Question

In Problems 11and 12use a numerical solver to obtain a numerical solution curve for the given initial-value problem. First use Euler’s method and then the RK4method. Use h = 0.25in each case. Superimpose both solution curves on the same coordinate axes. If possible, use a different color for each curve. Repeat, using h = 0.1and h = 0.05.

$\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathbf{(}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{x}\mathbf{)}\mathit{y}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

The Euler method has difficulty in approximating the solution with this particular step size while the fourth order Runge-Kutta does not have the same difficulty.

The methods have about the same approximation for the given different initial-value.

Step-by-step solution

Define Euler’s method.

The general form of the Euler’s method is given by,${\mathit{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathit{y}}_{\mathbf{n}}\mathbf{+}\mathit{h}\mathit{f}\mathbf{(}{\mathbf{x}}_{\mathbf{n}}\mathbf{,}{\mathbf{y}}_{\mathbf{n}}\mathbf{)}$ where${\mathit{x}}_{\mathbf{n}}\mathbf{=}{\mathit{x}}_{\mathbf{0}}\mathbf{+}\mathit{n}\mathit{h}$ and $\mathit{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}$.

Solve for constants.

Let the four constants for the fourth-order Runge-Kutta be,

$$\begin{gathered} k_{n1}=hf\left(x_n,y_n\right) \\ {k}_{n2}=hf\left(x_n+\frac{h}{2},y_n+\frac{1}{2}{k}_{n1}\right) \\ {k}_{n3}=hf\left(x_n+\frac{h}{2},y_n+\frac{1}{2}{k}_{n2}\right) \\ {k}_{n4}=hf\left(x_n+h,y_n+k_{n3}\right) \end{gathered}$$

Using the given step size, and then using the formula, the fourth-order Runge-Kutta approximation is,

$y_{n+1}=y_n+\frac{1}{6}\left(k_{n1}+2k_{n2}+2k_{n3}+k_{n4}\right)$

Obtain a numerical solution curve for h = 0.25.

Let the Euler’s method and the fourth-order Runge Kutta method approximation is shown in the graph be,

(Note that: The doted red line represents a Euler’s method and the blue line represents the fourth-order Runge Kutta method approximation.)

The Euler method has difficulty in approximating the solution with this particular step size while the fourth order Runge-Kutta does not have the same difficulty.

Obtain a numerical solution curve for h = 0.1.

Let the Euler’s method and the fourth-order Runge Kutta method approximation is shown in the graph be,

(Note that: The doted red line represents a Euler’s method and the blue line represents the fourth-order Runge Kutta method approximation.)

The methods have about the same approximation.

Obtain a numerical solution curve for h = 0.05.

Let the Euler’s method and the fourth-order Runge Kutta method approximation is shown in the graph be,

(Note that: The doted red line represents a Euler’s method and the blue line represents the fourth-order Runge Kutta method approximation.)

The methods have about the same approximation.