Question

In Problems 3 and 4 repeat the indicated problem using the RK4 method. First use h=0.2 and then use h=0.1 .

Approximatey(1.2) , wherey(x) is the solution of the given initial-value problem.

.${\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}\text{'}}}\mathbf{-}\mathbf{2}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{4}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{9}$

Find the analytic solution of the problem and compare the actual value of y (1.2) with y₂.

Short answer

The solution of the problem calculated using RK4 method is and the approximated value for y(0.2) is y₂=6.00 when h=0.1 and 6.00 when h=0.2.

Step-by-step solution

Define RK4 Method

To cancel out lower-order error terms, the RK4 technique of integrating differential equations uses a trial step at the middle of an interval. When paired with a clever adaptive step-size procedure, this method is relatively simple and resilient, making it an excellent general option for solving differential equations. The formula used in this method is,

$y_{n+1}=y_n+\frac{1}{6}{k}_1+\frac{1}{3}{k}_2+\frac{1}{3}{k}_3+\frac{1}{6}{k}_4+O\left(h^5\right)$

Find the first order differential equations.

We have the second order of differential equation,

$x^2{y}^{\text{'}\text{'}}-2x{y}^{\text{'}}+2y=0\text{(1)}$

with initial conditions

$y\left(1\right)=4,y^{\text{'}}\left(1\right)=9$

and we have to approximate $y\left(1.2\right)$using RK4 method as the following technique.

First we have to assume that,

$y^{\text{'}}=u\text{(a)}$

Then differentiate with respect to we have the following

$y^{\text{'}\text{'}}=u^{\text{'}}\text{(b)}$

Substitute (a),(b) in equation (1).

$x^2{u}^{\text{'}}-2xu+2y=0\text{(c)}$

Then from (a), (c) we can have the system of first order differential equations.

$\begin{array}{c}{y}^{\text{'}}=f(x,u,y)=u\\ u^{\text{'}}=g(x,u,y)=\frac{2}{x}u-\frac{2}{x^2}y\end{array}$

Hence the first order differential equations are

$u^{\text{'}}=g(x,u,y)=\frac{2}{x}u-\frac{2}{x^2}y,y^{\text{'}}=f(x,u,y)=u$

Find the value of y1,u1 for h=0.2

Using RK4 method using the step $h=0.2$along with the conditions

$\left(x_0,y_0\right)=(1,4)\text{,}\left(x_0,u_0\right)=(1,9)$we can have,

Find the approximate values using the formula,

$\begin{array}{c}{y}_{n+1}=y_n+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\text{(1a)}\\ u_{n+1}=u_n+\frac{h}{6}\left(m_1+2m_2+2m_3+m_4\right)\text{(2a)}\end{array}$

We have to find the constants $\left(k_1,k_2,k_3\text{,}{k}_4\right)$and $\left(m_1,m_2,m_3\text{,}{m}_4\right)$at $n=0\text{,}h=0.2$.

Since we have $u_0=9\text{,}{y}_0=4$we get,

$\begin{array}{c}{k}_1=f\left(x_0,u_0,y_0\right)\\ =u_0\\ =9\\ m_1=g\left(x_0,u_0,y_0\right)\\ =\frac{2}{x_0}{u}_0-\frac{2}{x^2}{y}_0\\ =\frac{2}{1}\times 9-\frac{2}{1^2}\times 4\\ =18-8\\ =10\end{array}$

$\begin{array}{c}{k}_2=f\left(x_0+\frac{1}{2}h,u_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)\\ =u_0+\frac{1}{2}h{m}_1\\ =9+\frac{1}{2}\times 0.2\times 10\\ =9+1\\ =10\end{array}$

