Question

Consider the initial-value problem${\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{2}\mathit{x}\mathbf{-}\mathbf{3}\mathit{y}\mathbf{+}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{5}$. The analytic solution is

$\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{9}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{3}}\mathit{x}\mathbf{+}\frac{\mathbf{38}}{\mathbf{9}}{\mathit{e}}^{\mathbf{-}\mathbf{3}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{)}}\mathbf{.}$

(a) Find a formula involvingandfor the local truncation error in the nth step if the RK4 method is used.

(b) Find a bound for the local truncation error in each step if h= 0.1is used to approximate

(c) Approximate y(1.5)using the RK4 method with h = 0.1 and h = 0.05 See Problem 3. You will need to carry more than six decimal places to see the effect of reducing the step size.

Short answer

(a)The truncation error in the nth step if the RK4 method is used is .

(b)The approximatebound for the local truncation error is .

(c) The approximate values for x = 1.5, h = 0.06 is y₅ = 2.0384 and for x = 1.50, h = 0.06 is y₁₀ = 2.05254

Step-by-step solution

Using Runge-Kutta Method

The Runge-Kutta method approximates the value of y for a given x. The Runge Kutta 2nd order method can only solve first-order ordinary differential equations.

Obtaining the truncation error if RK4 method is used

We have the initial value problem

$y^{\text{'}}=f(x,y)=2x-3y+1,y\left(1\right)=5$

With the analytical solution

$y=\frac{1}{9}+\frac{2}{3}x+\frac{38}{9}{e}^{-3(x-1)}$

and we have to obtain the required points as the following technique :

(a)

We can obtain the local truncation erroe in the nth if the RK4 method from the formula

$y^{\left(5\right)}\left(c\right)\frac{h^5}{5!}$ (a)

Where c= value of x.

After that, we have to find the fifth derivative of the given analytical solution as

$$\begin{gathered} y^{\text{'}}\left(x\right)=\frac{2}{3}-\frac{38}{3}{e}^{-3(x-1)} \\ {y}^{\text{'}\text{'}}\left(x\right)=38e^{-3(x-1)} \\ {y}^{\text{'}\text{'}\text{'}}\left(x\right)=-114e^{-3(x-1)} \\ {y}^{\left(4\right)}\left(x\right)=342e^{-3(x-1)} \\ {y}^{\left(5\right)}\left(x\right)=-1026e^{-3(x-1)} \end{gathered}$$

Then replace x by c, then we have

$y^{\left(5\right)}\left(c\right)=-1026e^{-3(c-1)}$

After that, substitute with y⁽⁵⁾(c) into (a) then we can have the genera local truncation error as

$$\begin{gathered} y^{\left(5\right)}\left(c\right)\frac{h^5}{5!}=\left|-1026e^{-3(c-1)}\right|\frac{h^5}{5!} \\ =1026e^{-3(c-1)}\frac{h^5}{120} \\ =8.55e^{-3(c-1)}{h}^5 \end{gathered}$$

The local truncation error is $8.55e^{-3(c-1)}{h}^5$ (b)

Therefore,the truncation error in the nth step if the RK4 method is used is $y^{\left(5\right)}\left(c\right)\frac{h^5}{5!}$

Obtaining the truncation error if RK4 method is used

(b)

Now, with we can obtain the truncation error in each step if using equation as

$$\begin{gathered} Error⩽8.55e^{-3(1-1)}(0.1{)}^5 \\ =8.55e^0\times \left(1\times {10}^{-5}\right) \\ =8.{5510}^{-5} \end{gathered}$$

Therefore,the approximatebound for the local truncation error is 8.5510^-5.

Obtaining the approximate values

(c)

We can obtain the approximate values for ywith 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 , k₃ and k₄ at n = 0 and h =0.1 as the following

Since we have x₀ = 1 and y₀ = 5 then we have

$$\begin{gathered} k_1=f\left(x_0,y_0\right) \\ =2x_0-3y_0+1 \\ =2\times 1-3\times 5+1 \\ =2-15+1 \\ =-12 \end{gathered}$$

$$\begin{gathered} k_2=f\left(x_0+\frac{1}{2}h,y_0+\frac{1}{2}h{k}_1\right) \\ =\left[2\left(x_0+\frac{1}{2}h\right)-3\left(y_0+\frac{1}{2}h{k}_1\right)+1\right] \\ =\left[2\left(1+\frac{1}{2}\times 0.1\right)-3\left(5+\frac{1}{2}\times 0.1\times (-12)\right)+1\right] \\ =2.1-13.2+1 \\ =-10.1 \end{gathered}$$

