3.1E: Euler’s Method (Exercises)

You may want to save the results of these exercises, since we will revisit in the next two sections. In Exercises [exer:3.1.1}– [exer: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.

[exer:3.1.1} \(y'=2x^2+3y^2-2,\quad y(2)=1;\quad h=0.05\)

[exer:3.1.2} \(y'=y+\sqrt{x^2+y^2},\quad y(0)=1;\quad h=0.1\)

[exer:3.1.3} \(y'+3y=x^2-3xy+y^2,\quad y(0)=2;\quad h=0.05\)

[exer:3.1.4} \(y'= {1+x\over1-y^2},\quad y(2)=3;\quad h=0.1\)

[exer:3.1.5} \(y'+x^2y=\sin xy,\quad y(1)=\pi;\quad h=0.2\)

[exer:3.1.6} 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 [table:3.1.1}.

[exer:3.1.7} 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 [table:3.1.1}.

[exer:3.1.8} 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 [table:3.1.1}.

[exer:3.1.9} 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 \[y'={2x+1\over5y^4+1},\quad y(2)=1. \eqno{\rm(A)}\] 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. 

[exer:3.1.10} You can see from Example [example:2.5.1} that \[x^4y^3+x^2y^5+2xy=4\] is an implicit solution of the initial value problem \[y'=-{4x^3y^3+2xy^5+2y\over3x^4y^2+5x^2y^4+2x},\quad y(1)=1. \eqno{\rm(A)}\] 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.

[exer:3.1.11} 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 \[(3y^2+4y)y'+2x+\cos x=0, \quad y(0)=1; \mbox{\; (Exercise~2.2.~\hspace*{-3pt}\ref{exer:2.2.13})}\] at \(x=0\), \(0.1\), \(0.2\), \(0.3\), …, \(1.0\).

[exer:3.1.12} 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 \[y'+{(y+1)(y-1)(y-2)\over x+1}=0, \quad y(1)=0 \mbox{\; (Exercise~2.2.~\hspace*{-3pt}\ref{exer:2.2.14})}\] at \(x=1.0\), \(1.1\), \(1.2\), \(1.3\), …, \(2.0\).

[exer:3.1.13} 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.

[exer:3.1.14} \(y'-2y= {1\over1+x^2},\quad y(2)=2\); \(h=0.1,0.05,0.025\) on \([2,3}\)

[exer:3.1.15} \(y'+2xy=x^2,\quad y(0)=3\) (Exercise 2.1. [exer:2.1.38}); \(h=0.2,0.1,0.05\) on \([0,2}\)

[exer:3.1.16} \( {y'+{1\over x}y={\sin x\over x^2},\quad y(1)=2}\); (Exercise 2.1. [exer:2.1.39}); \(h=0.2,0.1,0.05\) on \([1,3}\)

[exer:3.1.17} \( {y'+y={e^{-x}\tan x\over x},\quad y(1)=0}\); (Exercise 2.1. [exer:2.1.40}); \(h=0.05,0.025,0.0125\) on \([1,1.5}\)

[exer:3.1.18} \( {y'+{2x\over 1+x^2}y={e^x\over (1+x^2)^2}, \quad y(0)=1}\); (Exercise 2.1. [exer:2.1.41}); \(h=0.2,0.1,0.05\) on \([0,2}\)

[exer:3.1.19} \(xy'+(x+1)y=e^{x^2},\quad y(1)=2\); (Exercise 2.1. [exer:2.1.42}); \(h=0.05,0.025,0.0125\) on \([1,1.5}\)

[exer:3.1.20} \(y'+3y=xy^2(y+1),\quad y(0)=1\); \(h=0.1,0.05,0.025\) on \([0,1}\)

[exer:3.1.21} \( {y'-4y={x\over y^2(y+1)},\quad y(0)=1}\); \(h=0.1,0.05,0.025\) on \([0,1}\)

[exer:3.1.22} \( {y'+2y={x^2\over1+y^2},\quad y(2)=1}\); \(h=0.1,0.05,0.025\) on \([2,3}\)

[exer:3.1.23} 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 \[\int_a^bf(x)\,dx=F(b)-F(a). \eqno{\rm(A)}\] 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 \[\int_a^bf(x)\,dx\approx \sum_{i=0}^n c_if(x_i), \eqno{\rm(B)}\] 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.

1. 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\[y'=f(x),\quad y(a)=0.\]
2. The quadrature formula (C) is sometimes called the left rectangle rule. Draw a figure that justifies this terminology.
3. 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.
4. 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.