0.14803781$

$f1=−2×0.148037814.5=−0.0657946$

Thus, $f1=−0.0657946$ .

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

$y13=0.1875+0.524(9(−0.0657946)+19(−0.09375)−5(−0.1399417)+(−0.222222))=0.148001773$

At $t=4.5 and y13=0.148001773$

$f1=−2×0.1480017734.5=−0.0657786$

Thus, $f1=−0.0657786$ .

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

Substitute $i=1$ in equation (2),

$y20=y1+h24(55f1−59f0+37f−1−9f−2)$

Substitute values, $f1=−0.0657786$ $f0=−0.09375$ , $f−1=−0.1399417$ and $f−2=−0.222222$ in above equation, to get

$y20=0.148001773+0.524×(55(−0.0657786)−59(−0.09375)+37(−0.1399417)−9(−0.222222))=0.12165973$

At $t=5 and y20=0.12165973$ .

$f2=−2×0.121659735=−0.0486639$

Thus, $f2=−0.0486639$

Now error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

Calculate $εa$ as shown below,

$εa=|0.14760545−0.15279370.14760545|×100=3.5149%$

The true percent relative errors can be calculated as,

$εt=|0.148148−0.147605450.148148|×100=0.36622%$

Corrector formula is given as,

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$

Substitute $i=1 and j=1$ in above formula,

$y21=y1+h24(9f20+19f1−5f0+f−1)$

Substitute values, $f1=−0.0657786$ $f0=−0.09375$ , $f−1=−0.1399417$ and $f2=−0.0486639$ in above equation, to get

$y21=0.148001773+0.524(9(−0.0486639)+19(−0.0657786)−5(−0.09375)+(−0.1399417))=0.1196901$

At $t=5 and y21=0.1196901$

$f2=−2×0.11969015=−0.0478760$

Thus, $f2=−0.0478760$ .

Hence,

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

At $t=5 and y22=0.11983783$

$f2=−2×0.119837835=−0.0479351$

Thus, $f2=−0.0479351$ .

Hence,

$y23=0.148001773+0.524(9(−0.0479351)+19(−0.0657786)−5(−0.09375)+(−0.1399417))=0.11982675$

Now error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

Calculate $εa$ as shown below,

$εa=|0.1196901−0.121659730.1196901|×100=1.6456%$

The true percent relative errors can be calculated as,

$εt=|0.12−0.11969010.12|×100=0.25825%$

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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Answer 4

Textbook Question

Chapter 26, Problem 6P

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

$d y d t = − 2 y t$

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 $( ε S = 1 % )$ , and (b) the fourth-order Adams method $( ε S = 0.01 % )$ . [Note: The exact answers obtained analytically are $y ( 4.5 ) = 0.148148 and y ( 5 ) = 0.12$ .] Compute the true percent relative errors $ε t$ for your results.

(a)

Expert Solution

To determine

To calculate: The solution of differential equation, $dydt=−2yt$ from $t=4 to t=5$ using the non-self-starting Heun method with step size of 0.5 and initial values $y(2.5)=0.48, y(3)=0.333333, y(3.5)=0.244898 and y(4)=0.1875$ . 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:

$jyj+1jεa10.14726832.63%20.14769940.292%$

Second step:

The true error is $εt=0.39%$ .

Corrector yields:

$jyj+1jεa10.119531930.236%$

Explanation of Solution

Given Information:

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

Formula used:

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

$yi+10=yi−1m+f(ti,yim)2h$

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

$yi+1j=yim+f(ti,yim)+f(ti+1,yi+1j−1)2h$

(3) Predictor modifier is given by:

$yi+10=yi+1,u0+45(yi,um−yi,u0)$

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,

$yi+10=yi−1m+f(ti,yim)2h$ .. .. .. (1)

For the given differential equation, it can be written as

$yi+10=yi−1m+(−2yimti)2h$

Substitute, $i=0, m=0 and h=0.5$ in above equation,

$y10=y−10+(−2y00t0)2(0.5)=y−10+(−2y00t0)$

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

Now, substitute $y−10=0.244898 , y00=0.1875 and t0=4$ .

Thus,

$y10=0.244898+(−2×0.18754)=0.151148$

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

$yi+1j=yim+f(ti,yim)+f(ti+1,yi+1j−1)2h$

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

$yi+1j=yim+[(−2yimti)+(−2yi+1j−1ti+1)2]h$

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

$y11=y00+[(−2y00t0)+(−2y10t1)2]h$

Put $y00=0.1875, y10=0.151148, t0=4, t1=4.5 and h=0.5$

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

Similarly, for $i=0 and j=2$ corrector gives $y12=0.1476994$ .

