Chapter 8: Problem 1
Solve the following problems using Euler's method with stepsizes of \(h=0.2,0.1\), 0.05. Compute the error and relative error using the true answer \(Y(x) .\) For selected values of \(x\), observe the ratio by which the error decreases when \(h\) is halved. \(\begin{array}{ll}\text { (a) } \quad & Y^{\prime}(x)=[\cos (Y(x))]^{2}, \quad 0 \leq x \leq 10, \quad Y(0)=0 ; \\\ Y(x)=\tan ^{-1}(x) & & \\ \text { (b) } \quad Y^{\prime}(x)=\frac{1}{1+x^{2}}-2[Y(x)]^{2}, \quad 0 \leq x \leq 10, \quad Y(0)=0 \\ Y(x)=\frac{x}{1+x^{2}} & \\ \text { (c) } \quad Y^{\prime}(x)=\frac{1}{4} Y(x)\left[1-\frac{1}{20} Y(x)\right], \quad 0 \leq x \leq 20, \quad Y(0)=1 ; \\ Y(x)= & \frac{20}{1+19 e^{-x / 4}}\end{array}\) (d) \(Y^{\prime}(x)=-[Y(x)]^{2}, \quad 1 \leq x \leq 10, \quad Y(1)=1 ; \quad Y(x)=\frac{1}{x}\) (e) \(\quad Y^{\prime}(x)=x e^{-x}-Y(x), \quad 0 \leq x \leq 10, \quad Y(0)=1 ;\) \(y(x)=\left(1=\frac{1}{2} x^{2}\right) e^{-x}\) (f) \(\quad Y^{\prime}(x)=\frac{x^{3}}{Y(x)}, \quad 0 \leq x \leq 10, \quad Y(0)=1 ; \quad Y(x)=\sqrt{\frac{1}{2} x^{4}+1}\) (g) \(\quad Y^{\prime}(x)=\left(3 x^{2}-1\right) Y(x)^{2}, \quad 0 \leq x \leq 10, \quad Y(0)=1\); \(y(x)=\sqrt{3\left(x^{3}-x\right)+1}\)
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.