Question

Question: In Problems $\mathbf{5}\mathbf{-}\mathbf{8}$use the Adams-Bashforth-Moultonmethod to approximate $\mathbf{y}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{)}$, where$\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$is the solution of the given initial-value problem. First use$\mathbf{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}$and $\mathbf{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}$then use.Use the${\mathbf{RK}}_{\mathbf{4}}$method to compute${\mathbf{y}}_{\mathbf{1}}\mathbf{,}{\mathbf{y}}_{\mathbf{2}}$and${\mathit{y}}_{\mathbf{3}}$.

${\mathbf{y}}^{\mathbf{I}}\mathbf{=}\mathbf{xy}\mathbf{+}\sqrt{\mathbf{y}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{y}\left(0\right)\mathbf{=}\mathbf{1}$

Short answer

The value of $y(1.0)$calculated using Adams-Bashforth-Moulton method with$\mathrm{h}=0.2$ is$y_5=3.5523$

The value of $y(1.0)$calculated using Adams-Bashforth-Moulton method with$\mathrm{h}=0.1$ is$y_{10}=3.5196$

Step-by-step solution

Define Adams-Bashforth-Moulton Method

• The Adams–Bashforth-Moulton methods allow us to calculate the estimated solution at a given instant from prior instants' solutions explicitly.
• Each phase of the Adams–Moulton approach produces an algebraic matrix Riccati equation (AMRE), which is then solved using Newton's method.
• The predictive formula for this method is,
• $\begin{array}{rcl}y{*}_{n+1}& =& y{*}_4=y_n+\frac{h}{24}\left(9y_{n+1}^{\text{'}}+19y_n^{\text{'}}-5y_{n-1}^{\text{'}}-y_{n-2}^{\text{'}}\right)\end{array}$
  

Find the value of the constants with h=0.2.

We have the initial-value problem,

$y^I=xy+\sqrt{y},y\left(0\right)=1$

and we have to approximate the value$y\left(1.0\right)$with $h=0.2$using the following technique.

First we have to obtain three approximate values using the formula

$y_{n+1}=y_n+\frac{h}{6}\left(k_1+2k_2+2k_3+2k_4\right)\left(1\right)$

Before that we have to find the constants $\left(k_{1,}{k}_{2,}{k}_3,k_4\right)$at $n=0$and $h=0.2$as the following.

Since we have $x_0=0,y_0=1$, then we have

localid="1667993256587" $\begin{array}{rcl}{k}_1& =& f\left(x_0,y_0\right)\\ & =& x_0{y}_0+\sqrt{y_0}\\ & =& 0\times 1+\sqrt{1}\\ & =& 1\\ k_2& =& f\left(x_0+\frac{1}{2}h,y_0+\frac{1}{2}h{k}_1\right)\\ & =& \left(x_0+\frac{1}{2}h\right)\left(y_0+\frac{1}{2}h{k}_1\right)+\sqrt{y_0+\frac{1}{2}h{k}_1}\\ & =& \left(0+\frac{1}{2}\times 0.2\right)\left(0+\frac{1}{2}\times 0.2\times 1\right)+\sqrt{0+\frac{1}{2}\times 0.2\times 1}\\ & =& 0.1\times 1.1+\sqrt{1.1}\\ & =& 1.158809\end{array}$

$\begin{array}{rcl}{k}_3& =& f\left(x_0+\frac{1}{2}h,y_0+\frac{1}{2}h{k}_2\right)\\ & =& \left(x_0+\frac{1}{2}h\right)\left(y_0+\frac{1}{2}h{k}_2\right)+\sqrt{y_0+\frac{1}{2}h{k}_2}\\ & =& \left(0+\frac{1}{2}\times 0.2\right)\left(0+\frac{1}{2}\times 0.2\times 1.158809\right)+\sqrt{0+\frac{1}{2}\times 0.2\times 1.158809}\\ & =& 0.1\times 1.158809+\sqrt{1.158809}\\ & =& 1.167941\\ k_4& =& f\left(x_0+h,y_0+h{k}_3\right)\\ & =& \left(x_0+h\right)\left(y_0+h{k}_3\right)+\sqrt{y_0+h{k}_3}\\ & =& \left(0+0.2\right)\left(0+0.2\times 1.167941\right)+\sqrt{0+0.2\times 1.167941}\\ & =& 0.1\times 1.233588+\sqrt{1.233588}\\ & =& 1.234029\end{array}$

