Q5.3-16E - 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 for systems with $\mathit{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{175}$, approximate the solution to the initial value problem $\mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{;}\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}$ at $\mathbf{t}\mathbf{=}\mathbf{1}$. 

Compare this approximation to the actual solution.

Step-by-Step Solution

Answer

The solution is $y\left(1\right)=-423.48$ and $x\left(1\right)=127.77$.

Transform the equation

Write the equation as $x\text{'}=2x-y$ and $y\text{'}=3x+6y$

 

The transformation of the equation is:

$$\begin{gathered} x{\text{'}}_1\left(t\right)=x_2\left(t\right) \\ {x}_1\left(t\right)=y\left(t\right) \\ {x}_2\left(t\right)=2x-x_1 \\ x{\text{'}}_2\left(t\right)=y\text{'}(\left(t\right) \\ x{\text{'}}_2\left(t\right)=3x+6x_1\left(t\right) \end{gathered}$$

The initial conditions are:

$$\begin{gathered} \mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0} \\ \mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2} \end{gathered}$$

Apply Runge –Kutta method

For the solution, apply the Runge-Kutta method in MATLAB, and the solution is $\mathrm{y}\left(1\right)=-423.48$ and $\mathrm{x}\left(1\right)=127.77$.

Compare this approximation to the actual solution x ( t ) = e 5 t - e 3 t , y ( t ) = e 3 t - 3 e 5 t

By putting the value of $t=1$

$$\begin{gathered} x\left(1\right)=e^5-e^3 \\ =128.32 \\ y\left(1\right)=e^3-3e^5 \\ =-425.15 \end{gathered}$$

 

Therefore, the approximation solution is $x\left(1\right)=128.32$ and $y\left(1\right)=-425.15$.