Question

In Problems \(3\) and \(4\)use the Adams-Bashforth-Moulton method to approximate \(y(0.8)\), where \(y(x)\) is the solution of the given initial-value problem. Use \(h = 0.2\)and the \(RK4\) method to compute \({y_1},{y_2}\)and \({y_3}\).

\(y' = 4x - 2y,y(0) = 2.\)

Short answer

The value of\(y(0.8)\) calculated using Adams-Bashforth-Moulton method is \(1.2065.\)

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,
• \({y_{n + 1,p}} = {y_n} + \frac{h}{{24}}\left( {55y_n^\cent - 59y_{n - 1}^\cent + 37y_{n - 2}^\cent - 9y_{n - 3}^\cent} \right)\).

Find the value of the constants.

We have the initial-value problem,

\(y' = 4x - 2y,\quad y(0) = 2\)

and we have to approximate the value \(y(0.8)\) with \(h = 0.2\)using the following technique.

First we have to obtain three approximate values \({y_1},{y_2},{y_3}\)using the formula

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

Before that we have to find the constants \({k_1},{k_2},{k_3},{k_4}\)at \(n = 0\)and \(h = 0.2\)as the following.

Since we have \({x_0} = 0,{y_0} = 2\), then we have

\(\begin{aligned}{*{20}{c}}{{k_1} = f\left( {{x_0},{y_0}} \right)}\\{ = 4{x_0} - 2{y_0}}\\{ = 4 \times 0 - 2 \times 2}\\{ = 0 - 4}\\{ = - 4}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{{k_2} = f\left( {{x_0} + \frac{1}{2}h,{y_0} + \frac{1}{2}h{k_1}} \right)}\\{ = 4\left( {{x_0} + \frac{1}{2}h} \right) - 2\left( {{y_0} + \frac{1}{2}h{k_1}} \right)}\\{ = 4\left( {0 + \frac{1}{2} \times 0.2} \right) - 2\left( {2 + \frac{1}{2} \times 0.2 \times ( - 4)} \right)}\\{ = 0.4 - 3.2}\\{ = - 2.8}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{{k_3} = f\left( {{x_0} + \frac{1}{2}h,{y_0} + \frac{1}{2}h{k_2}} \right)}\\{ = 4\left( {{x_0} + \frac{1}{2}h} \right) - 2\left( {{y_0} + \frac{1}{2}h{k_2}} \right)}\\{ = 4\left( {0 + \frac{1}{2} \times 0.2} \right) - 2\left( {2 + \frac{1}{2} \times 0.2 \times ( - 2.8)} \right)}\\{ = 0.4 - 3.44}\\{ = - 3.04}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{{k_4} = f\left( {{x_0} + h,{y_0} + h{k_3}} \right)}\\{ = 4\left( {{x_0} + h} \right) - 2\left( {{y_0} + h{k_3}} \right)}\\{ = 4(0 + 0.2) - 2(2 + 0.2 \times ( - 3.4))}\\{ = 0.8 - 2.64}\\{ = - 1.84}\end{aligned}\)

Hence the values of the constants are \({k_1} = - 4,{k_2} = - 2.8,{k_3} = - 3.04,{k_4} = - 1.84\).

Find the value of\({y_1}\).

We need to find the value of \({y_1}\)by using \((1)\).

\(\begin{aligned}{*{20}{c}}{{y_1} = {y_0} + \frac{h}{6}\left( {{k_1} + 2{k_2} + 2{k_3} + {k_4}} \right)}\\{ = 2 + \frac{{0.2}}{6}[( - 4) + 2( - 2.8) + 2( - 3.04) + ( - 1.84)]}\\{ = 2 + \frac{{0.2}}{6} \times ( - 17.52)}\\{ = 1.416}\end{aligned}\)

After this we can obtain the value of \(y\)for each value of \(x\)until \(x = 0.6\)and \({y_3} = 1.1061\)using the below table.

