# 3.1E: Euler’s Method (Exercises)

- Page ID
- 18259

## Q3.1.1

You may want to save the results of these exercises, since we will revisit in the next two sections. In *Exercises 3.1.1-3.1.5* use Euler’s method to find approximate values of the solution of the given initial value problem at the points \(x_i=x_0+ih\), where \(x_0\) is the point where the initial condition is imposed and \(i=1\), \(2\), \(3\). The purpose of these exercises is to familiarize you with the computational procedure of Euler’s method.

1. \(y'=2x^2+3y^2-2,\quad y(2)=1;\quad h=0.05\)

2. \(y'=y+\sqrt{x^2+y^2},\quad y(0)=1;\quad h=0.1\)

3. \(y'+3y=x^2-3xy+y^2,\quad y(0)=2;\quad h=0.05\)

4. \(y'= {1+x\over1-y^2},\quad y(2)=3;\quad h=0.1\)

5. \(y'+x^2y=\sin xy,\quad y(1)=\pi;\quad h=0.2\)

## Q3.1.2

6. Use Euler’s method with step sizes \(h=0.1\), \(h=0.05\), and \(h=0.025\) to find approximate values of the solution of the initial value problem \[y'+3y=7e^{4x},\quad y(0)=2\] at \(x=0\), \(0.1\), \(0.2\), \(0.3\), …, \(1.0\). Compare these approximate values with the values of the exact solution \(y=e^{4x}+e^{-3x}\), which can be obtained by the method of Section 2.1. Present your results in a table like *Table 3.1.1*.

7. Use Euler’s method with step sizes \(h=0.1\), \(h=0.05\), and \(h=0.025\) to find approximate values of the solution of the initial value problem \[y'+{2\over x}y={3\over x^3}+1,\quad y(1)=1\] at \(x=1.0\), \(1.1\), \(1.2\), \(1.3\), …, \(2.0\). Compare these approximate values with the values of the exact solution \[y={1\over3x^2}(9\ln x+x^3+2),\] which can be obtained by the method of Section 2.1. Present your results in a table like *Table 3.1.1*.

8. Use Euler’s method with step sizes \(h=0.05\), \(h=0.025\), and \(h=0.0125\) to find approximate values of the solution of the initial value problem \[y'={y^2+xy-x^2\over x^2},\quad y(1)=2\] at \(x=1.0\), \(1.05\), \(1.10\), \(1.15\), …, \(1.5\). Compare these approximate values with the values of the exact solution \[y={x(1+x^2/3)\over1-x^2/3}\] obtained in Example [example:2.4.3}. Present your results in a table like *Table 3.1.1*.

9. In Example [example:2.2.3} it was shown that \[y^5+y=x^2+x-4\] is an implicit solution of the initial value problem \[y'={2x+1\over5y^4+1},\quad y(2)=1. \tag{A}\] Use Euler’s method with step sizes \(h=0.1\), \(h=0.05\), and \(h=0.025\) to find approximate values of the solution of (A) at \(x=2.0\), \(2.1\), \(2.2\), \(2.3\), …, \(3.0\). Present your results in tabular form. To check the error in these approximate values, construct another table of values of the residual \[R(x,y)=y^5+y-x^2-x+4\] for each value of \((x,y)\) appearing in the first table.

10. You can see from *Example 2.5.1* that \[x^4y^3+x^2y^5+2xy=4\] is an implicit solution of the initial value problem \[y'=-{4x^3y^3+2xy^5+2y\over3x^4y^2+5x^2y^4+2x},\quad y(1)=1. \tag{A}\] Use Euler’s method with step sizes \(h=0.1\), \(h=0.05\), and \(h=0.025\) to find approximate values of the solution of (A) at \(x=1.0\), \(1.1\), \(1.2\), \(1.3\), …, \(2.0\). Present your results in tabular form. To check the error in these approximate values, construct another table of values of the residual \[R(x,y)=x^4y^3+x^2y^5+2xy-4\] for each value of \((x,y)\) appearing in the first table.

11. Use Euler’s method with step sizes \(h=0.1\), \(h=0.05\), and \(h=0.025\) to find approximate values of the solution of the initial value problem \[(3y^2+4y)y'+2x+\cos x=0, \quad y(0)=1; \quad\text{(Exercise 2.2.13)}\] at \(x=0\), \(0.1\), \(0.2\), \(0.3\), …, \(1.0\).

12. Use Euler’s method with step sizes \(h=0.1\), \(h=0.05\), and \(h=0.025\) to find approximate values of the solution of the initial value problem \[y'+{(y+1)(y-1)(y-2)\over x+1}=0, \quad y(1)=0 \quad\text{(Exercise 2.2.14)}\] at \(x=1.0\), \(1.1\), \(1.2\), \(1.3\), …, \(2.0\).

