Question

Show that when Euler’s method is used to approximate the solution of the initial value problem ${\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}$,at x = 2, then the approximation with step size h is$\mathbf{3}\mathbf{(}\mathbf{1}\mathbf{-}\frac{\mathbf{h}}{\mathbf{2}}{\mathbf{)}}^{\frac{\mathbf{2}}{\mathbf{h}}}$ .

Short answer

Proved

Step-by-step solution

Apply Euler’s formula

Here ${\mathrm{y}}^{\text{'}}=-\frac{1}{2}\mathrm{y},\mathrm{y}\left(0\right)=3$

Apply the Euler’s formula

${\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}+\mathrm{hf}({\mathrm{x}}_{\mathrm{n}},{\mathrm{y}}_{\mathrm{n}})$

In our case of differential equation our equation is

${\mathrm{y}}_{\mathrm{n}}=3(1-\frac{\mathrm{h}}{2}{)}^{\mathrm{n}}$

The interval is

$$\begin{gathered} \frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}=\frac{2-0}{\mathrm{n}}=\frac{2}{\mathrm{n}} \\ \mathrm{n}=\frac{2}{\mathrm{h}} \end{gathered}$$

substitute in the Euler’s formula 

Now,

$$\begin{gathered} {\mathrm{y}}_1=3\left(1-\frac{\mathrm{h}}{2}\right) \\ {\mathrm{y}}_2=3{\left(1-\frac{\mathrm{h}}{2}\right)}^2 \\ . \\ . \\ . \\ {\mathrm{y}}_{\mathrm{n}}=3{\left(1-\frac{\mathrm{h}}{2}\right)}^{\mathrm{n}} \\ {\mathrm{y}}_{\mathrm{n}}=3{\left(1-\frac{\mathrm{h}}{2}\right)}^{\frac{2}{\mathrm{h}}} \end{gathered}$$

Hence it is proved that ${\mathbf{y}}_{\mathbf{n}}\mathbf{=}\mathbf{3}{\left(\mathbf{1}\mathbf{-}\frac{\mathbf{h}}{\mathbf{2}}\right)}^{\frac{\mathbf{2}}{\mathbf{h}}}$