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.

$\mathit{y}\mathbf{\text{'}}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{y}}\mathbf{,}\mathbf{}\mathit{y}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathit{y}\left(0.5\right)$

Short answer

The four decimal approximation of the indicated value $y(1.5)$using the RK4 method is 0.4055

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,

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

 Step 2: Determine the First order Runga-Kutta method

The given differential equation can be written as,

$\begin{array}{l}y=F(x,y)\\ F\left(x,y\right)=e^{-y_0}\end{array}$

The given boundary conditions can be written as,

$\begin{array}{l}{y}_0=\left(x_0\right)\\ x_0=0\\ y_0=0\\ h=0.1\end{array}$

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

$\begin{array}{l}{k}_1=hF(x_0,y_0)\\ =0.1\times \left(e^{-0}\right)\\ =0.1\times \left(1\right)\\ =0.1\end{array}$

The value of $k_2$is calculated as,

localid="1668171096573" $$\begin{gathered} k_2=hF(x_0+\frac{h}{2},y_0+k_1) \\ \begin{array}{l}=0.1\times F\left(0+\frac{0.1}{2},0+\frac{0.1}{2}\right)\\ =0.1\times F\left(0.05,0.05\right)\\ =0.0952\end{array} \end{gathered}$$

The value of $k_3$is calculated as,

$$\begin{gathered} k_3=hF(x_0+\frac{h}{2},y_0+k_2) \\ \begin{array}{l}=0.1\times F\left(0+\frac{0.1}{2},0+\frac{0.0952}{2}\right)\\ =0.1\times F\left(0.05,0.0476\right)\\ =0.095352\end{array} \end{gathered}$$

The value of $k_4$is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_0+h,y_0+k_3\right)\\ =0.1\times F\left(0+0.1,0+0.95352\right)\\ =0.1\times F\left(0.1,095352\right)\\ =0.0909052\end{array}$

Now substitute the value of $k_1,k_2,k_3,k_4$in $y_1$.

$\begin{array}{rcl}{y}_1& =& y_0+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0+\frac{1}{6}\left(0.1+2\left(0.0952\right)+2\left(-0.09532\right)+0.0909052\right)\\ & =& 0.0953\end{array}$

For $n=1$, the value of $x_1$is calculated as

$x_1=x_0+h=0+0.1=0.1$

Substitute the value of $x_1{y}_1$and $h$in the following equations,

$\begin{array}{c}{k}_1=hF(x_0,y_0)\\ =0.1\times 0.9091\\ =0.09091\end{array}$

The value of $k_2$is calculated as,

$$\begin{gathered} k_2=hF(x_0+\frac{h}{2},y_0+k_1) \\ \begin{array}{l}=0.1\times F\left(0.1+\frac{0.1}{2},0.0953+\frac{0.09091}{2}\right)\\ =0.1\times F\left(0.15,0.140755\right)\\ =0.086870\end{array} \end{gathered}$$

The value of$k_3$is calculated as,

$$\begin{gathered} k_3=hF(x_0+\frac{h}{2},y_0+k_2) \\ \begin{array}{l}=0.1\times F\left(0+\frac{0.1}{2},0.0953+\frac{0.086870}{2}\right)\\ =0.1\times F\left(0.15,0.13874\right)\\ =0.0870455\end{array} \end{gathered}$$

The value of$k_4$is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_0+h,y_0+k_3\right)\\ =0.1\times F\left(0.1+0.1,0.0953+0.0870455\right)\\ =0.1\times F\left(0.2,0.182345\right)\\ =0.083344\end{array}$

Now substitute the value of $k_1,k_2.k_3,k_4$in $y_2$.

role="math" localid="1668407618010" $\begin{array}{rcl}{y}_2& =& y_1+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0.0953+\frac{1}{6}\left(0.09091+2\left(0.086870\right)+2\left(0.0870455\right)+0.08334\right)\\ & =& 0.1823\end{array}$

Compute the second order Runga-Kutta method

For$n=2$, the value of$x_2$is calculated as,

$\begin{array}{l}{x}_2=x_1+h\\ =0.1+0.1\\ =0.2\end{array}$

Substitute the value of$x_2,y_2$and$h$in the following equations,

localid="1668407261466" $\begin{array}{l}{k}_1=hF(x_2,y_2)\\ =0.1\times \left(e^{-0.1823}\right)\\ =0.1\times \left(0.8334\right)\\ =0.08334\end{array}$

The value of $k_2$is calculated as,

$$\begin{gathered} k_2=hF(x_0+\frac{h}{2},y_0+k_1) \\ \begin{array}{l}=0.1\times F\left(0.2+\frac{0.1}{2},0.1823+\frac{0.08334}{2}\right)\\ =0.1\times F\left(0.25,0.4167\right)\\ =0.06592\end{array} \end{gathered}$$

The value of $k_3$is calculated as,

$$\begin{gathered} k_3=hF(x_2+\frac{h}{2},y_2+k_2) \\ \begin{array}{l}=0.1\times F\left(0.2+\frac{0.1}{2},0.1823+\frac{0.06592}{2}\right)\\ =0.1\times F\left(0.25,0.21526\right)\\ =0.080633\end{array} \end{gathered}$$

The value of role="math" localid="1668410359245" $k_4$is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_2+h,y_2+k_3\right)\\ =0.1\times F\left(0.2+0.1,0.1823+0.0806733\right)\\ =0.1\times F\left(0.3,0.98863\right)\\ =0.0372\end{array}$

Now substitute the value of $k_1,k_2,k_3,k_4$in $y_3$.