Error is calculated as,

$εa=|y12−y11y12|×100=|0.1476994−0.14726830.1476994|×100=0.2918766$

Also, the true error can be computed as,

$εt=|0.148148−0.14769940.148148|×100=0.3028$

Hence, the true error is $εt=0.3028%$ .

$jyj+1jεa10.14726832.63%20.14769940.292%$

Second step:

Predictor

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

$y20=y00+(−2y10t1)2h$

Substitute $y00=4.762673, y10=3.913253 and t1=2.5$ in above equation,

$y20=0.1875+(−2×0.1511484.5)1=0.12032311$

Predictor modifier is given by:

$yi+10=yi+1,u0+45(yi,um−yi,u0)$

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

Now, substitute $i=1 and m=2$ in above equation,

$y20=y2,u0+45(y1,u2−y1,u0)$

Substitute, $y2,u0=0.12032311, y1,u2=0.1476994 and y1,u0=0.151148$ in above equation,

$y20=0.12032311+45(0.1476994−0.151148)=0.11756423$

Corrector is given by:

$yi+1j=yim+(−2yimti)+(−2yi+1j−1ti+1)2h$

Substitute $i=1 j=1, m=2 and h=0.5$ , m= 2 in above formula as shown below,

$y21=y12+[(−2y12t1)+(−2y20t2)2]h$

Now, substitute $y12=0.1476994$ , $y20=0.11756423$ , $t1=4.5$ and $t2=5$ as shown below,

$y21=0.1476994+[(−2×0.14769944.5)+(−2×0.117564235)2]0.5=0.11953193$

Error can be computed as,

$εa=|y21−y20y20|×100=|0.11953193−0.117564230.11953193|×100=1.646%$

The true error can be computed as,

$εt=|0.12−0.119531930.12|×100=0.39%$

Hence, the true error is $εt=0.39%$ .

Corrector yields:

$jyj+1jεa10.119531930.236%$

(b)

Expert Solution

To determine

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

Formula used:

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

$yi+10=yim+h24(55fim−59fi−1m+37fi−2m−9fi−3m)$

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

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$

(3) Error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

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, $y−3=0.48$ . Substitute this value in equation (1)

Thus,

$f−3=−2×0.482.5=−0.384$

At $t=3$ , $y(3)=0.333333$ . So, $y−2=0.333333$ . Substitute this value in equation (1)

Thus,

$f−2=−2×0.3333333=−0.222222$

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

Thus,

$f−1=−2×0.2448983.5=−0.1399417$

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,

$yi+10=yim+h24(55fim−59fi−1m+37fi−2m−9fi−3m)$ .. .. .. (2)

The fourth order Adams formula as the corrector is,

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$ .. .. .. (3)

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

$y10=y00+h24(55f0−59f−1+37f−2−9f−3)$

Substitute values $f0=−0.09375$ , $f−1=−0.1399417$ , $f−2=−0.222222$ , $f−3=−0.384$ and $h=0.5$ .

$y10=0.1875+0.524×(55(−0.09375)−59(−0.1399417)+37(−0.222222)−9(−0.384))=0.1527937$

Hence, at $t=4.5, y10=0.1527937$

$f1=−2×0.15279374.5=−0.0679083$

Thus, $f1=−0.0679083$ .

Now, use corrector,

Put $i=0 , j=1$ in equation (3),

$y11=y00+h24(9f1+19f0−5f−1+f−2)$

Substitute values $f0=−0.09375$ , $f−1=−0.1399417$ , $f−2=−0.222222$ , $f1=−0.0679083$ and $h=0.5$ .

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

At $t=4.5 and y11=0.14760545$

$f1=−2×0.147605454.5=−0.0656024$

Thus, $f1=−0.0656024$ .

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

$y12=0.1875+0.524(9(−0.0656024)+19(−0.09375)−5(−0.1399417)+(−0.222222))=0.14803781$

Again,

At $t=4.5 and y12=0.14803781$

$f1=−2×0.148037814.5=−0.0657946$

Thus, $f1=−0.0657946$ .

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

$y13=0.1875+0.524(9(−0.0657946)+19(−0.09375)−5(−0.1399417)+(−0.222222))=0.148001773$

At $t=4.5 and y13=0.148001773$

$f1=−2×0.1480017734.5=−0.0657786$

Thus, $f1=−0.0657786$ .