Hence the values of the constants with $h=0.2$are

$k_1=1,k_2=1.158809,k_3=1.167941,k_4=1.234029$

Find the value of y1with h =0.2

We need to find the value of$y_1$by using $\left(1\right)$.

localid="1667994739282" $\begin{array}{rcl}{y}_1& =& y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 1+\frac{0.2}{6}\left(1+2\times 1.158809+2\times 1.167941+1.234029\right)\\ & =& 1+\frac{0.2}{6}\times 7.647709\\ & =& 1.254924\end{array}$

After this we can obtain the value of for each value of until $x=0.6$and $y_3=2.013419$using the below table.

	$h=0.2$
$x_n$	$k_1+2k_2+2k_3+k_4$	$y_n$
0.00	7.647709	0.00
0.20	9.380783	1.254924
0.40	13.374067	1.567617
0.60		2.01349

Hence the value of$y_1$with $h=0.2$is$1.254924$.

Find the value of the predictor y4

We have to find the value of $y\left(1.0\right)$using Adams-Bashforth-Moulton method. Since we have,localid="1667995253239" $\left(x_0=0,x_1=0.2,x_2=0.4,x_3=0.6\right)$and the value of from the table we can find,

$\begin{array}{rcl}y{\text{'}}_0& =& f\left(x_0,y_0\right)\\ & =& x_0{y}_0+\sqrt{y_0}\\ & =& 0\times 1+\sqrt{1}\\ & =& 1\\ y{\text{'}}_1& =& f\left(x_1,y_1\right)\\ & =& x_1{y}_1+\sqrt{y_1}\\ & =& 0.2\times 1.254924+\sqrt{1.254924}\\ & =& 1.371219\\ & & \end{array}$

$\begin{array}{rcl}y{\text{'}}_2& =& f\left(x_2,y_2\right)\\ & =& x_2{y}_3+\sqrt{y_2}\\ & =& 0.4\times 1.567617+\sqrt{1.567617}\\ & =& 1.879092\\ y{\text{'}}_3& =& f\left(x_3,y_3\right)\\ & =& x_3{y}_3+\sqrt{y_3}\\ & =& 0.6\times 2.013419+\sqrt{2.013419}\\ & =& 2.627\end{array}$

After these we can get the predictor$y_4*$using,

$\begin{array}{rcl}y{*}_{n+1}& =& y{*}_4=y_n+\frac{h}{24}\left(55y_n^{\text{'}}-59y_{n-1}^{\text{'}}+37y_{n-2}^{\text{'}}-9y_{n-3}^{\text{'}}\right)\\ & =& y_3+\frac{h}{24}\left(55y_3^{\text{'}}-59y_2^{\text{'}}+37y_1^{\text{'}}-9y_0^{\text{'}}\right)\\ & =& 2.031349+\frac{0.2}{24}\left(55\times 2.627-59\times 1.879092+37\times 1.37219-9\times 1\right)\\ & =& 2.031349+\frac{0.2}{24}\times 77.12375\\ & =& 2.656117\end{array}$

where $n=3$.

Hence the value of the predictor$y_4*$is $2.656117$.

