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.

$\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathit{x}\mathit{y}\mathbf{,}\mathbf{}\mathit{y}\left(1\right)\mathbf{=}\mathbf{1}\mathbf{;}\mathit{y}\left(1.5\right)$

Short answer

The indicated value for $h=0.1$is $y\left(1.5\right)\approx 3.806$, and for h = 0.05 is $y(1.5)\approx 3.325$.

Step-by-step solution

Define Euler’s method.

The general form of the Euler’s method is given by, ${\mathit{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathit{y}}_{\mathbf{n}}\mathbf{+}\mathit{h}\mathit{f}\mathbf{(}{\mathbf{x}}_{\mathbf{n}}\mathbf{,}{\mathbf{y}}_{\mathbf{n}}\mathbf{)}$ 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 indicated value for h = 0.1.

Let the given differential equation be,

$F\left(x,y\right)=\frac{dy}{dx}=2xy,y\left(1\right)=1,y\left(1.5\right).$

Solve the following by two methods: Numerical method and Exact method.

Let the given be h = 0.1.

role="math" localid="1664195090803" $$\begin{gathered} y_1=1+0.1\left(2\times 1\times 1\right)=1.2 \\ {y}_2=1.2+0.1\left(2\times 1.1\times 1.2\right)=1.464 \\ {y}_3=1.464+0.1\left(2\times 1.2\times 1.464\right)=1.815 \\ {y}_4=1.815+0.1\left(2\times 1.3\times 1.815\right)=2.287 \\ {y}_5=2.287+0.1\left(2\times 1.4\times 2.287\right)=2.928 \\ {y}_6=2.928+0.1\left(2\times 1.5\times 2.928\right)=3.806 \end{gathered}$$

Find the indicated value for h = 0.05.

Let the given be h = 0.05.

$$\begin{gathered} y_1=1+0.05\left(2\times 1\times 1\right)=1.1 \\ {y}_2=1.1+0.05\left(2\times 1.05\times 1.1\right)=1.215 \\ {y}_3=1.215+0.05\left(2\times 1.1\times 1.215\right)=1.349 \\ {y}_4=1.349+0.05\left(2\times 1.15\times 1.349\right)=1.504 \\ {y}_5=1.504+0.05\left(2\times 1.2\times 1.504\right)=1.685 \\ {y}_6=1.685+0.05\left(2\times 1.25\times 1.685\right)=1.727 \\ {y}_7=1.727+0.05\left(2\times 1.3\times 1.727\right)=1.951 \\ {y}_8=1.951+0.05\left(2\times 1.35\times 1.951\right)=2.215 \\ {y}_9=2.215+0.05\left(2\times 1.4\times 2.215\right)=2.525 \\ {y}_{10}=2.525+0.05\left(2\times 1.45\times 2.525\right)=2.891 \\ {y}_{11}=2.891+0.05\left(2\times 1.5\times 2.891\right)=3.325 \end{gathered}$$

Solve by exact solution.

Let the given differential solution be,

$\frac{dy}{dx}=2xy$

Separate the variables and integrate it.

$$\begin{gathered} \frac{dy}{y}=2xdx \\ \int \frac{dy}{y}=\int 2xdx \\ Iny=x^2+C \end{gathered}$$

Find the value of C at y (1) = 1.

$\begin{array}{rcl}In\left(1\right)& =& 1^2+C\\ C& =& -1\\ Iny& =& x^2-1\\ y& =& e^{x^2-1}\end{array}$

Find the actual value for h = 0.1.

$$\begin{gathered} y_0=1 \\ {y}_1=e^{1.1^2-1}=1.234 \\ {y}_2=e^{1.2^2-1}=1.553 \\ {y}_3=e^{1.3^2-1}=1.994 \\ {y}_4=e^{1.4^2-1}=2.611 \\ {y}_5=e^{1.5^2-1}=3.49 \end{gathered}$$

Find the actual value for h = 0.05.

$$\begin{gathered} y_0=1 \\ {y}_1=e^{1.{05}^2-1}=1.108 \\ {y}_2=e^{1.1^2-1}=1.234 \\ {y}_3=e^{1.{15}^2-1}=1.275 \\ {y}_4=e^{1.2^2-1}=1.553 \\ {y}_5=e^{1.2.5^2-1}=1.755 \\ {y}_6=e^{1.3^2-1}=1.994 \\ {y}_7=e^{1.{35}^2-1}=2.276 \\ {y}_8=e^{1.4^2-1}=2.611 \\ {y}_9=e^{1.{45}^2-1}=3.011 \\ {y}_{10}=e^{1.5^2-1}=3.49 \end{gathered}$$

Find the actual value and approximation value for h = 0.1.

Actual value and approximation for x₀ = 0.1 is

$$\begin{gathered} y_1=1.108 \\ {y}_1=1.1 \end{gathered}$$

Now,

$\begin{array}{rcl}Absoluteerror& =& \left|1.108-1.1\right|\\ & =& \left|0.008\right|\\ Precentagerelativeerror& =& \frac{0.008}{\left|1.108\right|}\times 100\\ & =& 0.722\%\end{array}$

Find the actual value and approximation value for h = 0.05.

Actual value and approximation for x₀ = 0.05 is

$$\begin{gathered} y_1=1.234 \\ {y}_1=1.2 \end{gathered}$$

Now,

$\begin{array}{rcl}Absoluteerror& =& \left|1.234-1.2\right|\\ & =& \left|0.034\right|\\ Precentagerelativeerror& =& \frac{0.034}{\left|1.234\right|}\times 100\\ & =& 2.755\%\end{array}$