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

Substitute $i=1$ in equation (2),

$y20=y1+h24(55f1−59f0+37f−1−9f−2)$

Substitute values, $f1=−0.0657786$ $f0=−0.09375$ , $f−1=−0.1399417$ and $f−2=−0.222222$ in above equation, to get

$y20=0.148001773+0.524×(55(−0.0657786)−59(−0.09375)+37(−0.1399417)−9(−0.222222))=0.12165973$

At $t=5 and y20=0.12165973$ .

$f2=−2×0.121659735=−0.0486639$

Thus, $f2=−0.0486639$

Now error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

Calculate $εa$ as shown below,

$εa=|0.14760545−0.15279370.14760545|×100=3.5149%$

The true percent relative errors can be calculated as,

$εt=|0.148148−0.147605450.148148|×100=0.36622%$

Corrector formula is given as,

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$

Substitute $i=1 and j=1$ in above formula,

$y21=y1+h24(9f20+19f1−5f0+f−1)$

Substitute values, $f1=−0.0657786$ $f0=−0.09375$ , $f−1=−0.1399417$ and $f2=−0.0486639$ in above equation, to get

$y21=0.148001773+0.524(9(−0.0486639)+19(−0.0657786)−5(−0.09375)+(−0.1399417))=0.1196901$

At $t=5 and y21=0.1196901$

$f2=−2×0.11969015=−0.0478760$

Thus, $f2=−0.0478760$ .

Hence,

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

At $t=5 and y22=0.11983783$

$f2=−2×0.119837835=−0.0479351$

Thus, $f2=−0.0479351$ .

Hence,

$y23=0.148001773+0.524(9(−0.0479351)+19(−0.0657786)−5(−0.09375)+(−0.1399417))=0.11982675$

Now error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

Calculate $εa$ as shown below,

$εa=|0.1196901−0.121659730.1196901|×100=1.6456%$

The true percent relative errors can be calculated as,

$εt=|0.12−0.11969010.12|×100=0.25825%$

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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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

To calculate: The solution of differential equation, $dydt=−2yt$ from $t=4 to t=5$ using the non-self-starting Heun method with step size of 0.5 and initial values $y(2.5)=0.48, y(3)=0.333333, y(3.5)=0.244898 and y(4)=0.1875$ . 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:

$jyj+1jεa10.14726832.63%20.14769940.292%$

Second step:

The true error is $εt=0.39%$ .

Corrector yields:

$jyj+1jεa10.119531930.236%$

Explanation of Solution

Given Information:

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

Formula used:

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

$yi+10=yi−1m+f(ti,yim)2h$

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

$yi+1j=yim+f(ti,yim)+f(ti+1,yi+1j−1)2h$

(3) Predictor modifier is given by:

$yi+10=yi+1,u0+45(yi,um−yi,u0)$

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,

$yi+10=yi−1m+f(ti,yim)2h$ .. .. .. (1)

For the given differential equation, it can be written as

$yi+10=yi−1m+(−2yimti)2h$

Substitute, $i=0, m=0 and h=0.5$ in above equation,

$y10=y−10+(−2y00t0)2(0.5)=y−10+(−2y00t0)$

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

Now, substitute $y−10=0.244898 , y00=0.1875 and t0=4$ .

Thus,

$y10=0.244898+(−2×0.18754)=0.151148$

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

$yi+1j=yim+f(ti,yim)+f(ti+1,yi+1j−1)2h$

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

$yi+1j=yim+[(−2yimti)+(−2yi+1j−1ti+1)2]h$

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

$y11=y00+[(−2y00t0)+(−2y10t1)2]h$

Put $y00=0.1875, y10=0.151148, t0=4, t1=4.5 and h=0.5$

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

Similarly, for $i=0 and j=2$ corrector gives $y12=0.1476994$ .

Error is calculated as,

$εa=|y12−y11y12|×100=|0.1476994−0.14726830.1476994|×100=0.2918766$

Also, the true error can be computed as,

$εt=|0.148148−0.14769940.148148|×100=0.3028$

Hence, the true error is $εt=0.3028%$ .

