We are given the first order differential equation,
\(\begin{aligned}{*{20}{c}}{\frac{{d{i_1}}}{{dt}} = - 20{i_1} + 10{i_3} + 100}\\{\frac{{d{i_3}}}{{dt}} = 10{i_1} - 20{i_3}}\end{aligned}\)
Using \(RK4\)method using the step \(h = 0.1\)along with the conditions we can have the value of current.
Find the approximate values using the formula,
\(\begin{aligned}{*{20}{c}}{{i_{1(n + 1)}} = {i_{1(n)}} + \frac{h}{6}\left( {{k_1} + 2{k_2} + 2{k_3} + {k_4}} \right)}\\{{i_{3(n + 1)}} = {i_{3(n)}} + \frac{h}{6}\left( {{m_1} + 2{m_2} + 2{m_3} + {m_4}} \right)}\end{aligned}\)
We have to find the constants \(\left( {{k_1},{k_2},{k_3}{\rm{\;,}}{k_4}} \right){\rm{\;}}\)and \(\left( {{m_1},{m_2},{m_3}{\rm{,\;}}{m_4}} \right)\)at \(n = 0{\rm{\;,\;}}h = 0.2\).
Since we have \({i_{1(0)}} = 0{\rm{\;,\;}}{i_{3(0)}} = 0\)we get,
\(\begin{aligned}{*{20}{c}}{{k_1} = f\left( {{t_0},{i_{1(0)}},{i_{3(0)}}} \right)}\\{ = - 20{i_{1(0)}} + 10{i_{3(0)}} + 100}\\{ = - 20 \times 0 + 10 \times 0 + 100}\\{ = 100}\\{{m_1} = g\left( {{t_0},{i_{1(0)}},{i_{3(0)}}} \right)}\\{ = 10{i_1} - 20{i_3}}\\{ = 10 \times 0 - 20 \times 0}\\{ = 0}\end{aligned}\)
\(\begin{aligned}{*{20}{c}}{{k_2} = f\left( {{t_0} + \frac{1}{2}h,{i_{1(0)}} + \frac{1}{2}h{k_1},{i_{3(0)}} + \frac{1}{2}h{m_1}} \right)}\\{ = - 20\left( {{i_{1(0)}} + \frac{1}{2}h{k_1}} \right) + 10\left( {{i_{3(0)}} + \frac{1}{2}h{m_1}} \right) + 100}\\{ = - 20\left( {0 + \frac{1}{2} \times 0.1 \times 100} \right) + 10\left( {0 + \frac{1}{2} \times 0.1 \times 0} \right) + 100}\\{ = - 100 + 0 + 100}\\{ = 0}\end{aligned}\)
\(\begin{aligned}{*{20}{c}}{{m_2} = g\left( {{t_0} + \frac{1}{2}h,{i_{1(0)}} + \frac{1}{2}h{k_1},{i_{3(0)}} + \frac{1}{2}h{m_1}} \right)}\\{ = 10\left( {{i_{1(0)}} + \frac{1}{2}h{k_1}} \right) - 20\left( {{i_{3(0)}} + \frac{1}{2}h{m_1}} \right)}\\{ = 10\left( {0 + \frac{1}{2} \times 0.1 \times 100} \right) - 20\left( {0 + \frac{1}{2} \times 0.1 \times 0} \right)}\\{ = 50 - 0}\\{ = 50}\end{aligned}\)
\(\begin{aligned}{*{20}{c}}{{k_3} = f\left( {{t_0} + \frac{1}{2}h,{i_{1(0)}} + \frac{1}{2}h{k_2},{i_{3(0)}} + \frac{1}{2}h{m_2}} \right)}\\{ = - 20\left( {{i_{1(0)}} + \frac{1}{2}h{k_2}} \right) + 10\left( {{i_{3(0)}} + \frac{1}{2}h{m_2}} \right) + 100}\\{ = - 20\left( {0 + \frac{1}{2} \times 0.1 \times 0} \right) + 10\left( {0 + \frac{1}{2} \times 0.1 \times 50} \right) + 100}\\{ = 0 + 25 + 100}\\{ = 125}\end{aligned}\)
