Chapter 7.2, Problem 23E

To determine

To calculate : the $y\left(0.5\right)$ .

Answer to Problem 23E

The value is $y\left(0.5\right)=1.76$ .

Explanation of Solution

Given information :

The initial value problem is $y\text{'}=y+xy,y\left(0\right)=1$ .

Formula used :

Euler’s method: approximate values for the solution of the initial-value problem

  $y\text{'}=F\left(x,y\right),y\left(x_0\right)=y_0$ with step size h , at $x_n=x_{n-1}+h$

  $y_n=y_{n-1}+hF\left(x_{n-1},y_{n-1}\right)$

Calculation : here since, $h=0.1,x_0=0,y_0=1$ and $F\left(x,y\right)=y+xy$ .

So,

  $\begin{array}{l}{y}_1=y_0+hF\left(x_0,y_0\right)\\ y_1=1+0.1F\left(0,1\right)\text{ [substitute ]}\\ y_1=1+0.1\left(1+0·1\right)\text{ [}F\left(x,y\right)=y+xy\text{ and substitute]}\\ y_1=1.1\text{ [simplify]}\end{array}$

  $\begin{array}{l}{y}_2=y_1+hF\left(x_1,y_1\right)\\ y_2=1.1+0.1F\left(0.1,1.1\right)\text{ [substitute }{x}_1\text{=0+0}\text{.1]}\\ y_2=1.1+0.1\left(1.1+1.1·0.1\right)\text{ [}F\left(x,y\right)=y+xy\text{ and substitute]}\\ y_2=1.221\text{ [simplify]}\end{array}$

  $\begin{array}{l}{y}_3=y_2+hF\left(x_2,y_2\right)\\ y_3=1.221+0.1F\left(0.2,1.221\right)\text{ [substitute }{x}_2\text{=0}\text{.1+0}\text{.1]}\\ y_3=1.221+0.1\left(1.221+1.221·0.2\right)\text{ [}F\left(x,y\right)=y+xy\text{ and substitute]}\\ y_3=1.36\text{ [simplify]}\end{array}$

  $\begin{array}{l}{y}_4=y_3+hF\left(x_3,y_3\right)\\ y_4=1.36+0.1F\left(0.3,1.36\right)\text{ [substitute }{x}_3\text{=0}\text{.2+0}\text{.1]}\\ y_4=1.36+0.1\left(1.36+1.36·0.3\right)\text{ [}F\left(x,y\right)=y+xy\text{ and substitute]}\\ y_4=1.53\text{ [simplify]}\end{array}$

  $\begin{array}{l}{y}_5=y_4+hF\left(x_4,y_4\right)\\ y_5=1.53+0.1F\left(0.4,1.53\right)\text{ [substitute }{x}_4\text{=0}\text{.3+0}\text{.1]}\\ y_5=1.53+0.1\left(1.53+1.53·0.4\right)\text{ [}F\left(x,y\right)=y+xy\text{ and substitute]}\\ y_5=1.76\text{ [simplify]}\end{array}$

This means that if $y\left(x\right)$ is exact solution, then $y\left(0.5\right)=1.76$ .

Thus the value is $y\left(0.5\right)=1.76$ .