$jyj+1jεa10.14726832.63%20.14769940.292%$

Second step:

Predictor

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

$y20=y00+(−2y10t1)2h$

Substitute $y00=4.762673, y10=3.913253 and t1=2.5$ in above equation,

$y20=0.1875+(−2×0.1511484.5)1=0.12032311$

Predictor modifier is given by:

$yi+10=yi+1,u0+45(yi,um−yi,u0)$

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

Now, substitute $i=1 and m=2$ in above equation,

$y20=y2,u0+45(y1,u2−y1,u0)$

Substitute, $y2,u0=0.12032311, y1,u2=0.1476994 and y1,u0=0.151148$ in above equation,

$y20=0.12032311+45(0.1476994−0.151148)=0.11756423$

Corrector is given by:

$yi+1j=yim+(−2yimti)+(−2yi+1j−1ti+1)2h$

Substitute $i=1 j=1, m=2 and h=0.5$ , m= 2 in above formula as shown below,

$y21=y12+[(−2y12t1)+(−2y20t2)2]h$

Now, substitute $y12=0.1476994$ , $y20=0.11756423$ , $t1=4.5$ and $t2=5$ as shown below,

$y21=0.1476994+[(−2×0.14769944.5)+(−2×0.117564235)2]0.5=0.11953193$

Error can be computed as,

$εa=|y21−y20y20|×100=|0.11953193−0.117564230.11953193|×100=1.646%$

The true error can be computed as,

$εt=|0.12−0.119531930.12|×100=0.39%$

Hence, the true error is $εt=0.39%$ .

Corrector yields:

$jyj+1jεa10.119531930.236%$

(b)

Expert Solution

To determine

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

Formula used:

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

$yi+10=yim+h24(55fim−59fi−1m+37fi−2m−9fi−3m)$

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

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$

(3) Error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

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, $y−3=0.48$ . Substitute this value in equation (1)

Thus,

$f−3=−2×0.482.5=−0.384$

At $t=3$ , $y(3)=0.333333$ . So, $y−2=0.333333$ . Substitute this value in equation (1)

Thus,

$f−2=−2×0.3333333=−0.222222$

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

Thus,

$f−1=−2×0.2448983.5=−0.1399417$

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,

$yi+10=yim+h24(55fim−59fi−1m+37fi−2m−9fi−3m)$ .. .. .. (2)

The fourth order Adams formula as the corrector is,

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$ .. .. .. (3)

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

$y10=y00+h24(55f0−59f−1+37f−2−9f−3)$

Substitute values $f0=−0.09375$ , $f−1=−0.1399417$ , $f−2=−0.222222$ , $f−3=−0.384$ and $h=0.5$ .

$y10=0.1875+0.524×(55(−0.09375)−59(−0.1399417)+37(−0.222222)−9(−0.384))=0.1527937$

Hence, at $t=4.5, y10=0.1527937$

$f1=−2×0.15279374.5=−0.0679083$

Thus, $f1=−0.0679083$ .

Now, use corrector,

Put $i=0 , j=1$ in equation (3),

$y11=y00+h24(9f1+19f0−5f−1+f−2)$

Substitute values $f0=−0.09375$ , $f−1=−0.1399417$ , $f−2=−0.222222$ , $f1=−0.0679083$ and $h=0.5$ .

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

At $t=4.5 and y11=0.14760545$

$f1=−2×0.147605454.5=−0.0656024$

Thus, $f1=−0.0656024$ .

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

$y12=0.1875+0.524(9(−0.0656024)+19(−0.09375)−5(−0.1399417)+(−0.222222))=0.14803781$

Again,

At $t=4.5 and y12=0.14803781$

$f1=−2×0.148037814.5=−0.0657946$

Thus, $f1=−0.0657946$ .

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

$y13=0.1875+0.524(9(−0.0657946)+19(−0.09375)−5(−0.1399417)+(−0.222222))=0.148001773$

At $t=4.5 and y13=0.148001773$

$f1=−2×0.1480017734.5=−0.0657786$

Thus, $f1=−0.0657786$ .

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

Substitute $i=1$ in equation (2),

$y20=y1+h24(55f1−59f0+37f−1−9f−2)$

Substitute values, $f1=−0.0657786$ $f0=−0.09375$ , $f−1=−0.1399417$ and $f−2=−0.222222$ in above equation, to get

$y20=0.148001773+0.524×(55(−0.0657786)−59(−0.09375)+37(−0.1399417)−9(−0.222222))=0.12165973$

At $t=5 and y20=0.12165973$ .