\(\begin{aligned}{*{20}{c}}{{m_3} = g\left( {{t_0} + \frac{1}{2}h,{i_{1(0)}} + \frac{1}{2}h{k_2},{i_{3(0)}} + \frac{1}{2}h{m_2}} \right)}\\{ = 10\left( {{i_{1(0)}} + \frac{1}{2}h{k_2}} \right) - 20\left( {{i_{3(0)}} + \frac{1}{2}h{m_2}} \right)}\\{ = 10\left( {0 + \frac{1}{2} \times 0.1 \times 0} \right) - 20\left( {0 + \frac{1}{2} \times 0.1 \times 50} \right)}\\{ = 0 - 50}\\{ = - 50}\end{aligned}\)
\(\begin{aligned}{*{20}{c}}{{k_4} = f\left( {{t_0} + h,{i_{1(0)}} + h{k_3},{i_{3(0)}} + h{m_3}} \right)}\\{ = - 20\left( {{i_{1(0)}} + h{k_3}} \right) + 10\left( {{i_{3(0)}} + h{m_3}} \right) + 100}\\{ = - 20(0 + 0.1 \times 125) + 10(0 + 0.1 \times ( - 50)) + 100}\\{ = - 250 - 50 + 100}\\{ = - 200}\end{aligned}\)
\(\begin{aligned}{*{20}{c}}{{m_4} = g\left( {{t_0} + h,{i_{1(0)}} + h{k_3},{i_{3(0)}} + h{m_3}} \right)}\\{ = 10\left( {{i_{1(0)}} + h{k_3}} \right) - 20\left( {{i_{3(0)}} + h{m_3}} \right)}\\{ = 10(0 + 0.1 \times 125) - 20(0 + 0.1 \times ( - 50))}\\{ = 125 + 100}\\{ = 225}\end{aligned}\)
After that when \(n = 0\), substitute the value of constants to get the value of \({i_{1(1)}}{\rm{,\;}}{i_{3(1)}}\).
\(\begin{aligned}{*{20}{c}}{{i_{1(1)}} = {i_{1(0)}} + \frac{h}{6}\left( {{k_1} + 2{k_2} + 2{k_3} + {k_4}} \right)}\\{ = 0 + \frac{{0.1}}{6}[100 + 2(0) + 2(125) + ( - 200)]}\\{ = 0 + \frac{{0.1}}{6} \times (150)}\\{ = 0 + 2.5}\\{ = 2.5}\end{aligned}\)\(\)
and
\(\begin{aligned}{*{20}{c}}{{i_{3(1)}} = {i_{3(0)}} + \frac{h}{6}\left( {{m_1} + 2{m_2} + 2{m_3} + {m_4}} \right)}\\{ = 0 + \frac{{0.1}}{6}[0 + 2(50) + 2( - 50) + 225]}\\{ = 0 + \frac{{0.1}}{6} \times (225)}\\{ = 0 + 3.75}\\{ = 3.75}\end{aligned}\)
\(RK4\)Method |
\(h = 0.1\) |
\({x_n}\) | \({k_1} + 2{k_2} + 2{k_3} + {k_4}\) | \({m_1} + 2{m_2} + 2{m_3} + {m_4}\) | \({y_n}\) | \({u_n}\) |
\(0.00\) | \(150.00\) | \(225.00\) | \(0.00\) | \(0.00\) |
\(0.10\) | \(18.750\) | \(121.88\) | \(2.500\) | \(3.750\) |
\(0.20\) | \( - 44.531\) | \(97.266\) | \(2.8125\) | \(5.7813\) |
\(0.30\) | \(59.501\) | \(106.27\) | \(2.0703\) | \(7.4023\) |
\(0.40\) | \(82.833\) | \(64.340\) | \(3.0620\) | \(9.1734\) |
\(0.50\) | | | \(4.4425\) | \(10.2457\) |
Hence the values of the current \({i_{1(5)}} = 4.4425{\rm{\;,\;}}{i_{3(5)}} = 10.2457\).