13. Use Euler’s method and the Euler semilinear method with step sizes \(h=0.1\), \(h=0.05\), and \(h=0.025\) to find approximate values of the solution of the initial value problem \[y'+3y=7e^{-3x},\quad y(0)=6\]

at \(x=0\), \(0.1\), \(0.2\), \(0.3\), …, \(1.0\). Compare these approximate values with the values of the exact solution \(y=e^{-3x}(7x+6)\), which can be obtained by the method of Section 2.1. Do you notice anything special about the results? Explain.

## Q3.1.3

The linear initial value problems in *Exercises 3.1.14–3.1.19* can’t be solved exactly in terms of known elementary functions. In each exercise, use Euler’s method and the Euler semilinear methods with the indicated step sizes to find approximate values of the solution of the given initial value problem at 11 equally spaced points (including the endpoints) in the interval.

14. \(y'-2y= {1\over1+x^2},\quad y(2)=2\); \(h=0.1,0.05,0.025\) on \([2,3]\)

15. \(y'+2xy=x^2,\quad y(0)=3 \quad\text{(Exercise 2.1.38)};\quad\) \(h=0.2,0.1,0.05\) on \([0,2]\)

16. \( {y'+{1\over x}y={\sin x\over x^2},\quad y(1)=2}; \quad\text{(Exercise 2.1.39)};\quad\) \(h=0.2,0.1,0.05\) on \([1,3]\)

17. \( {y'+y={e^{-x}\tan x\over x},\quad y(1)=0}; \quad\text{(Exercise 2.1.40)};\quad\) \(h=0.05,0.025,0.0125\) on \([1,1.5]\)

18. \( {y'+{2x\over 1+x^2}y={e^x\over (1+x^2)^2}, \quad y(0)=1};\quad\text{(Exercise 2.1.41)};\quad\) \(h=0.2,0.1,0.05\) on \([0,2]\)

19. \(xy'+(x+1)y=e^{x^2},\quad y(1)=2; \quad\text{(Exercise 2.1.42)};\quad\) \(h=0.05,0.025,0.0125\) on \([1,1.5]\)

## Q3.1.4

In *Exercises 3.1.20-3.1.22*, use Euler’s method and the Euler semilinear method with the indicated step sizes to find approximate values of the solution of the given initial value problem at 11 equally spaced points (including the endpoints) in the interval.

20. \(y'+3y=xy^2(y+1),\quad y(0)=1\); \(h=0.1,0.05,0.025\) on \([0,1]\)

21. \( {y'-4y={x\over y^2(y+1)},\quad y(0)=1}\); \(h=0.1,0.05,0.025\) on \([0,1]\)

22. \( {y'+2y={x^2\over1+y^2},\quad y(2)=1}\); \(h=0.1,0.05,0.025\) on \([2,3]\)

## Q3.1.5

23. Numerical Quadrature. The fundamental theorem of calculus says that if \(f\) is continuous on a closed interval \([a,b]\) then it has an antiderivative \(F\) such that \(F'(x)=f(x)\) on \([a,b]\) and \[\int_a^bf(x)\,dx=F(b)-F(a). \tag{A}\] This solves the problem of evaluating a definite integral if the integrand \(f\) has an antiderivative that can be found and evaluated easily. However, if \(f\) doesn’t have this property, (A) doesn’t provide a useful way to evaluate the definite integral. In this case we must resort to approximate methods. There’s a class of such methods called *numerical quadrature*, where the approximation takes the form \[\int_a^bf(x)\,dx\approx \sum_{i=0}^n c_if(x_i), \tag{B}\] where \(a=x_0<x_1<\cdots<x_n=b\) are suitably chosen points and \(c_0\), \(c_1\), …, \(c_n\) are suitably chosen constants. We call (B) a *quadrature formula*.

- Derive the quadrature formula \[\int_a^bf(x)\,dx\approx h\sum_{i=0}^{n-1}f(a+ih) \tag{C}\] where \(h=(b-a)/n)\) by applying Euler’s method to the initial value problem\[y'=f(x),\quad y(a)=0.\]
- The quadrature formula (C) is sometimes called
*the left rectangle rule*. Draw a figure that justifies this terminology. - For several choices of \(a\), \(b\), and \(A\), apply (C) to \(f(x)=A\) with \(n = 10,20,40,80,160,320\). Compare your results with the exact answers and explain what you find.
- For several choices of \(a\), \(b\), \(A\), and \(B\), apply (C) to \(f(x)=A+Bx\) with \(n=10\), \(20\), \(40\), \(80\), \(160\), \(320\). Compare your results with the exact answers and explain what you find.