Use Euler’s method (4) with h= 0.1 to approximate the solution to the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{20}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$, on the interval$\mathbf{0}\mathbf{⩽}\mathbf{x}\mathbf{⩽}\mathbf{1}$ (that is, at x= 0, 0.1, . . . , 1.0).Compare your answers with the actual solutiony = e^-20x.What went wrong? Next, try the step size h= 0.025 and also h= 0.2. What conclusions can you draw concerning the choice of step size?

Here f(x,y)=-20y then

$$\begin{gathered} {\mathrm{y}}_{\mathrm{n}}={\mathrm{y}}_{\mathrm{n}-1}+\mathrm{h}(-20{\mathrm{y}}_{\mathrm{n}-1}) \\ =(1-20\mathrm{h}){\mathrm{y}}_{\mathrm{n}-1} \\ . \\ . \\ . \\ =(1-20\mathrm{h}{)}^{\mathrm{n}}{\mathrm{y}}_{\mathrm{o}} \\ ({\mathrm{y}}_0=1\left)\right(1-20\mathrm{h}{)}^{\mathrm{n}} \end{gathered}$$

Apply initial points ${\mathrm{x}}_{\mathrm{n}}={\mathrm{x}}_{\mathrm{o}}+\mathrm{nh},{\mathrm{x}}_{\mathrm{o}}=0$for

$$\begin{gathered} \mathrm{h}=0.1 \\ {\mathrm{x}}_{\mathrm{n}}=0.1\mathrm{n},{\mathrm{y}}_{\mathrm{n}}=(-1{)}^{\mathrm{n}},\mathrm{n}=1,2,.....10 \\ \mathrm{h}=0.025 \\ {\mathrm{x}}_{\mathrm{n}}=0.025\mathrm{n}, \\ {\mathrm{y}}_{\mathrm{n}}=(0.5{)}^{\mathrm{n}}, \\ \mathrm{n}=1,2,...40 \\ \mathrm{h}=0.2 \\ {\mathrm{x}}_{\mathrm{n}}=0.2\mathrm{n},{\mathrm{y}}_{\mathrm{n}}=(-3{)}^{\mathrm{n}},\mathrm{n}=1,2,...5 \end{gathered}$$

Where $\mathrm{n}=1,2,3,....\frac{1}{\mathrm{h}}$

$$\begin{gathered} ({\mathrm{x}}_1=0.1,{\mathrm{y}}_1=-1) \\ ({\mathrm{x}}_2=0.2,{\mathrm{y}}_2=1) \\ ({\mathrm{x}}_3=0.3,{\mathrm{y}}_3=-1) \\ ({\mathrm{x}}_4=0.4,{\mathrm{y}}_4=1) \\ ({\mathrm{x}}_5=0.5,{\mathrm{y}}_5=-1) \\ ({\mathrm{x}}_6=0.6,{\mathrm{y}}_6=1) \\ ({\mathrm{x}}_7=0.7,{\mathrm{y}}_7=-1) \\ ({\mathrm{x}}_8=0.8,{\mathrm{y}}_8=1) \\ ({\mathrm{x}}_9=0.9,{\mathrm{y}}_9=-1) \\ ({\mathrm{x}}_{10}=1,{\mathrm{y}}_{10}=1) \end{gathered}$$

$$\begin{gathered} ({\mathrm{x}}_1=0.1,{\mathrm{y}}_1=0.0625) \\ ({\mathrm{x}}_2=0.2,{\mathrm{y}}_2=0.003906) \\ ({\mathrm{x}}_3=0.3,{\mathrm{y}}_3=0.000244) \\ ({\mathrm{x}}_4=0.4,{\mathrm{y}}_4=0.000015) \\ ({\mathrm{x}}_5=0.5,{\mathrm{y}}_5=0.000001) \\ ({\mathrm{x}}_6=0.6,{\mathrm{y}}_6=0) \\ ({\mathrm{x}}_7=0.7,{\mathrm{y}}_7=0) \\ ({\mathrm{x}}_8=0.8,{\mathrm{y}}_8=0) \\ ({\mathrm{x}}_9=0.9,{\mathrm{y}}_9=0) \\ ({\mathrm{x}}_{10}=1,{\mathrm{y}}_{10}=0) \end{gathered}$$

$$\begin{gathered} ({\mathrm{x}}_2=0.2,{\mathrm{y}}_2=-3) \\ ({\mathrm{x}}_4=0.4,{\mathrm{y}}_4=9) \\ ({\mathrm{x}}_6=0.6,{\mathrm{y}}_6=-27) \\ ({\mathrm{x}}_8=0.8,{\mathrm{y}}_8=81) \\ ({\mathrm{x}}_{10}=1,{\mathrm{y}}_{10}=-243) \end{gathered}$$

Hence the solution is

The result is for h=0.1 the values are 1 and -1. The values for h=0.2 the results are increasing and decreasing with different signs. For h=0.025 is the best approximation result.