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Expert-verified Found in: Page 47 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # Question: Consider the initial value problem $\frac{\mathbf{dy}}{\mathbf{dx}}{\mathbf{+}}\sqrt{\mathbf{1}\mathbf{+}{\mathbf{sin}}^{\mathbf{2}}\mathbf{xy}}{\mathbf{=}}{\mathbf{x}}{\mathbf{,}}{\mathbf{y}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}{\mathbf{=}}{\mathbf{2}}$.(a) Using definite integration, show that the integrating factor for the differential equation can be written as ${\mathbf{\mu }}\left(x\right){\mathbf{=}}\left({\int }_{0}^{x}\sqrt{1+{\mathrm{sin}}^{2}\mathrm{tdt}}\right)$ and that the solution to the initial value problem is ${\mathbf{y}}\left(x\right){\mathbf{=}}\frac{\mathbf{1}}{\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{{\mathbf{\int }}}_{{\mathbf{0}}}^{{\mathbf{x}}}{\mathbf{\mu }}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{sds}\mathbf{+}\frac{\mathbf{2}}{\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}}$(b) Obtain an approximation to the solution at x = 1 by using numerical integration (such as Simpson’s rule, Appendix C) in a nested loop to estimate values of ${\mathbf{\mu }}\left(x\right)$and, thereby, the value of ${{\mathbf{\int }}}_{{\mathbf{0}}}^{{\mathbf{1}}}{\mathbf{\mu }}\left(s\right){\mathbf{ds}}$.[Hint: First, use Simpson’s rule to approximate ${\mathbf{\mu }}\left(x\right)$ at x = 0.1, 0.2, . . . , 1. Then use these values and apply Simpson’s rule again to approximate ${{\mathbf{\int }}}_{{\mathbf{0}}}^{{\mathbf{1}}}{\mathbf{\mu }}\mathbf{\left(}\mathbf{s}\mathbf{\right)}{\mathbf{ds}}$](c) Use Euler’s method (Section 1.4) to approximate the solution at x = 1, with step sizes h = 0.1 and 0.05. [A direct comparison of the merits of the two numerical schemes in parts (b) and (c) is very complicated, since it should take into account the number of functional evaluations in each algorithm as well as the inherent accuracies.]

1. 0.9960
2. 0.9486
3. 0.9729
See the step by step solution

## Step 1(a): Find the value of y(x)

Since $\mathrm{P}\left(\mathrm{t}\right)=\sqrt{1+{sin}^{2}\mathrm{t}}$ is a continuous real valued function on an open interval which contains points 0 we can use fundamental theorem of calculus to obtain

$\frac{\mathrm{d}}{\mathrm{dx}}\left[{\int }_{0}^{\mathrm{x}}\mathrm{P}\left(\mathrm{t}\right)\mathrm{dt}\right]=\mathrm{P}\left(\mathrm{x}\right)={\int }_{0}^{\mathrm{x}}\mathrm{P}\left(\mathrm{t}\right)\mathrm{dt}=\int \mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}$

We can write integrating factor as $\mathrm{\mu }\left(\mathrm{x}\right)=\mathrm{exp}\left({\int }_{0}^{\mathrm{x}}\sqrt{1+{\mathrm{sin}}^{2}\mathrm{t}}\mathrm{dt}\right)\right)$

Multiplying by $\mu \left(x\right)$

$\frac{\mathrm{d}}{\mathrm{dx}}\overline{)\left(\mathrm{\mu }\left(\mathrm{x}\right)\mathrm{y}\left(\mathrm{x}\right)\right)}=\mathrm{x\mu }\left(\mathrm{x}\right)=\mathrm{d\mu }\left(\mathrm{x}\right)\mathrm{y}\left(\mathrm{x}\right)=\mathrm{x\mu }\left(\mathrm{x}\right)\mathrm{y}\left(\mathrm{x}\right)$

