Question

Show that when the improved Euler’s method is used to approximate the solution of the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{4}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}$, at$\mathbf{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$ , then the approximation with step size his$\frac{\mathbf{1}}{\mathbf{3}}{\left(1+4h+8h^2\right)}^{\frac{\mathbf{1}}{\mathbf{2}\mathbf{h}}}$ .

Short answer

Proved

Step-by-step solution

Find the value of  yn

Here f(x, y) = 4y. Apply the formula

$$\begin{gathered} {\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}+\frac{\mathrm{h}}{2}\left[\mathrm{f}({\mathrm{x}}_{\mathrm{n}},{\mathrm{y}}_{\mathrm{n}})+\mathrm{f}({\mathrm{x}}_{\mathrm{n}}+\mathrm{h},{\mathrm{y}}_{\mathrm{n}}+\mathrm{hf}({\mathrm{x}}_{\mathrm{n}},{\mathrm{y}}_{\mathrm{n}})\right] \\ ={\mathrm{y}}_{\mathrm{n}}+\frac{\mathrm{h}}{2}\left[4{\mathrm{y}}_{\mathrm{n}})+\mathrm{f}({\mathrm{x}}_{\mathrm{n}}+\mathrm{h},{\mathrm{y}}_{\mathrm{n}}+4{\mathrm{hy}}_{\mathrm{n}})\right] \\ ={\mathrm{y}}_{\mathrm{n}}+\frac{\mathrm{h}}{2}\left[4{\mathrm{y}}_{\mathrm{n}})+4({\mathrm{y}}_{\mathrm{n}}+4{\mathrm{hy}}_{\mathrm{n}})\right] \\ ={\mathrm{y}}_{\mathrm{n}}+\frac{\mathrm{h}}{2}\left[8{\mathrm{y}}_{\mathrm{n}})+16\mathrm{h}{\mathrm{y}}_{\mathrm{n}})\right] \\ ={\mathrm{y}}_{\mathrm{n}}+4{\mathrm{hy}}_{\mathrm{n}}+8{\mathrm{h}}^2{\mathrm{y}}_{\mathrm{n}} \\ {\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}(1+4\mathrm{h}+8{\mathrm{h}}^2) \end{gathered}$$

Evaluate the approximation value for x=12 .

Since $\mathrm{y}\left(0\right)=\frac{1}{3}$

$$\begin{gathered} {\mathrm{y}}_1=\frac{1}{3}(1+4\mathrm{h}+8{\mathrm{h}}^2) \\ {\mathrm{y}}_2=\frac{1}{3}(1+4\mathrm{h}+8{\mathrm{h}}^2{)}^2 \\ {\mathrm{y}}_3=\frac{1}{3}(1+4\mathrm{h}+8{\mathrm{h}}^2{)}^3 \\ . \\ . \\ . \\ {\mathrm{y}}_{\mathrm{n}}=\frac{1}{3}(1+4\mathrm{h}+8{\mathrm{h}}^2{)}^{\mathrm{n}} \end{gathered}$$

Since ${\mathrm{x}}_{\mathrm{o}}=0,\mathrm{x}=\frac{1}{2}$ then

$$\begin{gathered} \frac{1}{2}={\mathrm{x}}_{\mathrm{o}}+\mathrm{nh}=\mathrm{nh} \\ \mathrm{n}=\frac{1}{2\mathrm{h}} \end{gathered}$$

Then

${\mathrm{y}}_{\mathrm{n}}=\frac{1}{3}(1+4\mathrm{h}+8{\mathrm{h}}^2{)}^{\frac{1}{2\mathrm{h}}}$

Hence it is proved that role="math" localid="1664233218531" ${\mathbf{y}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{4}\mathbf{h}\mathbf{+}\mathbf{8}{\mathbf{h}}^{\mathbf{2}}{\mathbf{)}}^{\frac{\mathbf{1}}{\mathbf{2}\mathbf{h}}}$