Chapter 7.2, Problem 25E

a.

To determine

To write: A program to use Euler’s method.

Answer to Problem 25E

  $\begin{array}{l}\#include<stdio.h>\hfill \\ \#include<math.h>\hfill \\ \hfill \\ floatmain()\hfill \\ \{\hfill \\ floath,y,x,F;\hfill \\ inti;\hfill \\ printf\left("Putthevalueofh\backslash n"\right);\hfill \\ scanf\left("\%f",\&h\right);\hfill \\ y=3;\hfill \\ x=0;\hfill \\ for\left(i=1;i<=\left(1/h\right);i=i+1\right)\hfill \\ \{\hfill \\ F=6*\left(x*x\right)-3*y*\left(x*x\right);\hfill \\ x=x+h;\hfill \\ y=y+h*F;\hfill \\ \}\hfill \\ printf\left("\%f",y\right);\hfill \\ \}\hfill \end{array}$

Explanation of Solution

Given information:

  $\frac{dy}{dx}+3x^{2}y=6x^2$

  $y\left(0\right)=3$

  $h=1,h=0.1,h=0.01,\text{ h=0}\text{.001}$

Concept used :

Here, we use C language to program a computer.

Calculation:

  $\begin{array}{l}\frac{dy}{dx}+3x^{2}y=6x^2\\ \Rightarrow \frac{dy}{dx}=6x^2-3x^{2}y\end{array}$

The required program can be written as following,

#include 
#include 

float main()
{
  float h,y,x,F;
  int i;
  printf("Put the value of h\n");
  scanf("%f",&h);
  y=3;
  x=0;
  for(i=1;i<=(1/h);i=i+1)
  {
    F=6*(x*x)-3*y*(x*x);
    x=x+h;
    y=y+h*F;
  }
    printf("%f",y);
}

We put the given step size $h$ and get the corresponding result.

b.

To determine

To verify: $y=2+e^{-x^3}$ is the exact solution of the differential equation.

Explanation of Solution

Given information:

Differential equation: $\frac{dy}{dx}+3x^{2}y=6x^2$

  $y=2+e^{-x^3}$

Formula used:

  $\frac{d}{dx}\left(e^x\right)=e^x$

  $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Calculation:

  $\begin{array}{l}y=2+e^{-x^3}\\ \Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(2\right)+\frac{d}{dx}\left(e^{-x^3}\right)=-3x^2{e}^{-x^3}\end{array}$

Substituting $\frac{dy}{dx}=-3x^2{e}^{-x^3}$ to the right hand side of the given differential equation, we get,

  $\begin{array}{l}-3x^2{e}^{-x^3}+3x^{2}y\\ \Rightarrow 3x^2\left(-3e^{-x^3}+y\right)\\ \Rightarrow 3x^2\left(-e^{-x^3}+2+e^{-x^3}\right)\left[\because y=2+e^{-x^3}\right]\\ \Rightarrow 6x^2\left[\text{verified}\right]\end{array}$

c.

To determine

To find: The error in Euler’s method to compute $y\left(1\right)$

Answer to Problem 25E

  $h=1$

$y\left(1\right)$ obtained using Euler’s method	Exact value of $y\left(1\right)$	error
3	2.367879	$-0.632121$

  $h=0.1$

$y\left(1\right)$ obtained using Euler’s method	Exact value of $y\left(1\right)$	error
2.392794	2.367879	$-0.024915$

  $h=0.01$

$y\left(1\right)$ obtained using Euler’s method	Exact value of $y\left(1\right)$	error
2.370111	2.367879	$-0.002232$

  $h=0.001$

$y\left(1\right)$ obtained using Euler’s method	Exact value of $y\left(1\right)$	error
2.369212	2.367879	$-0.001333$

As the step size is divided by $10$ the mod value of error decrease.

Explanation of Solution

Given information:

The exact solution of the differential equation is $y=2+e^{-x^3}$

Formula used :

  $\text{error }=\text{ exact value }-\text{ value obtained}$

We use the computer program to get the estimates of Euler’s method.

Calculation:

  $h=1$

$y\left(1\right)$ obtained using Euler’s method	Exact value of $y\left(1\right)$	error
3	2.367879	$-0.632121$

  $h=0.1$

$y\left(1\right)$ obtained using Euler’s method	Exact value of $y\left(1\right)$	error
2.392794	2.367879	$-0.024915$

  $h=0.01$

$y\left(1\right)$ obtained using Euler’s method	Exact value of $y\left(1\right)$	error
2.370111	2.367879	$-0.002232$

  $h=0.001$

$y\left(1\right)$ obtained using Euler’s method	Exact value of $y\left(1\right)$	error
2.369212	2.367879	$-0.001333$

As the step size is divided by $10$ the mod value of error decrease.