Now,

$\mu \left(x\right){\overline{)y\left(x\right)}}_{0}^{s}={\int }_{0}^{s}\left(x\mu \left(x\right)\right)=\mu \left(s\right)y\left(s\right)-\mu \left(0\right)y\left(0\right)={\int }_{0}^{s}x\mu \left(x\right)\phantom{\rule{0ex}{0ex}}\mu \left(s\right)y\left(s\right)={\int }_{0}^{s}x\mu \left(x\right)dx+2\phantom{\rule{0ex}{0ex}}y\left(s\right)=\frac{1}{\mu \left(s\right)}{\int }_{0}^{s}x\mu \left(x\right)dx+\frac{2}{\mu \left(s\right)}$

## Step 2: Find the values of μ(x)

Here $y\left(s\right)=\frac{1}{\mu \left(s\right)}{\int }_{0}^{s}x\mu \left(x\right)dx+\frac{2}{\mu \left(s\right)}$

Put x=1 then $y\left(1\right)=\frac{1}{\mu \left(1\right)}{\int }_{0}^{s}x\mu \left(s\right)dx+\frac{2}{\mu \left(1\right)}$

We find the value of${\int }_{0}^{1}\mu \left(s\right)sdx$.

Using Simpson’s rule

${\int }_{a}^{b}f\left(x\right)dx=\frac{∆x}{3}\sum _{k=1}^{n}\left[f\left({x}_{2k-2}\right)+4f\left({x}_{2k-2}\right)+f\left({x}_{2k}\right)\right]\phantom{\rule{0ex}{0ex}}∆x=\frac{b-a}{2n},{x}_{k}=a+k∆x,k=0.1,0.2,...,1$

So, $\mathrm{\mu }\left(\mathrm{x}\right)=\mathrm{exp}\left({\int }_{0}^{\mathrm{x}}\sqrt{1+{\mathrm{sin}}^{2}\mathrm{t}}\mathrm{dt}\right)\right)$ at x=0.1,0.2, ……1

The values are

x = 0, =1

x = 0.1, =1.105354

x = 0.2, =1.223010

x = 0 .3, =1.355761

x = 0 .4, =1.506975

x = 0.5, =1.680635

x = 0 .6, =1.881401

x = 0 .7, =2.114679

x = 0 .8, =2.386713

x = 0 .9, =2.70670

x = 1, =3.076723

Now, using the previous conclusions

$∆x=\frac{1}{2n}=0.1=n=5$

So,

role="math" localid="1663932388162" $\begin{array}{rcl}{\int }_{0}^{1}s\mu \left(s\right)ds& =& \frac{0.1}{3}\left[0·\mu \left(0\right)+4\left(0.1\right)·\mu \left(0.1\right)+2\left(0.2\right)·\mu \left(0.2\right)+4\left(0.3\right)·\mu \left(0.3\right)+2\left(0.4\right)·\mu \left(0.4\right)+4\left(0.5\right)·\mu \left(0.5\right)+2\left(0.6\right)·\mu \left(0.6\right)\\ & & +4\left(0.7\right)·\mu \left(0.7\right)+2\left(0.8\right)·\mu \left(0.8\right)+4\left(0.9\right)·\mu \left(0.9\right)+1·\mu \left(1\right)\right]\\ & =& 1.064539\end{array}$role="math" localid="1663932537871" $\begin{array}{rcl}y\left(1\right)& =& \frac{1}{\mu \left(1\right)}{\int }_{0}^{s}x\mu \left(s\right)dx+\frac{2}{\mu \left(1\right)}\\ & =& 0.9960\end{array}$

Therefore, ${\mathbit{y}}\left(1\right){\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{9960}}$

## Step 3(b): Find the value of  y(1) at h=0.1

The differential equation is $y\text{'}=x-\sqrt{1+{\mathrm{sin}}^{2}xy}$

Use the recursive formula

localid="1663933001609" $\begin{array}{rcl}{x}_{n+1}& =& {x}_{n}+h\\ {y}_{n+1}& =& {y}_{n}+hf\left({x}_{n}{y}_{n}\right)\\ & =& {y}_{n}+hf\left({x}_{n}-\sqrt{1+{\mathrm{sin}}^{2}{x}_{n}{y}_{n}}\right)\end{array}$