$$\begin{gathered} k_3=f\left(x_0+\frac{1}{2}h,y_0+\frac{1}{2}h{k}_2\right) \\ =\left[2\left(x_0+\frac{1}{2}h\right)-3\left(y_0+\frac{1}{2}h{k}_2\right)+1\right] \\ =\left[2\left(1+\frac{1}{2}\times 0.1\right)-3\left(5+\frac{1}{2}\times 0.1\times (-10.1)\right)+1\right] \\ =2.1-13.485+1 \\ =-10.385 \end{gathered}$$

$$\begin{gathered} k_4=f\left(x_0+h,y_0+h{k}_3\right) \\ =\left[2\left(x_0+h\right)-3\left(y_0+h{k}_3\right)+1\right] \\ =\left[2\right(1+0.1)-3(5+0.1\times (-10.385))+1] \\ =2.2-11.8845+1 \\ =-8.6845 \end{gathered}$$

After that when n = 0 since we have y₀ = 5 then by substituting with the four constants we can obtain y₁ from equation (1) as

$$\begin{gathered} y_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =5+\frac{0.1}{6}\left[\right(-12)+2(-10.1)+2(-10.385)+(-8.6845\left)\right] \\ =5+\frac{0.1}{6}\times (-65.6545) \\ =3.9724 \end{gathered}$$

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 = 1.5 we can havey₅ = 2.0384

	h = 0.06
Xn	k1 + 2k2 + 2k3 + k4	Yn
1.00	-65.6545	5.00
1.10	-44.6393	3.9724
1.20	-32.9192	3.2284
1.30	-22.4178	2.6797
1.40	-16.0629	2.3061
1.50		2.0384

After that, we can obtain the approximate values y with two strps using the formla

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

But before that, we have to find the constant k₁,k₂ , k₃ and k₄ at n = 0 and h = 0.1 as the following

Since we have x₀ = 1 and y₀ =5 then we have

$$\begin{gathered} k_1=f\left(x_0,y_0\right) \\ =2x_0-3y_0+1 \\ =2\times 1-3\times 5+1 \\ =2-15+1 \\ =-12 \end{gathered}$$

$$\begin{gathered} k_2=f\left(x_0+\frac{1}{2}h,y_0+\frac{1}{2}h{k}_1\right) \\ =\left[2\left(x_0+\frac{1}{2}h\right)-3\left(y_0+\frac{1}{2}h{k}_1\right)+1\right] \\ =\left[2\left(1+\frac{1}{2}\times 0.05\right)-3\left(5+\frac{1}{2}\times 0.05\times (-12)\right)+1\right] \\ =2.05-14.1+1 \\ =-11.05 \end{gathered}$$

$$\begin{gathered} k_3=f\left(x_0+\frac{1}{2}h,y_0+\frac{1}{2}h{k}_2\right) \\ =\left[2\left(x_0+\frac{1}{2}h\right)-3\left(y_0+\frac{1}{2}h{k}_2\right)+1\right] \\ =\left[2\left(1+\frac{1}{2}\times 0.05\right)-3\left(5+\frac{1}{2}\times 0.05\times (-11.05)\right)+1\right] \\ =2.05-14.17125+1 \\ =-11.12125 \end{gathered}$$

$$\begin{gathered} k_4=f\left(x_0+h,y_0+h{k}_3\right) \\ =\left[2\left(x_0+h\right)-3\left(y_0+h{k}_1\right)+1\right] \\ =\left[2\right(1+0.05)-3(5+0.05\times (-11.12125))+1] \\ =2.1-13.3318125+1 \\ =-10.2318125 \end{gathered}$$

After that when n = 0 since we have y₀ = 5 then by substituting with the four constants we can obtain y₁ from equation (1) as

$$\begin{gathered} y_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =5+\frac{0.05}{6}\left[\right(-12)+2(-11.05)+2(-11.12125)+(-10.2318125\left)\right] \\ =5+\frac{0.05}{6}\times (-66.5743) \\ =4.4452 \end{gathered}$$

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 = 1.5 we can have y₁₀ = 2.05254

h = 0.06
Xn	k1+2k2+2k3+k4	Yn
1.00	-65.6545	5.00
1.050	-52.7437	4.44520
1.10	-48.5183	3.97240
1.15	-40.9666	3.56798
1.20	-34.6982	3.22659
1.25	-29.3129	2.93744
1.30	-24.6727	2.69317
1.35	-20.6788	2.48756
1.40	17.2413	2.31524
1.45	-14.2826	2.17156
1.50		2.05254

Therefore the approximate values for x = 1.5, h = 0.06 is y₅ = 2.0384 and for x = 1.50, h = 0.06 is y₁₀ = 2.05254