Q5E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Use Euler’s method with step size h = 0.1 to approximate the solution to the initial value problem 

 $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}$, y (1) = 0 at the points $\mathbf{x}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{,} \mathbf{1}\mathbf{.}\mathbf{2}\mathbf{,} \mathbf{1}\mathbf{.}\mathbf{3}\mathbf{,} \mathbf{1}\mathbf{.}\mathbf{4} \text{and} \mathbf{1}\mathbf{.}\mathbf{5}$ .

Step-by-Step Solution

Answer

$x_n$	1.1	1.2	1.3	1.4	1.5
$y_n$	0.1	0.209	0.325	0.444	0.564

Write the recursive formula

For the solution use the Euler’s formula $y_{n+1}=y_n+h.f\left(x_n,y_n\right)$

Apply recursive formula

We have, $f\left(x,y\right)=x-y^2, x_0=1, y_0=0, h=0.1$

Then, $$\begin{gathered} y_{n+1}=y_n+h.f\left(x_n,y_n\right) \\ =y_n+\left(0.1\right)\left(x-y^2\right) \end{gathered}$$

Put n = 0 to find y 1

Now, find the value of  $y_1$

$$\begin{gathered} y_1=y_0+\left(0.1\right)\left(x_0-y_0^2\right) \\ =0+\left(0.1\right)\left(1\right) \\ =0.1 \end{gathered}$$

Hence, the value of  ${\mathbf{y}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}$ when  ${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{1}$

Put n = 1 to find y 2

The value of $y_2$ is

$$\begin{gathered} y_2=y_1+\left(0.1\right)\left(x_1-y_1^2\right) \\ =0.1+\left(0.1\right)\left(1.1-0.01\right) \\ =0.1+0.109 \\ =0.209 \end{gathered}$$

Consequently, the value is  ${\mathbf{y}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{209}$  when  ${\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2}$

Put n = 2 to find y 3

 Now the value of  $y_3$ 

$$\begin{gathered} y_3=y_2+\left(0.1\right)\left(x_2-y_2^2\right) \\ =0.209+\left(0.1\right)\left(1.2-0.043\right) \\ =0.209+0.116 \\ =0.325 \end{gathered}$$

So, the value of  ${\mathbf{y}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{325}$  when  ${\mathbf{x}}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{3}$

Put n = 3 to find y 4

The value of $y_4$ is

$$\begin{gathered} y_4=y_3+\left(0.1\right)\left(x_3-y_3^2\right) \\ =0.325+\left(0.1\right)\left(1.3-0.106\right) \\ =0.325+0.119 \\ =0.444 \end{gathered}$$

Thus, the value is  ${\mathbf{y}}_{\mathbf{4}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{444}$  when  ${\mathbf{x}}_{\mathbf{4}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{4}$

Put n = 4 to find y 5

The value of $y_5$ is

$$\begin{gathered} y_5=y_4+\left(0.1\right)\left(x_4-y_4^2\right) \\ =0.444+\left(0.1\right)\left(1.4-0.197\right) \\ =0.444+0.12 \\ =0.564 \end{gathered}$$

Therefore, the value is  ${\mathbf{y}}_{\mathbf{5}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{564}$ when  ${\mathbf{x}}_{\mathbf{5}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5}$

Therefore the solution is 

${\mathit{x}}_{\mathbf{n}}$	1.1	1.2	1.3	1.4	1.5
${\mathit{y}}_{\mathbf{n}}$	0.1	0.209	0.325	0.444	0.564