Q5.3-14E - 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 with h = 0.5, approximate the solution to the initial value $\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}\mathbf{ty}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\frac{\mathbf{2}}{\mathbf{3}}$ problemat t = 8.

Compare this approximation to the actual solution $\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\frac{\mathbf{5}}{\mathbf{3}}}\mathbf{-}\mathbf{t}$.

Step-by-Step Solution

Answer

The solution is 0.153081544.

Transform the equation

Here $h=0.25$ on $\left[1,8\right]$

Write the equation as $y\text{'}\text{'}=\frac{5ty\text{'}-5y}{3t^2}$.

The equations can be written as:

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

The transformation of the equation is:

$$\begin{gathered} x{\text{'}}_1\left(t\right)=x_2\left(t\right) \\ x{\text{'}}_2\left(t\right)=\frac{5t{x}_2-5x_1}{3t^2} \end{gathered}$$

The initial conditions are:

$$\begin{gathered} x_1\left(1\right)=y_1\left(1\right) \\ =0 \\ {x}_2\left(1\right)=y\text{'}\left(1\right) \\ =\frac{2}{3} \end{gathered}$$

Apply Runge –Kutta method

For the solution apply the Runge-Kutta method in MATLAB, and the solution is 0.153081544

Compare this approximation to the actual solution y ( t ) = t 5 3 - t .

By putting the value of $t=1$

$$\begin{gathered} y\left(1\right)=1-1 \\ =0 \end{gathered}$$

Therefore, the approximation solution is $y\left(1\right)=0$.