$f2=−2×0.121659735=−0.0486639$

Thus, $f2=−0.0486639$

Now error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

Calculate $εa$ as shown below,

$εa=|0.14760545−0.15279370.14760545|×100=3.5149%$

The true percent relative errors can be calculated as,

$εt=|0.148148−0.147605450.148148|×100=0.36622%$

Corrector formula is given as,

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$

Substitute $i=1 and j=1$ in above formula,

$y21=y1+h24(9f20+19f1−5f0+f−1)$

Substitute values, $f1=−0.0657786$ $f0=−0.09375$ , $f−1=−0.1399417$ and $f2=−0.0486639$ in above equation, to get

$y21=0.148001773+0.524(9(−0.0486639)+19(−0.0657786)−5(−0.09375)+(−0.1399417))=0.1196901$

At $t=5 and y21=0.1196901$

$f2=−2×0.11969015=−0.0478760$

Thus, $f2=−0.0478760$ .

Hence,

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

At $t=5 and y22=0.11983783$

$f2=−2×0.119837835=−0.0479351$

Thus, $f2=−0.0479351$ .

Hence,

$y23=0.148001773+0.524(9(−0.0479351)+19(−0.0657786)−5(−0.09375)+(−0.1399417))=0.11982675$

Now error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

Calculate $εa$ as shown below,

$εa=|0.1196901−0.121659730.1196901|×100=1.6456%$

The true percent relative errors can be calculated as,

$εt=|0.12−0.11969010.12|×100=0.25825%$

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:

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

Answer 8

(a)

Expert Solution

To determine

To calculate: The solution of differential equation, $dydt=−2yt$ from $t=4 to t=5$ using the non-self-starting Heun method with step size of 0.5 and initial values $y(2.5)=0.48, y(3)=0.333333, y(3.5)=0.244898 and y(4)=0.1875$ . 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:

$jyj+1jεa10.14726832.63%20.14769940.292%$

Second step:

The true error is $εt=0.39%$ .

Corrector yields:

$jyj+1jεa10.119531930.236%$

Explanation of Solution

Given Information:

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

Formula used:

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

$yi+10=yi−1m+f(ti,yim)2h$

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

$yi+1j=yim+f(ti,yim)+f(ti+1,yi+1j−1)2h$

(3) Predictor modifier is given by:

$yi+10=yi+1,u0+45(yi,um−yi,u0)$

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,

$yi+10=yi−1m+f(ti,yim)2h$ .. .. .. (1)

For the given differential equation, it can be written as

$yi+10=yi−1m+(−2yimti)2h$

Substitute, $i=0, m=0 and h=0.5$ in above equation,

$y10=y−10+(−2y00t0)2(0.5)=y−10+(−2y00t0)$

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

Now, substitute $y−10=0.244898 , y00=0.1875 and t0=4$ .

Thus,

$y10=0.244898+(−2×0.18754)=0.151148$

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

$yi+1j=yim+f(ti,yim)+f(ti+1,yi+1j−1)2h$

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

$yi+1j=yim+[(−2yimti)+(−2yi+1j−1ti+1)2]h$

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

$y11=y00+[(−2y00t0)+(−2y10t1)2]h$

Put $y00=0.1875, y10=0.151148, t0=4, t1=4.5 and h=0.5$

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

Similarly, for $i=0 and j=2$ corrector gives $y12=0.1476994$ .

Error is calculated as,

$εa=|y12−y11y12|×100=|0.1476994−0.14726830.1476994|×100=0.2918766$

Also, the true error can be computed as,

$εt=|0.148148−0.14769940.148148|×100=0.3028$

Hence, the true error is $εt=0.3028%$ .

$jyj+1jεa10.14726832.63%20.14769940.292%$

Second step:

Predictor

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

$y20=y00+(−2y10t1)2h$

Substitute $y00=4.762673, y10=3.913253 and t1=2.5$ in above equation,

$y20=0.1875+(−2×0.1511484.5)1=0.12032311$

Predictor modifier is given by:

$yi+10=yi+1,u0+45(yi,um−yi,u0)$

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

Now, substitute $i=1 and m=2$ in above equation,

$y20=y2,u0+45(y1,u2−y1,u0)$

Substitute, $y2,u0=0.12032311, y1,u2=0.1476994 and y1,u0=0.151148$ in above equation,

$y20=0.12032311+45(0.1476994−0.151148)=0.11756423$

Corrector is given by:

$yi+1j=yim+(−2yimti)+(−2yi+1j−1ti+1)2h$

Substitute $i=1 j=1, m=2 and h=0.5$ , m= 2 in above formula as shown below,

$y21=y12+[(−2y12t1)+(−2y20t2)2]h$

Now, substitute $y12=0.1476994$ , $y20=0.11756423$ , $t1=4.5$ and $t2=5$ as shown below,

$y21=0.1476994+[(−2×0.14769944.5)+(−2×0.117564235)2]0.5=0.11953193$

Error can be computed as,

$εa=|y21−y20y20|×100=|0.11953193−0.117564230.11953193|×100=1.646%$

The true error can be computed as,

$εt=|0.12−0.119531930.12|×100=0.39%$

Hence, the true error is $εt=0.39%$ .

