Question

Repeat Problem 16 using the initial-value problem${\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{+}\mathit{x}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$. The analytic solution is$\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{x}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{+}\frac{\mathbf{5}}{\mathbf{4}}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathbf{.}$

Short answer

The initial value problem is given by the table

Step-by-step solution

Runga-Kutta method:

The solution of a linear differential equation of the form is given as follows:

$$\begin{gathered} y_{n+1}=y_n+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)\text{where} \\ {k}_1=f\left(x_n,y_n\right) \\ {k}_2=f\left(\left(x_n+\frac{h}{2}\right),\left(y_n+\frac{h{k}_1}{2}\right)\right) \\ {k}_3=f\left(\left(x_n+\frac{h}{2}\right),\left(y_n+\frac{h{k}_2}{2}\right)\right) \\ {k}_4=f\left(\left(x_n+h\right),\left(y_n+h{k}_3\right)\right) \end{gathered}$$

Applying First and initial approximation method

We have the initial value problem

$y^{\text{'}}=f(x,y)=x-2y,y\left(0\right)=1$with the analytical solution

$y=\frac{1}{2}x-\frac{1}{4}+\frac{5}{4}{e}^{-2x}$and we have to obtain the required points as the following :

a) We can obtain the approximate value y (0.1) with one step using the formula

$y_{n+1}=y_n+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)$But before that, we have to find the constant k₁,k₂ , k₃and k₄ at and as the following

Since we have x₀ = 0 and y₀ = 1 then we have

localid="1663956628492" $$\begin{gathered} k_1=f\left(x_0,y_0\right) \\ =0-2 \\ =-2 \\ {k}_2=f\left(x_0+\frac{1}{2}h,y_0+\frac{1}{2}h{k}_1\right) \\ =\left(x_0+\frac{1}{2}h\right)-2\left(y_0+\frac{1}{2}h{k}_1\right) \\ =\left(0+\frac{1}{2}\times 0.1\right)-2\left(1+\frac{1}{2}\times 0.1\times (-2)\right) \\ =0.05-2(0.9) \\ =-1.75 \end{gathered}$$

$$\begin{gathered} k_3=f\left(x_0+\frac{1}{2}h,y_0+\frac{1}{2}h{k}_2\right) \\ =\left(x_0+\frac{1}{2}h\right)-2\left(y_0+\frac{1}{2}h{k}_2\right) \\ =\left(0+\frac{1}{2}\times 0.1\right)-2\left(1+\frac{1}{2}\times 0.1\times (-1.75)\right) \\ =0.05-2(0.9125)=-1.775 \\ {k}_4=f\left(x_0+h,y_0+h{k}_3\right) \\ =\left(x_0+h\right)-2\left(y_0+h{k}_3\right) \\ =(0+0.1)-2(1+0.1\times (-1.775\left)\right) \\ =0.1-2(0.8225) \\ =-1.545 \end{gathered}$$

After that when, since we have, then by substituting with the four constants we can obtainfrom equation (1) as

$$\begin{gathered} y_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =1+\frac{0.1}{6}\left[\right(-2)+2(-1.75)+2(-1.775)+(-1.545\left)\right) \\ =1+\frac{0.1}{6}\times (-10.595) \\ =0.8234167 \end{gathered}$$

b) We can obtain a bound for the local truncation error in  as the following technique:

first, we have to obtain the fifth derivative the given solution as

$y^{\left(5\right)}\left(x\right)=-40e^{-2x}$

After that, since we have, then we can obtain the local truncation error as

$\left|y^{\left(5\right)}\left(c\right)\frac{h^5}{5!}\right|⩽40e^{-0.2}\frac{{(0.1)}^5}{120}=3.333\times {10}^{-6}$

 Calculate for  h = 0.5 :

c) From the table shown above in point a), the error between the actual value for and the value using RK4 as 3.259 x 10^-6 which is smaller than the local truncation error.

d) We can obtain the approximate value y(0.1) with two steps using the formula

$y_{n+1}=y_n+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)$

But before that, we have to find the constant k₁,k₂ , k₃and k₄ at n = 0 and h =0.5 as the following

Since we have x₀ = 0 and y₀ = 1 then we have

$$\begin{gathered} k_1=f\left(x_0,y_0\right) \\ =x_0-2y_0 \\ =0-2\times 1 \\ =-2 \\ {k}_2=f\left(x_0+\frac{1}{2}h,y_0+\frac{1}{2}h{k}_1\right) \\ =\left(x_0+\frac{1}{2}h\right)-2\left(y_0+\frac{1}{2}h{k}_1\right) \\ =\left(0+\frac{1}{2}\times 0.05\right)-2\left(1+\frac{1}{2}\times 0.05\times (-2)\right) \\ =0.025-2(0.95) \\ =-1.875 \end{gathered}$$

$$\begin{gathered} k_3=f\left(x_0+\frac{1}{2}h,y_0+\frac{1}{2}h{k}_2\right) \\ =\left(x_0+\frac{1}{2}h\right)-2\left(y_0+\frac{1}{2}h{k}_2\right) \\ =\left(0+\frac{1}{2}\times 0.05\right)-2\left(1+\frac{1}{2}\times 0.05\times (-1.875)\right) \\ =0.025-2(0.953125) \\ =-1.88125 \\ {k}_4=f\left(x_0+h,y_0+h{k}_3\right) \\ =\left(x_0+h\right)-2\left(y_0+\frac{1}{2}h{k}_2\right) \\ =(0+0.05)-2(1+0.05\times (-1.88125\left)\right) \\ =0.05-2(0.9059375) \\ =-1.761875 \end{gathered}$$

After that when n=0$, since we have y₀ = 1 then by substituting with the four constants we can obtain y₁ from equation (1) as$$\begin{gathered} y_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =1+\frac{0.05}{6}\left[\right(-2)+2(-1.875)+2(-1.88125)+(-1.761875\left)\right] \\ =1+\frac{0.05}{6}\times (-11.2744) \\ =0.9060469 \end{gathered}$$$$\begin{gathered} y_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =1+\frac{0.05}{6}\left[\right(-2)+2(-1.875)+2(-1.88125)+(-1.761875\left)\right] \\ =1+\frac{0.05}{6}\times (-11.2744) \\ =0.9060469 \end{gathered}$$

After that, we can obtain the value of y for each value of x with another four constants in the table shown below until x = 0.1 , we can have y₂= 0.8234136