Q5.3-15E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 14–24, you will need a computer and a programmed version of the vectorized classical fourth-order Runge–Kutta algorithm. (At the instructor’s discretion, other algorithms may be used.)†

Using the vectorized Runge–Kutta algorithm, approximate the solution to the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ at  $\mathbf{t}\mathbf{=}\mathbf{1}$. Starting with $\mathbf{h}\mathbf{=}\mathbf{1}$, continue halving the step size until two successive approximations of both $\mathbf{y}\left(\mathbf{1}\right)$ and $\mathbf{y}\mathbf{\text{'}}\left(\mathbf{1}\right)$ differ by at most 0.1.

Step-by-Step Solution

Answer

The solution is $y\left(1\right)=1.69$ and $y\text{'}\left(1\right)=1.82$.

Transform the equation

Write the equation as$y\text{'}\text{'}=t^2+y^2$

 

The equations can be written as:

$$\begin{gathered} x_1\left(t\right)=y\left(t\right) \\ {x}_2\left(t\right)=y\text{'}\left(t\right) \\ =x{\text{'}}_1 \end{gathered}$$

The transformation of the equation is:

$$\begin{gathered} \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}{{\mathbf{x}}^{\mathbf{2}}}_{\mathbf{1}} \end{gathered}$$

The initial conditions are:

$$\begin{gathered} x_1\left(1\right)=y_1\left(1\right)=1 \\ {x}_2\left(1\right)=y\text{'}\left(1\right)=0 \end{gathered}$$

Apply Runge –Kutta method

For the solution, apply the Runge-Kutta method in MATLAB, and the solution is $y\left(1\right)=1.69$ and  $y\text{'}\left(1\right)=1.82$.

Find that y 1 and y ' ( 1 ) differ by at most 0.01.

Subtracting the values of $u\left(1\right)$ and $v\left(1\right)$ then 

$$\begin{gathered} y\text{'}\left(1\right)-y\left(1\right)=1.82-1.69 \\ =0.13 \end{gathered}$$

 

Therefore, $y\left(1\right)$ and $y\text{'}\left(1\right)$ differ by at most 0.1.