Corrector yields:

$jyj+1jεa10.119531930.236%$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

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

Answer 13

Answer to Problem 6P

Solution:

First step:

The true error is $εt=0.3028%$ .

Corrector yields:

$jyj+1jεa10.14726832.63%20.14769940.292%$

Second step:

The true error is $εt=0.39%$ .

Corrector yields:

$jyj+1jεa10.119531930.236%$

Answer 14

Explanation of Solution

Given Information:

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

Formula used:

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

$yi+10=yi−1m+f(ti,yim)2h$

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

$yi+1j=yim+f(ti,yim)+f(ti+1,yi+1j−1)2h$

(3) Predictor modifier is given by:

$yi+10=yi+1,u0+45(yi,um−yi,u0)$

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,

$yi+10=yi−1m+f(ti,yim)2h$ .. .. .. (1)

For the given differential equation, it can be written as

$yi+10=yi−1m+(−2yimti)2h$

Substitute, $i=0, m=0 and h=0.5$ in above equation,

$y10=y−10+(−2y00t0)2(0.5)=y−10+(−2y00t0)$

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

Now, substitute $y−10=0.244898 , y00=0.1875 and t0=4$ .

Thus,

$y10=0.244898+(−2×0.18754)=0.151148$

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

$yi+1j=yim+f(ti,yim)+f(ti+1,yi+1j−1)2h$

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

$yi+1j=yim+[(−2yimti)+(−2yi+1j−1ti+1)2]h$

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

$y11=y00+[(−2y00t0)+(−2y10t1)2]h$

Put $y00=0.1875, y10=0.151148, t0=4, t1=4.5 and h=0.5$

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

Similarly, for $i=0 and j=2$ corrector gives $y12=0.1476994$ .

Error is calculated as,

$εa=|y12−y11y12|×100=|0.1476994−0.14726830.1476994|×100=0.2918766$

Also, the true error can be computed as,

$εt=|0.148148−0.14769940.148148|×100=0.3028$

Hence, the true error is $εt=0.3028%$ .

$jyj+1jεa10.14726832.63%20.14769940.292%$

Second step:

Predictor

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

$y20=y00+(−2y10t1)2h$

Substitute $y00=4.762673, y10=3.913253 and t1=2.5$ in above equation,

$y20=0.1875+(−2×0.1511484.5)1=0.12032311$

Predictor modifier is given by:

$yi+10=yi+1,u0+45(yi,um−yi,u0)$

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

Now, substitute $i=1 and m=2$ in above equation,

$y20=y2,u0+45(y1,u2−y1,u0)$

Substitute, $y2,u0=0.12032311, y1,u2=0.1476994 and y1,u0=0.151148$ in above equation,

$y20=0.12032311+45(0.1476994−0.151148)=0.11756423$

Corrector is given by:

$yi+1j=yim+(−2yimti)+(−2yi+1j−1ti+1)2h$

Substitute $i=1 j=1, m=2 and h=0.5$ , m= 2 in above formula as shown below,

$y21=y12+[(−2y12t1)+(−2y20t2)2]h$

Now, substitute $y12=0.1476994$ , $y20=0.11756423$ , $t1=4.5$ and $t2=5$ as shown below,

$y21=0.1476994+[(−2×0.14769944.5)+(−2×0.117564235)2]0.5=0.11953193$

Error can be computed as,

$εa=|y21−y20y20|×100=|0.11953193−0.117564230.11953193|×100=1.646%$

The true error can be computed as,

$εt=|0.12−0.119531930.12|×100=0.39%$

Hence, the true error is $εt=0.39%$ .

