Question

Consider the initial-value problem ${\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{3}}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}$. See Problem 12 in Exercises 9.1.

1. Compare the results obtained from using the RK4 method over the interval [1,1.4] with step sizes h = 0.1and h = 0.05.
2. Use a numerical solver to graph the solution of the initial value problem on the interval [1,1.4].

Short answer

1. The results obtained from using the RK4 method over the interval [1,1.4] with step sizes h = 0.1 and h = 0.05 is half of the found-out values.
2. The graph is plotted for the problem with the initial value on the interval [1,1.4].

Step-by-step solution

Runga-Kutta method:

The solution of a linear differential equation of the form y = f(x, y) is given as follows:

$$\begin{gathered} y_{n+1}=y_n+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\text{where} \\ {k}_1=f\left(x_n,y_n\right)\backslash \mathrm{hfill} \\ {k}_2=f\left(\left(x_n+\frac{h}{2}\right),\left(y_n+\frac{h{k}_1}{2}\right)\right) \\ {k}_3=f\left(\left(x_n+\frac{h}{2}\right),\left(y_n+\frac{h{k}_2}{2}\right)\right) \\ {k}_4=f\left(\left(x_n+h\right),\left(y_n+h{k}_3\right)\right) \end{gathered}$$

a) Applying First and initial approximation method

We have the initial value problem

and we have to obtain the approximate values of the solution at using RK4 method with and 0.05 as the following:

We can obtain the following result for using the formula

-------(1)

But before that, we have to find the constant and at as the following

First, for, :

Since we have x₀ = 1 and y₀ = 1 , then we have

$k_{1}f\left(x_0,y_0\right)=x_0^2+y_0^3$

$$\begin{gathered} =1+1 \\ =2 \\ {k}_2=f\left(x_0+\frac{1}{2}h,y_0+\frac{1}{2}h{k}_1\right) \\ ={\left(x_0+\frac{1}{2}h\right)}^2+{\left(y_0+\frac{1}{2}h{k}_1\right)}^3 \\ ={\left(1+\frac{1}{2}\times 0.1\right)}^2+{\left(1+\frac{1}{2}\times 0.1\times 2\right)}^3 \\ =(1.1025)+(1.331) \\ =2.4335 \\ {k}_3=f\left(x_0+\frac{1}{2}h,y_0+\frac{1}{2}h{k}_2\right) \\ ={\left(x_0+\frac{1}{2}h\right)}^2+{\left(y_0+\frac{1}{2}h{k}_2\right)}^3 \end{gathered}$$

$$\begin{gathered} k_3=f\left(x_0+\frac{1}{2}h,y_0+\frac{1}{2}h{k}_2\right) \\ ={\left(x_0+\frac{1}{2}h\right)}^2+{\left(y_0+\frac{1}{2}h{k}_2\right)}^3 \\ ={\left(1+\frac{1}{2}\times 0.1\right)}^2+{\left(1+\frac{1}{2}\times 0.1\times 2.4335\right)}^3 \\ =(1.1025)+(1.4112) \\ =2.5137 \\ {k}_4=f\left(x_0+h,y_0+h{k}_3\right) \\ ={\left(x_0+h\right)}^2+{\left(y_0+h{k}_3\right)}^3 \\ =(1+0.1{)}^2+(1+0.1\times 2.5137{)}^3 \\ =(1.21)+(1.9596) \\ =3.1696 \end{gathered}$$

After that when n= 0, since we have y₀ = 1, then by substituting with the four constants we can obtain from equation (1) as

$$\begin{gathered} y_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =1+\frac{0.1}{6}[2+2(2.4335)+2(2.5137)+3.1696] \\ =1+\frac{0.1}{6}\times (15.064) \\ =1.251 \end{gathered}$$

After that, we can obtain the value of for each value of t with another four constants in the table shown below until x = 1.4, we can have y₄ = 904.46

 Calculate for h = 0.05

Since we have x = 1 and y = 1, then we have

$$\begin{gathered} k_1=f\left(x_0,y_0\right)n \\ =x_0^2+y_0^3 \\ =1+1 \\ =2k_2=f\left(x_0+\frac{1}{2}h,y_0+\frac{1}{2}h{k}_1\right) \\ ={\left(x_0+\frac{1}{2}h\right)}^2+{\left(y_0+\frac{1}{2}h{k}_1\right)}^3 \\ ={\left(1+\frac{1}{2}\times 0.05\right)}^2+{\left(1+\frac{1}{2}\times 0.05\times 2\right)}^3 \\ =(1.050625)+(1.157625) \\ =2.2083 \end{gathered}$$

$$\begin{gathered} k_3=f\left(x_0+\frac{\perp }{2}h,y_0+\frac{\perp }{2}h{k}_2\right) \\ ={\left(x_0+\frac{1}{2}h\right)}^2+{\left(y_0+\frac{1}{2}h{k}_2\right)}^3 \\ ={\left(1+\frac{1}{2}\times 0.05\right)}^2+{\left(1+\frac{1}{2}\times 0.05\times 2.2083\right)}^3 \\ =(1.050625)+(1.17493)=2.2256 \\ {k}_4=f\left(x_0+h,y_0+h{k}_3\right) \\ ={\left(x_0+h\right)}^2+{\left(y_0+h{k}_3\right)}^3 \\ =(1+0.1{)}^2+(1+0.05\times 2.2256{)}^3 \\ =(1.21)+(1.3723)=2.5823 \end{gathered}$$

After that when n = 0, since we have y₀ = 1, then by substituting with the four constants we can obtain y₁ from equation (1) as

$$\begin{gathered} y_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =1+\frac{0.05}{6}[2+2(2.2083)+2(2.2256)+2.5823] \\ =1+\frac{0.05}{6}\times (13.45) \\ =1.112 \end{gathered}$$

After that, we can obtain the value of y for each value t of with another four constants in the table shown below until x = 1.4, we can have y₈ = 1.0987 x 10¹⁵