Find the value of y(1.0)

After that using the predictor and $x_4=0.8$we can find,

localid="1667999767819" $\begin{array}{rcl}y{\text{'}}_4& =& f\left(x_4,y_4\right)\\ & =& x_4{y}_4+\sqrt{y_4}\\ & =& 0.6\times 2.656117+\sqrt{2.656117}\\ & =& 3.754563\end{array}$

Now we can find the approximate value of $y\left(0.8\right)$as

$\begin{array}{rcl}y{*}_{n+1}& =& y{*}_4=y_n+\frac{h}{24}\left(9y_{n+1}^{\text{'}}+19y_n^{\text{'}}-5y_{n-1}^{\text{'}}-y_{n-2}^{\text{'}}\right)\\ & =& y_3+\frac{h}{24}\left(9y_4^{\text{'}}+19y_3^{\text{'}}-y_2^{\text{'}}+y_1^{\text{'}}\right)\\ & =& 2.013419+\frac{0.2}{24}\left(9\times 375463+19\times 2.627-5\times 1.87092+1.37219\right)\\ & =& 2.013419+\frac{0.2}{24}\times 75.680639\\ & =& 2.644091\end{array}$

where $n=3$.

After that we can obtain the value of which equals $3.5523$as shown in the table below.

	$\mathrm{h}=0.2$
$x_n$	$y_{n+1}^{*}$	$y_{n+1}^{}$
0.80	2.656117	2.644091
0.10	3.54911	3.55231

Hence the value of $y(1.0)$is $3.5523$.

Find the value of the constants with h=0.1.

We have the initial-value problem,

$y^I=xy+\sqrt{y},y\left(0\right)=1$

and we have to approximate the value $y\left(1.0\right)$with$h=0.1$using the following technique.

First we have to obtain three approximate values using the formula

$y_{n+1}=y_n+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)$

Before that we have to find the constants $\left(k_{1,}{k}_{2,}{k}_3,k_4\right)$at $n=0$and $h=0$as the following.

Since we have $x_0=0,y_0=1$, then we have

$\begin{array}{rcl}{k}_1& =& f\left(x_0,y_0\right)\\ & =& x_0{y}_0+\sqrt{y_0}\\ & =& 0\times 1+\sqrt{1}\\ & =& 1\\ k_2& =& f\left(x_0+\frac{1}{2}h,y_0+\frac{1}{2}h{k}_1\right)\\ & =& \left(x_0+\frac{1}{2}h\right)\left(y_0+\frac{1}{2}h{k}_1\right)+\sqrt{y_0+\frac{1}{2}h{k}_1}\\ & =& \left(0+\frac{1}{2}\times 0.1\right)\left(0+\frac{1}{2}\times 0.1\times 1\right)+\sqrt{0+\frac{1}{2}\times 0.1\times 1}\\ & =& 0.05\times 1.05+\sqrt{1.05}\\ & =& 1.077195\end{array}$

role="math" localid="1667998249723" $\begin{array}{rcl}{k}_3& =& f\left(x_0+\frac{1}{2}h,y_0+\frac{1}{2}h{k}_2\right)\\ & =& \left(x_0+\frac{1}{2}h\right)\left(y_0+\frac{1}{2}h{k}_2\right)+\sqrt{y_0+\frac{1}{2}h{k}_2}\\ & =& \left(0+\frac{1}{2}\times 0.1\right)\left(0+\frac{1}{2}\times 0.1\times 1.007195\right)+\sqrt{0+\frac{1}{2}\times 0.1\times 1.007915}\\ & =& 0.1\times 1.053860+\sqrt{1.053860}\\ & =& 1.079720\\ k_4& =& f\left(x_0+h,y_0+h{k}_3\right)\\ & =& \left(x_0+h\right)\left(y_0+h{k}_3\right)+\sqrt{y_0+h{k}_3}\\ & =& \left(0+0.1\right)\left(0+0.1\times 1.079720\right)+\sqrt{0+0.1\times 1.079720}\\ & =& 0.1\times 1.107927+\sqrt{1.107927}\\ & =& 1.107977\end{array}$

Hence the values of the constants are$k_1=1,k_2=1.077915,k_3=1.079270,k_4=1.107977$ .

Find the value of y1.

We need to find the value of by using $y_1$.

