Question

Question: In Problems 3-12 use the RK4 method with $\mathbf{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}$ to obtain a four decimal approximation of the indicated value.

role="math" localid="1668085278924" $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{+}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{y}\left(1\right)\mathbf{=}\mathbf{5}\mathbf{;}\mathbf{}\mathbf{y}\left(1.5\right)$

Short answer

The four decimal approximation of the indicated value $y(1.5]$ using the RK4 method is 2.0525

Step-by-step solution

Runga-Kutta method

The general formulas used in the given problem are,

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

And,

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

Determine the First order Runga-Kutta method

The given differential equation is $y\text{'}=2x-3y+1$.

The given differential equation can be written as,

$\begin{array}{l}y\text{'}=F\left(x,y\right)\\ F\left(x,y\right)=2x-3y+1\end{array}$

The given boundary conditions can be written as,

$\begin{array}{l}{y}_0=f\left(x_0\right)\\ x_0=1\\ y_0=5\\ h=0.1\end{array}$

For $n=0$, substitute the value of $x_0,y_0$and $h$in the following equations to get the values of $k_1,k_2,k_3,k_4$,

The value of $k_1$ is calculated as,

$\begin{array}{l}k=h=\{x,y_0,0_0)\\ =0.1\times \left(2\times 1-3\times 5+1\right)\\ =0.1\times \left(-12\right)\\ =-1.2\end{array}$

The value of $k_2$is calculated as,

$k_2=hF(x_0+\frac{h}{2},y_0+k_3)$

$\begin{array}{l}=0.1\times F\left(1+\frac{0.1}{2},5+\frac{-1.2}{2}\right)\\ =0.1\times F\left(1.05,4.4\right)\\ =-1.01\end{array}$

The value of $k_3$is calculated as,

localid="1668143165481" $\begin{array}{l}{k}_3=hF\left(x_0+\frac{h}{2},y_0+\frac{k_2}{2}\right)\\ =0.1\times F\left(1+\frac{0.1}{2},5+\frac{-1.01}{2}\right)\\ =0.1\times F\left(1.05,4.495\right)\\ =-1.0385\end{array}$

The value of $k_4$is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_0+h,y_0+k_3\right)\\ =0.1\times F\left(1+0.1,5+\left(-1.0385\right)\right)\\ =0.1\times F\left(1.01,3.9615\right)\\ =-0.88645\end{array}$

Now substitute the value of k and in .