Chapter 7, Problem 4RE

a.

To determine

To calculate : the value of $y\left(0.4\right)$

Answer to Problem 4RE

The value of $y\left(0.4\right)\approx 1.08$

Explanation of Solution

Given information : The question is

Use Euler’s method with step size $0.2$ where $y\left(x\right)$ is the solution of the initial value problem

  $y\text{'}=2x{y}^2\text{ ;}y\left(0\right)=1$

Calculation :

By Euler’s method $y_n=y_{n-1}+hF\left(x_{n-1},y_{n-1}\right);n=1,2,3,\mathrm{.......},$ with $F\left(x,y\right)=y\left(x,y\right)\&x_n=x_{n-1}+h$

Use $h=0.2,F\left(x,y\right)=2x{y}^2,x_0=0\&y_0=1$

  $\begin{array}{l}{y}_1=1+0.2\times \left(2\times 0\times 1^2\right)=1\\ y_2=1+0.2\times \left(2\times 0.2\times 1^2\right)=1.8\\ y\left(0.4\right)\approx 1.08\end{array}$

b.

To determine

To repeat : the part (a) with step size $0.1$

Answer to Problem 4RE

The value is $y\left(0.4\right)\approx 1.1292$

Explanation of Solution

Given information : The question is

Use Euler’s method with step size $0.1$ where $y\left(x\right)$ is the solution of the initial value problem

  $y\text{'}=2x{y}^2\text{ ;}y\left(0\right)=1$

Calculation :

By Euler’s method $y_n=y_{n-1}+hF\left(x_{n-1},y_{n-1}\right);n=1,2,3,\mathrm{.......},$ with $F\left(x,y\right)=y\left(x,y\right)\&x_n=x_{n-1}+h$

Use $h=0.2,F\left(x,y\right)=2x{y}^2,x_0=0\&y_0=1$

  $\begin{array}{l}{y}_1=1+0.1\times \left(2\times 0\times 1^2\right)=1\\ y_2=1+0.1\times \left(2\times 0.1\times 1^2\right)=1.02\\ y_3=1.02+0.1\times \left(2\times 0.2\times {1.02}^2\right)\approx 1.0616\\ y_4=1.0616+0.1\times \left(2\times 0.3\times {1.0616}^2\right)\approx 1.1292\end{array}$

c.

To determine

To find : the exact value of differential equation

Answer to Problem 4RE

The exact value is $y\left(0.4\right)\approx 1.19048$

Explanation of Solution

Given information : the question is

Compare the value of $0.4$ with part (a) and part (b)

Calculation :

Compare the value

  $\begin{array}{l}\frac{dy}{dx}=2x{y}^2\\ \frac{dy}{y^2}=2xdx\\ {\displaystyle \int \frac{dy}{y^2}={\displaystyle \int 2xdx}}\\ -\frac{1}{y}=x^2+C\\ y\left(0\right)=1\\ -\frac{1}{1}={\left(0\right)}^2+C\\ -\frac{1}{y}=x^2-1\\ \frac{1}{y}=1-x^2\\ y=\frac{1}{1-x^2}\\ \text{at }x=0.4\\ y\left(0.4\right)=\frac{1}{1-{\left(0.4\right)}^2}=\frac{1}{0.84}\approx 1.19048\\ y\left(0.4\right)\approx 1.19048\end{array}$