$\begin{array}{rcl}{y}_1& =& y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 1+\frac{0.1}{6}\left(1+2\times 1.077915+2\times 1.079270+1.107977\right)\\ & =& 1+\frac{0.1}{6}\times 6.4209\\ & =& 1.107015\end{array}$

After this we can obtain the value of for each value of with another four constants as shown in the table below

	$h=0.1$
$x_n$	$k_1+2k_2+2k_3+k_4$	$y_n$
0.00	6.420907	0.00
0.10	7.542059	1.107015
0.20	8.815014	1.123716
0.30		1.379633

Hence the value of$y\left(1\right)$ is$1.107015$

Find the value of the predictor y4*

We have to find the value of using Adams-Bashforth-Moulton method. Since we have, $\left(x_0=0,x_1=0.1,x_2=0.2,x_3=0.3\right)$and the value of from the table we can find,

$\begin{array}{rcl}y{\text{'}}_0& =& f\left(x_0,y_0\right)\\ & =& x_0{y}_0+\sqrt{y_0}\\ & =& 0\times 1+\sqrt{1}\\ & =& 1\\ y{\text{'}}_1& =& f\left(x_1,y_1\right)\\ & =& x_1{y}_1+\sqrt{y_1}\\ & =& 0.1\times 1.107015+\sqrt{1.107015}\\ & =& 1.162849\end{array}$

$\begin{array}{rcl}y{\text{'}}_2& =& f\left(x_2,y_2\right)\\ & =& x_2{y}_2+\sqrt{y_2}\\ & =& 0.2\times 1.232716+\sqrt{1.232716}\\ & =& 1.356821\\ y{\text{'}}_3& =& f\left(x_3,y_3\right)\\ & =& x_3{y}_3+\sqrt{y_3}\\ & =& 0.3\times 1.379633+\sqrt{1.379633}\\ & =& 1.588648\end{array}$

After these we can get the $y_4^{*}$predictor using,

$\begin{array}{rcl}y{*}_{n+1}& =& y{*}_4=y_n+\frac{h}{24}\left(55y_n^{\text{'}}-59y_{n-1}^{\text{'}}+37y_{n-2}^{\text{'}}-9y_{n-3}^{\text{'}}\right)\\ & =& y_3+\frac{h}{24}\left(55y_3^{\text{'}}-59y_2^{\text{'}}+37y_1^{\text{'}}-9y_0^{\text{'}}\right)\\ & =& 1.379633+\frac{0.1}{24}\left(55\times 1.588468-59\times 1.356821+37\times 1.162849-9\times 1\right)\\ & =& 2.031349+\frac{0.1}{24}\times 41.338778\\ & =& 1.551878\end{array}$

where $n=3$.

Hence the value of the predictor$y_4^{*}$ is $1.551878$.

Find the value of y(1)

After that using the predictor and $x_4=0.4$we can find,

$\begin{array}{rcl}y{\text{'}}_4& =& f\left(x_4,y_4\right)\\ & =& x_{4}y{*}_4+\sqrt{y{*}_4}\\ & =& 0.6\times 1.551878+\sqrt{1.551878}\\ & =& 1.866495\end{array}$

Now we can find the approximate value of $y(1.0)$as

$\begin{array}{rcl}y{*}_{n+1}& =& y{*}_4=y_n+\frac{h}{24}\left(9y_{n+1}^{\text{'}}+19y_n^{\text{'}}-5y_{n-1}^{\text{'}}-y_{n-2}^{\text{'}}\right)\\ & =& y_3+\frac{h}{24}\left(9y_4^{\text{'}}+19y_3^{\text{'}}-y_2^{\text{'}}+y_1^{\text{'}}\right)\\ & =& 1.379663+\frac{0.1}{24}\left(9\times 1.866495+19\times 1.588468-5\times 1.356821+1.1628\right)\\ & =& 1,379663+\frac{0.2}{24}\times 41.358092\\ & =& 1.5551958\end{array}$

where $n=3$.

After that we can obtain the value of which equals as shown in the table below.

	$h=0.1$
$x_n$	$y{*}_{n+1}$	$y_{n+1}$
0.40	1.551878	1.551958
0.50	1.762143	1.755102
0.60	1.996385	1.95581
0.70	2.279135	2.280473
0.80	2.620727	2.620432
0.90	3.028231	3.028226
1.00	3.518999	3.51953

Hence the value $y\left(1.0\right)$of is$3.5196$ .