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3.1E: Euler’s Method (Exercises)

• • Contributed by William F. Trench
• Andrew G. Cowles Distinguished Professor Emeritus (Mathamatics) at Trinity University

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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.