Where,

${x}_{0}=0$ and ${y}_{0}=0$, h=0.1, N=10 steps at x=1

The values are

${x}_{1}=0.1,{y}_{1}=1.8\phantom{\rule{0ex}{0ex}}{x}_{2}=0.2,{y}_{2}=1.6921\phantom{\rule{0ex}{0ex}}{x}_{3}=0.3,{y}_{3}=1.4830\phantom{\rule{0ex}{0ex}}{x}_{4}=0.4,{y}_{4}=1.3584\phantom{\rule{0ex}{0ex}}{x}_{5}=0.5,{y}_{5}=1.2526\phantom{\rule{0ex}{0ex}}{x}_{6}=0.6,{y}_{6}=1.1637\phantom{\rule{0ex}{0ex}}{x}_{7}=0.7,{y}_{7}=1.09\phantom{\rule{0ex}{0ex}}{x}_{8}=0.8,{y}_{8}=1.0304\phantom{\rule{0ex}{0ex}}{x}_{9}=0.9,{y}_{9}=0.9836\phantom{\rule{0ex}{0ex}}{x}_{10}=1,{y}_{10}=0.9586$

Therefore, ${\mathbit{y}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{9586}}$

## Step 4 (c): Determine the value of y(1) at h=0.05

${x}_{0}=0$and${y}_{0}=0$, h=0.1, N=20 steps at x=1

${x}_{1}=0.05,{y}_{1}=1.9\phantom{\rule{0ex}{0ex}}{x}_{2}=0.1,{y}_{2}=1.8074\phantom{\rule{0ex}{0ex}}{x}_{3}=0.15,{y}_{3}=1.4830\phantom{\rule{0ex}{0ex}}{x}_{4}=0.2,{y}_{4}=1.642\phantom{\rule{0ex}{0ex}}{x}_{5}=0.25,{y}_{5}=1.5683\phantom{\rule{0ex}{0ex}}{x}_{6}=0.3,{y}_{6}=1.5\phantom{\rule{0ex}{0ex}}{x}_{7}=0.35,{y}_{7}=1.4368\phantom{\rule{0ex}{0ex}}{x}_{8}=0.4,{y}_{8}=1.3784\phantom{\rule{0ex}{0ex}}{x}_{9}=0.45,{y}_{9}=1.3244\phantom{\rule{0ex}{0ex}}{x}_{10}=0.5,{y}_{10}=1.2747\phantom{\rule{0ex}{0ex}}{x}_{11}=0.55,{y}_{11}=1.229\phantom{\rule{0ex}{0ex}}{x}_{12}=0.6,{y}_{12}=1.187\phantom{\rule{0ex}{0ex}}{x}_{13}=0.65,{y}_{13}=1.149\phantom{\rule{0ex}{0ex}}{x}_{14}=0.7,{y}_{14}=1.1144\phantom{\rule{0ex}{0ex}}{x}_{15}=0.75,{y}_{15}=1.0831\phantom{\rule{0ex}{0ex}}{x}_{16}=0.8,{y}_{16}=1.0551\phantom{\rule{0ex}{0ex}}{x}_{17}=0.85,{y}_{17}=1.0301\phantom{\rule{0ex}{0ex}}{x}_{18}=0.9,{y}_{18}=1.0082\phantom{\rule{0ex}{0ex}}{x}_{19}=0.95,{y}_{19}=0.9892\phantom{\rule{0ex}{0ex}}{x}_{20}=1,{y}_{20}=0.9729$

Therefore, ${\mathbit{y}}\left(1\right){\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{9729}}$ ${\mathbit{y}}\left(1\right){\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{9729}}$ ### Want to see more solutions like these? 