Corrector yields:

$jyj+1jεa10.119531930.236%$

Answer 15

(b)

Expert Solution

To determine

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

Formula used:

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

$yi+10=yim+h24(55fim−59fi−1m+37fi−2m−9fi−3m)$

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

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$

(3) Error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

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, $y−3=0.48$ . Substitute this value in equation (1)

Thus,

$f−3=−2×0.482.5=−0.384$

At $t=3$ , $y(3)=0.333333$ . So, $y−2=0.333333$ . Substitute this value in equation (1)

Thus,

$f−2=−2×0.3333333=−0.222222$

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

Thus,

$f−1=−2×0.2448983.5=−0.1399417$

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,

$yi+10=yim+h24(55fim−59fi−1m+37fi−2m−9fi−3m)$ .. .. .. (2)

The fourth order Adams formula as the corrector is,

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$ .. .. .. (3)

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

$y10=y00+h24(55f0−59f−1+37f−2−9f−3)$

Substitute values $f0=−0.09375$ , $f−1=−0.1399417$ , $f−2=−0.222222$ , $f−3=−0.384$ and $h=0.5$ .

$y10=0.1875+0.524×(55(−0.09375)−59(−0.1399417)+37(−0.222222)−9(−0.384))=0.1527937$

Hence, at $t=4.5, y10=0.1527937$

$f1=−2×0.15279374.5=−0.0679083$

Thus, $f1=−0.0679083$ .

Now, use corrector,

Put $i=0 , j=1$ in equation (3),

$y11=y00+h24(9f1+19f0−5f−1+f−2)$

Substitute values $f0=−0.09375$ , $f−1=−0.1399417$ , $f−2=−0.222222$ , $f1=−0.0679083$ and $h=0.5$ .

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

At $t=4.5 and y11=0.14760545$

$f1=−2×0.147605454.5=−0.0656024$

Thus, $f1=−0.0656024$ .

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

$y12=0.1875+0.524(9(−0.0656024)+19(−0.09375)−5(−0.1399417)+(−0.222222))=0.14803781$

Again,

At $t=4.5 and y12=0.14803781$

$f1=−2×0.148037814.5=−0.0657946$

Thus, $f1=−0.0657946$ .

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

$y13=0.1875+0.524(9(−0.0657946)+19(−0.09375)−5(−0.1399417)+(−0.222222))=0.148001773$

At $t=4.5 and y13=0.148001773$

$f1=−2×0.1480017734.5=−0.0657786$

Thus, $f1=−0.0657786$ .

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

Substitute $i=1$ in equation (2),

$y20=y1+h24(55f1−59f0+37f−1−9f−2)$

Substitute values, $f1=−0.0657786$ $f0=−0.09375$ , $f−1=−0.1399417$ and $f−2=−0.222222$ in above equation, to get

$y20=0.148001773+0.524×(55(−0.0657786)−59(−0.09375)+37(−0.1399417)−9(−0.222222))=0.12165973$

At $t=5 and y20=0.12165973$ .

$f2=−2×0.121659735=−0.0486639$

Thus, $f2=−0.0486639$

Now error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

Calculate $εa$ as shown below,

$εa=|0.14760545−0.15279370.14760545|×100=3.5149%$

The true percent relative errors can be calculated as,

$εt=|0.148148−0.147605450.148148|×100=0.36622%$

Corrector formula is given as,

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$

Substitute $i=1 and j=1$ in above formula,

$y21=y1+h24(9f20+19f1−5f0+f−1)$

Substitute values, $f1=−0.0657786$ $f0=−0.09375$ , $f−1=−0.1399417$ and $f2=−0.0486639$ in above equation, to get

$y21=0.148001773+0.524(9(−0.0486639)+19(−0.0657786)−5(−0.09375)+(−0.1399417))=0.1196901$

At $t=5 and y21=0.1196901$

$f2=−2×0.11969015=−0.0478760$

Thus, $f2=−0.0478760$ .

Hence,

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

At $t=5 and y22=0.11983783$

$f2=−2×0.119837835=−0.0479351$

Thus, $f2=−0.0479351$ .

