Question

Use a numerical solver and Euler’s method to approximate y(1.0), where y(x)is the solution to y' = 2xy², y(0) = 1. First use h = 0.1and then use h = 0.05. Repeat, using the RK4 method. Discuss what might cause the approximations to y(1.0)to differ so greatly.

Short answer

The indicated values are $y(1.0)\approx 3.82,y(1.0)\approx 43,y(1.0)\approx 5.94,$ and $y\left(1.0\right)\approx 84.01$.

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 $\mathit{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}$.

Find the Euler’s approximation.

Let the given differential equation be,

$y\text{'}=2x{y}^2,y\left(0\right)=1$

The Euler’s approximation with step size is tabulated below:

Now, based on the given differential equation to get:

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

Find the Fourth order Runge-Kutta approximation.

Let the given step size be h = 0.1 . By using the above formula, the fourth order Runge-Kutta approximation is tabulated below:

Find the Euler’s approximation.

Let the given differential equation be,

$y\text{'}=2x{y}^2,y\left(0\right)=1$

The Euler’s approximation with step size h = 0.05 is tabulated below:

Now, based on the given differential equation to get:

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

Find the Fourth order Runge-Kutta approximation.

Let the given step size be h = 0.1. By using the above formula, the fourth order Runge-Kutta approximation is tabulated below:

Graph the Euler’s approximation and the Fourth order Runge-Kutta approximation.

Let the Euler’s approximation be red color, and the fourth order Runge-Kutta approximation be blue color.

From table 1 and 2,

From table 3 and 4,

Since $x\to 1$, specifically for $0.9\le x<1$, then the slope of y(x) increases very quickly, because y(x) has a vertical asymptote at x = 1. Euler method does not catch this attribute of the function because only the slope at ( x_n, y_n ) is used to approximate ( x_n+1, y_n+1 ). Fourth-order Runge-Kutta on the other hand use four slopes between x_n and x_n+1 for the same approximation. So, when the slope of y(x) changes rapidly it is better able to capture the feature.