$\begin{array}{c}{m}_2=g\left(x_0+\frac{1}{2}h,u_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)\\ =\frac{2}{x_0+\frac{1}{2}h}\left(u_0+\frac{1}{2}h{m}_1\right)-\frac{2}{{\left(x_0+\frac{1}{2}h\right)}^2}\left(y_0+\frac{1}{2}h{k}_1\right)\\ =\frac{2}{1+\frac{1}{2}\times 0.2}\left(9+\frac{1}{2}\times 0.2\times 10\right)-\frac{2}{{\left(1+\frac{1}{2}\times 0.2\right)}^2}\left(4+\frac{1}{2}\times 0.2\times 9\right)\\ =\frac{2}{1.1}\left(10\right)-\frac{2}{{\left(1.1\right)}^2}\left(4.9\right)\\ =18.18182-8.09917\\ =10.0826\end{array}$

$\begin{array}{c}{k}_3=f\left(x_0+\frac{1}{2}h,u_0+\frac{1}{2}h{m}_2,y_0+\frac{1}{2}h{k}_2\right)\\ =u_0+\frac{1}{2}h{m}_2\\ =9+\frac{1}{2}\times 0.2\times 10.0826\\ =9+1.00826\\ =10.00826\end{array}$

$\begin{array}{c}{m}_3=g\left(x_0+\frac{1}{2}h,u_0+\frac{1}{2}h{m}_2,y_0+\frac{1}{2}h{k}_2\right)\\ =\frac{2}{x_0+\frac{1}{2}h}\left(u_0+\frac{1}{2}h{m}_2\right)-\frac{2}{{\left(x_0+\frac{1}{2}h\right)}^2}\left(y_0+\frac{1}{2}h{k}_2\right)\\ =\frac{2}{1+\frac{1}{2}\times 0.2}\left(9+\frac{1}{2}\times 0.2\times 10.0826\right)-\frac{2}{{\left(1+\frac{1}{2}\times 0.2\right)}^2}\left(4+\frac{1}{2}\times 0.2\times 10\right)\\ =\frac{2}{1.1}\left(10.00826\right)-\frac{2}{{\left(1.1\right)}^2}\left(5\right)\\ =18.19684-8.26446\\ =9.9324\end{array}$

$\begin{array}{c}{k}_4=f\left(x_0+h,u_0+h{m}_3,y_0+h{k}_3\right)\\ =u_0+h{m}_3\\ =9+0.2\times 9.9324\\ =9+1.9865\\ =10.9865\end{array}$

$\begin{array}{c}{m}_4=g\left(x_0+h,u_0+h{m}_3,y_0+h{k}_3\right)\\ =\frac{2}{x_0+h}\left(u_0+h{m}_3\right)-\frac{2}{{\left(x_0+h\right)}^2}\left(y_0+h{k}_3\right)\\ =\frac{2}{1+0.2}(9+0.2\times 9.9324)-\frac{2}{{(1+0.2)}^2}(4+0.2\times 10.00826)\\ =\frac{2}{1.2}\left(10.9865\right)-\frac{2}{{\left(1.2\right)}^2}\left(6.001652\right)\\ =18.3108-10.00275\\ =8.308\end{array}$

After that when n=0 , substitute the value of constants

$\begin{array}{c}{y}_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ =4+\frac{0.2}{6}[9+2(10)+2(10.00826)+10.9865]\\ =4+\frac{0.2}{6}\times \left(60.00112\right)\\ =4+2.00\\ =6\end{array}$

and

$\begin{array}{c}{u}_1=u_0+\frac{h}{6}\left(m_1+2m_2+2m_3+m_4\right)\\ =9+\frac{0.2}{6}[10+2(10.0826)+2(9.9324)+8.308]\\ =9+\frac{0.2}{6}\times \left(58.338\right)\\ =9+1.9446\\ =10.9446\end{array}$

Find the values of y2  for h=0.1

$\begin{array}{c}{k}_1=f\left(x_0,u_0,y_0\right)\\ =u_0\\ =9\\ m_1=g\left(x_0,u_0,y_0\right)\\ =\frac{2}{x_0}{u}_0-\frac{2}{x^2}{y}_0\\ =\frac{2}{1}\times 9-\frac{2}{1^2}\times 4\\ =18-8\\ =10\end{array}$

$\begin{array}{c}{k}_2=f\left(x_0+\frac{1}{2}h,u_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)\\ =u_0+\frac{1}{2}h{m}_1\\ =9+\frac{1}{2}\times 0.1\times 10\\ =9+0.5\\ =9.5\end{array}$

