Question

In Problems 3 and 4 use Euler’s method to obtain a four-decimal approximation of the indicated value. First use h = 0.1and then use h = 0.05. Find an explicit solution for each initial-value problem and then construct tables similar to Tables 2.6.3 and 2.6.4.

$\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathit{y}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{;}\mathit{y}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{)}$

Short answer

The indicated value for h = 0.1 is $y\left(1.0\right)\approx 2.5937$, and for h = 0.05 is $y\left(1.0\right)\approx 2.6533$.

Step-by-step solution

Define Euler’s method.

The general form of the Euler’s method is given by,${\mathit{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathit{y}}_{\mathbf{n}}\mathbf{+}\mathit{h}\mathit{f}\left(x_n,y_n\right)$ where${\mathit{x}}_{\mathbf{n}}\mathbf{=}{\mathit{x}}_{\mathbf{0}}\mathbf{+}\mathit{n}\mathit{h}$ and $\mathit{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}$.

Find the value of y.

Let the given differential equation be,

$$\begin{gathered} \frac{dy}{dx}=y \\ \frac{dy}{y}=dx \end{gathered}$$

Integrate both sides of the equation.

$$\begin{gathered} \int \frac{dy}{y}=\int dx \\ In\left|y\right|=x+c \\ y=e^{x+c}\left(e^{a+b}=e^a{e}^b\right) \\ y=e^x{e}^c\left(e^c=c_1\right) \\ y=e^x{c}_1 \end{gathered}$$… (1)

Use the initial condition $y\left(0\right)=1$to get c₁.

$\begin{array}{rcl}y\left(0\right)& =& e^0{c}_1\left(e^0=1\right)\\ 1& =& c_1\end{array}$

Substitute the value c₁ = 1 in the equation (1).

y = e^x

Find the indicated value for h = 0.1.

Let the given be$h=0.1,x_0=0$and $y_0=1$.

When n = 0, then

role="math" localid="1664189857311" $$\begin{gathered} y_{0+1}=y_0+hf\left(x_0,y_0\right) \\ {y}_1=y_0+h{e}^{x_0} \\ {y}_1=y_0+h{y}_0 \\ {y}_1=1+0.1.1 \\ {y}_1=1.1 \end{gathered}$$

When n = 1, then

$\begin{array}{rcl}{x}_1& =& x_0+1\left(0.1\right)\\ & =& 0+0.1\\ & =& 0.1\end{array}$

Hence, by Euler’s method, it is given by

role="math" localid="1664190003633" width="151" height="148">y1+1=y1+hfx1,y1y2=y1+hex1y2=y1+hy1y2=1.1+0.1.1.1y2=1.21

Tabulate the actual value, approximation value and indicated values for h = 0.1..

Actual value and approximation for $x_0=0.1$is

$$\begin{gathered} y_1=e^{0.1}=1.1052 \\ {y}_1=1.1 \end{gathered}$$

Now,

$\begin{array}{rcl}Absoluteerror& =& \left|1.1052-1.1\right|\\ & =& \left|0.0052\right|\\ Precentagerelativeerror& =& \frac{0.0052}{\left|1.1052\right|}\times 100\\ & =& 0.47\%\end{array}$

Actual value and approximation for x₀ = 0.2 is

$$\begin{gathered} y_2=e^{0.2}=1.2214 \\ {y}_2=1.21 \end{gathered}$$

Now,

$\begin{array}{rcl}Absoluteerror& =& \left|1.2214-1.21\right|\\ & =& \left|0.0114\right|\\ Precentagerelativeerror& =& \frac{0.0114}{\left|1.2214\right|}\times 100\\ & =& 0.93\%\end{array}$

Further continuing using the numerical solver and Euler’s method, the results are given in the table.

Find the indicated value for h = 0.05.

Let the given be h = 0.05, x₀ = 0 and y₀ = 1.

When n = 0, then

$$\begin{gathered} y_{0+1}=y_0+hf\left(x_0,y_0\right) \\ {y}_1=y_0+h{e}^{x_0} \\ {y}_1=y_0+h{y}_0 \\ {y}_1=1+0.5.1 \\ {y}_1=1.05 \end{gathered}$$

When n = 1, then

$\begin{array}{rcl}{x}_1& =& x_0+1\left(0.05\right)\\ & =& 0+0.05\\ & =& 0.05\end{array}$

Hence, by Euler’s method, it is given by

$$\begin{gathered} y_{1+1}=y_1+hf\left(x_1,y_1\right) \\ {y}_2=y_1+h{e}^{x_1} \\ {y}_2=y_1+h{y}_1 \\ {y}_2=1.05+0.05.1.05 \\ {y}_2=1.1025 \end{gathered}$$

Tabulate the actual value, approximation value and indicated values for h = 0.05.

Actual value and approximation for x₀ = 0.05 is

$$\begin{gathered} y_1=e^{0.05}=1.051 \\ {y}_1=1.05 \end{gathered}$$

Now,

$\begin{array}{rcl}Absoluteerror& =& \left|1.0513-1.05\right|\\ & =& \left|0.0013\right|\\ Precentagerelativeerror& =& \frac{0.0013}{\left|1.0513\right|}\times 100\\ & =& 0.12\%\end{array}$

Actual value and approximation forx₀ = 0.1 is

$$\begin{gathered} y_2=e^{0.1}=1.1052 \\ {y}_2=1.1025 \end{gathered}$$

Now,

$\begin{array}{rcl}Absoluteerror& =& \left|1.1052-1.1025\right|\\ & =& \left|0.0027\right|\\ Precentagerelativeerror& =& \frac{0.0027}{\left|1.1052\right|}\times 100\\ & =& 0.24\%\end{array}$

Further continuing using the numerical solver and Euler’s method, the results are given in the table.