Hence,

$y23=0.148001773+0.524(9(−0.0479351)+19(−0.0657786)−5(−0.09375)+(−0.1399417))=0.11982675$

Now error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

Calculate $εa$ as shown below,

$εa=|0.1196901−0.121659730.1196901|×100=1.6456%$

The true percent relative errors can be calculated as,

$εt=|0.12−0.11969010.12|×100=0.25825%$

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:

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

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

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

Answer 20

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:

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

Answer 21

Explanation of Solution

Given Information:

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

Formula used:

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

$yi+10=yim+h24(55fim−59fi−1m+37fi−2m−9fi−3m)$

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

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$

(3) Error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

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, $y−3=0.48$ . Substitute this value in equation (1)

Thus,

$f−3=−2×0.482.5=−0.384$

At $t=3$ , $y(3)=0.333333$ . So, $y−2=0.333333$ . Substitute this value in equation (1)

Thus,

$f−2=−2×0.3333333=−0.222222$

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

Thus,

$f−1=−2×0.2448983.5=−0.1399417$

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,

$yi+10=yim+h24(55fim−59fi−1m+37fi−2m−9fi−3m)$ .. .. .. (2)

The fourth order Adams formula as the corrector is,

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$ .. .. .. (3)

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

$y10=y00+h24(55f0−59f−1+37f−2−9f−3)$

Substitute values $f0=−0.09375$ , $f−1=−0.1399417$ , $f−2=−0.222222$ , $f−3=−0.384$ and $h=0.5$ .

$y10=0.1875+0.524×(55(−0.09375)−59(−0.1399417)+37(−0.222222)−9(−0.384))=0.1527937$

Hence, at $t=4.5, y10=0.1527937$

$f1=−2×0.15279374.5=−0.0679083$

Thus, $f1=−0.0679083$ .

Now, use corrector,

Put $i=0 , j=1$ in equation (3),

$y11=y00+h24(9f1+19f0−5f−1+f−2)$

Substitute values $f0=−0.09375$ , $f−1=−0.1399417$ , $f−2=−0.222222$ , $f1=−0.0679083$ and $h=0.5$ .

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

At $t=4.5 and y11=0.14760545$

$f1=−2×0.147605454.5=−0.0656024$

Thus, $f1=−0.0656024$ .

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

$y12=0.1875+0.524(9(−0.0656024)+19(−0.09375)−5(−0.1399417)+(−0.222222))=0.14803781$

Again,

At $t=4.5 and y12=0.14803781$

$f1=−2×0.148037814.5=−0.0657946$

Thus, $f1=−0.0657946$ .

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

$y13=0.1875+0.524(9(−0.0657946)+19(−0.09375)−5(−0.1399417)+(−0.222222))=0.148001773$

At $t=4.5 and y13=0.148001773$

$f1=−2×0.1480017734.5=−0.0657786$

Thus, $f1=−0.0657786$ .

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

Substitute $i=1$ in equation (2),

$y20=y1+h24(55f1−59f0+37f−1−9f−2)$

Substitute values, $f1=−0.0657786$ $f0=−0.09375$ , $f−1=−0.1399417$ and $f−2=−0.222222$ in above equation, to get

$y20=0.148001773+0.524×(55(−0.0657786)−59(−0.09375)+37(−0.1399417)−9(−0.222222))=0.12165973$

At $t=5 and y20=0.12165973$ .

$f2=−2×0.121659735=−0.0486639$

Thus, $f2=−0.0486639$

Now error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

Calculate $εa$ as shown below,

$εa=|0.14760545−0.15279370.14760545|×100=3.5149%$

The true percent relative errors can be calculated as,

$εt=|0.148148−0.147605450.148148|×100=0.36622%$

Corrector formula is given as,

$yi+1j=yim+h24(9fi+1j−1+19fim−5fi−1m+fi−2m)$

Substitute $i=1 and j=1$ in above formula,

$y21=y1+h24(9f20+19f1−5f0+f−1)$

Substitute values, $f1=−0.0657786$ $f0=−0.09375$ , $f−1=−0.1399417$ and $f2=−0.0486639$ in above equation, to get

$y21=0.148001773+0.524(9(−0.0486639)+19(−0.0657786)−5(−0.09375)+(−0.1399417))=0.1196901$

At $t=5 and y21=0.1196901$

$f2=−2×0.11969015=−0.0478760$

Thus, $f2=−0.0478760$ .

Hence,

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

At $t=5 and y22=0.11983783$

$f2=−2×0.119837835=−0.0479351$

Thus, $f2=−0.0479351$ .

Hence,

$y23=0.148001773+0.524(9(−0.0479351)+19(−0.0657786)−5(−0.09375)+(−0.1399417))=0.11982675$

Now error is given as,

$εa=|yi+1j−yi+1j−1yi+1j|×100$

Calculate $εa$ as shown below,

$εa=|0.1196901−0.121659730.1196901|×100=1.6456%$

The true percent relative errors can be calculated as,

$εt=|0.12−0.11969010.12|×100=0.25825%$

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:

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

Answer 22

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