$\begin{array}{rcl}{y}_3& =& y_2+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0.1823+\frac{1}{6}\left(0.08344+2\left(0.06592\right)+2\left(0.080633\right)+0.0372\right)\\ & =& 0.2624\end{array}$

Compute the third order of Runga-Kutta method

For $n=3$, the value of $x_3$is calculated as,

$\begin{array}{l}{x}_3=x_2+h\\ =0.2+0.1\\ =0.3\end{array}$

Substitute the value of $x_3,y_3$and$h$in the following equations get the values of $k_1,k_2,k_3,k_4$,

The value of $k_1$is calculated as,

$\begin{array}{l}{k}_1=hF(x_3,y_3)\\ =0.1\times \left(e^{-0.2624}\right)\\ =0.1\times \left(0.769203\right)\\ =0.059167\end{array}$

The value of$k_2$ is calculated as,

$$\begin{gathered} k_2=hF(x_3+\frac{h}{2},y_3+k_1) \\ \begin{array}{l}=0.1\times F\left(0.3+\frac{0.1}{2},0.2624+\frac{0.059167}{2}\right)\\ =0.1\times F\left(0.35,0.2819835\right)\\ =0.074678\end{array} \end{gathered}$$

The value of $k_3$is calculated as,

$$\begin{gathered} k_3=hF(x_3+\frac{h}{2},y_3+k_2) \\ \begin{array}{l}=0.1\times F\left(0.3+\frac{0.1}{2},0.2624+\frac{0.074678}{2}\right)\\ =0.1\times F\left(0.35,0.299739\right)\\ =0.0741012\end{array} \end{gathered}$$

The value of$k_4$ is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_3+h,y_3+k_3\right)\\ =0.1\times F\left(0.3+0.1,0.2624+0.0741012\right)\\ =0.1\times F\left(0.4,0.3365\right)\\ =0.0714265\end{array}$

Now substitute the value of $k_1,k_2,k_3,k_4$in $y_4$.

$\begin{array}{rcl}{y}_4& =& y_3+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0.2634+\frac{1}{6}\left(0.059167+2\left(0.074678\right)+2\left(0.0741012\right)+0.071426\right)\\ & =& 0.3365\end{array}$

Compute the fourth order of Runga-Kutta method

For $n=4$, the value of $x_4$is calculated as,

$\begin{array}{l}{x}_4=x_3+h\\ =0.3+0.1\\ =0.4\end{array}$

Substitute the value of $x_4,y_4$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,

$\begin{array}{l}{k}_1=hF(x_4,y_4)\\ =0.1\times \left(e^{-0.3365}\right)\\ =0.1\times \left(0.714266\right)\\ =0.0714266\end{array}$

The value of $k_2$is calculated as,

$$\begin{gathered} k_2=hF(x_4+\frac{h}{2},y_4+k_1) \\ \begin{array}{l}=0.1\times F\left(0.4+\frac{0.1}{2},0.3365+\frac{0.0714266}{2}\right)\\ =0.1\times F\left(0.45,0.372213\right)\\ =0.068921\end{array} \end{gathered}$$

The value of $k_3$is calculated as,

$$\begin{gathered} k_3=hF(x_4+\frac{h}{2},y_4+k_2) \\ \begin{array}{l}=0.1\times F\left(0.4+\frac{0.1}{2},0.3365+\frac{0.068921}{2}\right)\\ =0.1\times F\left(0.45,0.3709605\right)\\ =0.06900\end{array} \end{gathered}$$

The value of$k_4$is calculated as,

$\begin{array}{l}{k}_4=hF\left(x_4+h,y_4+k_3\right)\\ =0.1\times F\left(0.4+0.1,0.3365+0.06900\right)\\ =0.1\times F\left(0.5,0.4055\right)\\ =0.06667\end{array}$

Now substitute the value of $k_1,k_2,k_3,k_4$in $y_5$.

$\begin{array}{rcl}{y}_5& =& y_4+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right)\\ & =& 0.3365+\frac{1}{6}\left(0.0714266+2\left(0.069821\right)+2\left(0.06900\right)+0.06667\right)\\ & =& 0.4055\end{array}$

For $n=5$, the value of $x_5$is calculated as,

$\begin{array}{l}{x}_5=x_4+h\\ =0.4+0.1\\ =0.5\end{array}$

Therefore, the four decimal approximation of the indicated value $y\left(0.5\right)$using the RK4 method is 0.4055