$\begin{array}{c}{m}_2=g\left(x_0+\frac{1}{2}h,u_0+\frac{1}{2}h{m}_1,y_0+\frac{1}{2}h{k}_1\right)\\ =\frac{2}{x_0+\frac{1}{2}h}\left(u_0+\frac{1}{2}h{m}_1\right)-\frac{2}{{\left(x_0+\frac{1}{2}h\right)}^2}\left(y_0+\frac{1}{2}h{k}_1\right)\\ =\frac{2}{1+\frac{1}{2}\times 0.1}\left(9+\frac{1}{2}\times 0.1\times 10\right)-\frac{2}{{\left(1+\frac{1}{2}\times 0.1\right)}^2}\left(4+\frac{1}{2}\times 0.1\times 9\right)\\ =\frac{2}{1.05}\left(9.5\right)-\frac{2}{{\left(1.05\right)}^2}\left(4.45\right)\\ =18.09524-8.07256\\ =10.02268\end{array}$

$\begin{array}{c}{k}_3=f\left(x_0+\frac{1}{2}h,u_0+\frac{1}{2}h{m}_2,y_0+\frac{1}{2}h{k}_2\right)\\ =u_0+\frac{1}{2}h{m}_2\\ =9+\frac{1}{2}\times 0.1\times 10.02268\\ =9+0.71\\ =9.50113\end{array}$

$\begin{array}{c}{m}_3=g\left(x_0+\frac{1}{2}h,u_0+\frac{1}{2}h{m}_2,y_0+\frac{1}{2}h{k}_2\right)\\ =\frac{2}{x_0+\frac{1}{2}h}\left(u_0+\frac{1}{2}h{m}_2\right)-\frac{2}{{\left(x_0+\frac{1}{2}h\right)}^2}\left(y_0+\frac{1}{2}h{k}_2\right)\\ =\frac{2}{1+\frac{1}{2}\times 0.1}\left(9+\frac{1}{2}\times 0.1\times 10.02268\right)-\frac{2}{{\left(1+\frac{1}{2}\times 0.1\right)}^2}\left(4+\frac{1}{2}\times 0.1\times 9.5\right)\\ =\frac{2}{1.05}\left(9.50113\right)-\frac{2}{{\left(1.05\right)}^2}\left(4.475\right)\\ =18.09740-8.11791\\ =9.97949\end{array}$

$\begin{array}{c}{k}_4=f\left(x_0+h,u_0+h{m}_2,y_0+h{k}_2\right)\\ =u_0+h{m}_3\\ =9+0.1\times 9.97749\\ =9+0.99775\\ =9.99775\end{array}$

$\begin{array}{c}{m}_4=g\left(x_0+h,u_0+h{m}_3,y_0+h{k}_3\right)\\ =\frac{2}{x_0+h}\left(u_0+h{m}_3\right)-\frac{2}{{\left(x_0+h\right)}^2}\left(y_0+h{k}_3\right)\\ =\frac{2}{1+0.1}(9+0.1\times 9.99775)-\frac{2}{{(1+0.1)}^2}(4+0.1\times 9.97949)\\ =\frac{2}{1.1}\left(9.99977\right)-\frac{2}{{\left(1.1\right)}^2}\left(4.99795\right)\\ =18.18141-8.26107\\ =9.92034\end{array}$

$\begin{array}{c}{y}_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ =4+\frac{0.1}{6}[9+2(9.5)+2(9.50113)+9.99775]\\ =4+\frac{0.1}{6}\times \left(57.00\right)\\ =4+0.95\\ =4.95\end{array}$

$\begin{array}{c}{u}_1=u_0+\frac{h}{6}\left(m_1+2m_2+2m_3+m_4\right)\\ =9+\frac{0.1}{6}[10+2(10.02268)+2(9.97949)+9.92034]\\ =9+\frac{0.1}{6}\times \left(59.9247\right)\\ =9+0.9987\\ =9.9987\end{array}$

After that we can find the value of x and u using other constants as shown in the table below. Then we get the value of y₂=6.00.