\(h = 0.2\)
\({x_n}\)	\({k_1} + 2{k_2} + 2{k_3} + {k_4}\)	\({y_n}\)
\(0.00\)	\( - 17.520\)	\(2.00\)
\(0.20\)	\( - 7.9342\)	\(1.4160\)
\(0.4\)	\( - 1.3639\)	\(1.1515\)
\(0.6\)		\(1.1061\)

Hence the value of \({y_1} = 1.416\).

Find the value of \(y(0.4)\)

We have to find the value of \(y(0.4)\)using Adams-Bashforth-Moulton method. Since we have, \({x_0} = 0,{x_1} = 0.2,{x_2} = 0.4{\rm{,\;}}{x_3} = 0.6\)and the value of \(y\)from the table we can find,

\(\begin{aligned}{*{20}{c}}{y_0^\cent = f\left( {{x_0},{y_0}} \right)}\\{ = 4{x_0} - 2{y_0} + 1}\\{ = 0 - 2 \times 2}\\{ = - 4}\\{y_1^\cent = f\left( {{x_0},{y_0}} \right)}\\{ = 4{x_1} - 2{y_1}}\\{ = 4 \times 0.2 - 2 \times 1.416}\\{ = - 2.032}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{y_2^\cent = f\left( {{x_0},{y_0}} \right)}\\{ = 2{x_2} - 3{y_2} + 1}\\{ = 4 \times 0.4 - 2 \times 1.1515}\\{ = - 0.7030528}\\{y_3^\cent = f\left( {{x_0},{y_0}} \right)}\\{ = 4{x_3} - 2{y_3}}\\{ = 4 \times 0.6 - 2 \times 1.1061}\\{ = 0.1878734}\end{aligned}\)

After these we can get the predictor \({y_4}\)using,

\(\begin{aligned}{*{20}{c}}{y_{n + 1}^{} = y_4^{} = {y_n} + \frac{h}{{24}}\left( {55y_n^\prime - 59y_{n - 1}^\prime + 37y_{n - 2}^\prime - 9y_{n - 3}^\prime } \right)}\\{ = {y_3} + \frac{h}{{24}}\left( {55y_3^\prime - 59y_2^\prime + 37y_1^\prime - 9y_0^\prime } \right)}\\{ = 1.1061 + \frac{{0.2}}{{24}}[55 \times 0.1878734 - 59 \times ( - 0.7030528) + 37 \times ( - 2.032) - 9 \times ( - 4)]}\\{ = 1.1061 + \frac{{0.2}}{{24}} \times (12.6292)}\\{ = 1.2113}\end{aligned}\)

where \(n = 3\).

Hence the value of the predictor \({y_4} = 1.2113\).

Find the value of \(y(0.8).\)

After that using the predictor and \({x_4} = 0.8\)we can find,

\(\begin{aligned}{*{20}{c}}{y_4^\prime = f\left( {{x_4},y_4^{}} \right)}\\{ = 4{x_4} - 23y_4^{}}\\{ = 4 \times 0.8 - 2 \times 1.2113}\\{ = 0.7774}\end{aligned}\)

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

\(\begin{aligned}{*{20}{c}}{{y_{n + 1}} = {y_4} = {y_n} + \frac{h}{{24}}\left( {9y_{n + 1}^\cent + 19y_n^\cent - 5y_{n - 1}^\cent + y_{n - 2}^\cent} \right)}\\{ = {y_3} + \frac{h}{{24}}\left( {9y_4^\cent + 19y_3^\cent - 5y_2^\cent + y_1^\cent} \right)}\\{ = 1.1061 + \frac{{0.2}}{{24}}[9 \times 0.7774 + 19 \times 0.1878734 - 5 \times ( - 0.7030528) + ( - 2.032)]}\\{ = 1.1061 + \frac{{0.2}}{{24}}(12.0496)}\\{ = 1.2065}\end{aligned}\)

where \(n = 3\).

Hence the value of \(y(0.8)\)is \(1.2065.\)