Question

Use the RK4 method with$\mathbf{h}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{1}$to approximate$\mathbf{y}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{)}$, where$\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$is the solution of the initial-value problem${\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{2}}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}$. Compare this approximate value with the actual value obtained in Problem 11 in Exercises 9.1.

Short answer

The approximated and actual value of $y(0.5)$for step size $h=0.1$is 3.9078 and 3.9082, respectively.

Step-by-step solution

Runga-Kutta method

The solution of a linear differential equation of the form ${\mathrm{y}}^{\text{'}}=\mathrm{f}(\mathrm{x},\mathrm{y})$ is given as follows: $y_{n+1}=y_n+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)$, here

$$\begin{gathered} k_1=f\left(x_n,y_n\right) \\ {k}_2=f\left(\left(x_n+\frac{h}{2}\right),\left(y_n+\frac{h{k}_1}{2}\right)\right) \\ {k}_3=f\left(\left(x_n+\frac{h}{2}\right),\left(y_n+\frac{h{k}_2}{2}\right)\right) \\ {k}_4=f\left(\left(x_n+h\right),\left(y_n+h{k}_3\right)\right) \end{gathered}$$

The linear differential equation is given as follows:

$y^{\text{'}}=(x+y-1{)}^2$

It is given that for the Runga-Kutta method of the fourth order, the value of h is 0.1

As per the Runga-Kutta method of the fourth order, the value of $y_{n+1}$is given as follows:

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

Here, the value of $k_1,k_2,k_3$and $k_4$is given as follows:

$$\begin{gathered} k_1=f\left(x_n,y_n\right) \\ {k}_2=f\left(\left(x_n+\frac{h}{2}\right),\left(y_n+\frac{h{k}_1}{2}\right)\right) \\ {k}_3=f\left(\left(x_n+\frac{h}{2}\right),\left(y_n+\frac{h{k}_2}{2}\right)\right) \\ {k}_4=f\left(\left(x_n+h\right),\left(y_n+h{k}_3\right)\right) \end{gathered}$$

The differential equation $y^{\text{'}}=(x+y-1{)}^2$is of the form $y^{\text{'}}=f(x,y)$.

This implies that $f(x,y)=(x+y-1{)}^2$.

It is given that the initial value is$y\left(0\right)=2$.

So, the value of $x_0$is 0 and the value of $y_0$is 2.

To calculate the value of $y_1$substitute $n=0$in equation (1).

$y_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)$...(2)

Calculate the value of k1,k2,k3 and k4.

Calculate the value of $k,k_2,k_3$and $k_4$as follows:

$k_1=f\left(x_0,y_0\right)=(0+2-1{)}^2=(1{)}^2=1$

Therefore, the value of $k_1$is 1

$$\begin{gathered} k_2=f\left(\left(x_0+\frac{0.1}{2}\right),\left(y_0+\frac{(0.1)\left(1\right)}{2}\right)\right) \\ =f\left(\right(0.05),(2.05\left)\right) \\ =(0.05+2.05-1{)}^2 \\ =1.21 \end{gathered}$$

Therefore, the value of $k_2$is 1.21.

$$\begin{gathered} k_3=f\left(\left(x_0+\frac{0.1}{2}\right),\left(y_0+\frac{(1.21)·(0.1)}{2}\right)\right) \\ =f\left(\right(0.05),(2.0605\left)\right) \\ =(0.05+2.0605-1{)}^2 \\ =1.23321 \end{gathered}$$

Therefore, the value of $k_4$is 1.23321.

$$\begin{gathered} k_4=f\left(\left(x_0+0.1\right),\left(y_0+(0.1)(1.23321)\right)\right) \\ =f\left(\right(0.1),(2.123321\left)\right) \\ =(0.1+2.123321-1{)}^2 \\ =1.496514 \end{gathered}$$

Substitute the value of $k_1,k_2,k_3\text{and}{k}_4$in equation (2).

$$\begin{gathered} y_1=2+\frac{0.1}{6}(1+2·(1.21)+2·(1.23321)+1.496514) \\ =2+\frac{0.1}{6}(7.382934) \\ =2+0.1230489 \\ =2.1230489 \end{gathered}$$

Therefore, the value of $y_1$is 2.1230489.

Obtain the exact solution of the differential equation

Now, to obtain the exact solution of the differential equation $y^{\text{'}}=(x+y-1{)}^2$, assume as a variable t then the expression becomes as shown below.

$$\begin{gathered} \frac{dy}{dt}-1=t^2 \\ \frac{dy}{dt}=t^2+1 \end{gathered}$$

Now, integrate the above expression by rearranging the similar variable terms on one side of the equation.

$$\begin{gathered} \int \frac{dt}{t^2+1}=\int dx \\ {\tan }^{-1}t=x+\text{c} \end{gathered}$$

Back-substitute the value of t as $(x+y-1)$in the above expression and simplify.

${\tan }^{-1}t=x+\text{c}{\tan }^{-1}(x+y-1)=x+\text{c}x+y-1=\tan (x+\text{c})y=1-x+\tan (x+\text{c})$

Substitute the initial value $y\left(0\right)=2$in the above equation to solve for integration constant c.

$$\begin{gathered} 2=1-0+\tan (0+\text{c}) \\ 2-1=\tan \text{c} \end{gathered}$$

$$\begin{gathered} \tan c=1 \\ c=0.7854 \end{gathered}$$

Hence, the solution of the given differential equation is obtained, as shown below.

$y=1-x+\tan (x+0.7854)$

Substitute 0.1 for x to obtain the actual value of $y(0.10)$.

$y=1-x+\tan (x+0.7854)=1-0.1+\tan (0.1+0.7854)=0.9+1.2230=2.1230$

Therefore, the actual value of $y_1$or $y(0.1)$is 2.1230.

Similarly, calculate the actual and approximated value of $y(0.2),y(0.3),y(0.4)$and $y(0.5)$as shown in Table 1.

The approximated and actual value of y(0.5) for step size $h=0.1$is $\underset{¯}{3.9078}$and $\underset{¯}{3.9082}$respectively.