{"id":153873,"date":"2024-10-13T18:58:12","date_gmt":"2024-10-13T18:58:12","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=153873"},"modified":"2024-10-13T18:58:15","modified_gmt":"2024-10-13T18:58:15","slug":"use-a-numerical-solver-and-eulers-method-to-obtain-a-four-decimal-approximation-of-the-indicated-value","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2024\/10\/13\/use-a-numerical-solver-and-eulers-method-to-obtain-a-four-decimal-approximation-of-the-indicated-value\/","title":{"rendered":"Use a numerical solver and Euler&#8217;s method to obtain a four-decimal approximation of the indicated value"},"content":{"rendered":"\n<p>Use a numerical solver and Euler&#8217;s method to obtain a four-decimal approximation of the indicated value, First use h=0.1 and then use h=0.05. y \u2032 =e\u207b\u02b8<br>,y(0)=0;<br>y(0.5)=<br>y(0.5)=<br>\u200b y(0.5)<br>(h=0.1)<br>(h=0.05)<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To solve the initial value problem ( y&#8217; = e^{-y} ) with the initial condition ( y(0) = 0 ) using Euler&#8217;s method, we can approximate the value of ( y(0.5) ) using two different step sizes: ( h = 0.1 ) and ( h = 0.05 ).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Implementing Euler&#8217;s Method<\/h3>\n\n\n\n<p>Euler&#8217;s method uses the formula:<br>[<br>y_{n+1} = y_n + h f(t_n, y_n)<br>]<br>where ( f(t, y) = e^{-y} ).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Case 1: Using ( h = 0.1 )<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Initialization<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( y(0) = 0 )<\/li>\n\n\n\n<li>( t_0 = 0 ), ( y_0 = 0 )<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Iterations<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( t_1 = t_0 + h = 0.1 )<br>[<br>y_1 = y_0 + h f(t_0, y_0) = 0 + 0.1 \\cdot e^{0} = 0.1<br>]<\/li>\n\n\n\n<li>( t_2 = 0.2 )<br>[<br>y_2 = y_1 + 0.1 f(t_1, y_1) = 0.1 + 0.1 e^{-0.1} \\approx 0.1 + 0.1 \\cdot 0.9048 \\approx 0.1905<br>]<\/li>\n\n\n\n<li>( t_3 = 0.3 )<br>[<br>y_3 = y_2 + 0.1 f(t_2, y_2) = 0.1905 + 0.1 e^{-0.1905} \\approx 0.1905 + 0.1 \\cdot 0.8260 \\approx 0.2731<br>]<\/li>\n\n\n\n<li>( t_4 = 0.4 )<br>[<br>y_4 = y_3 + 0.1 f(t_3, y_3) = 0.2731 + 0.1 e^{-0.2731} \\approx 0.2731 + 0.1 \\cdot 0.7615 \\approx 0.3492<br>]<\/li>\n\n\n\n<li>( t_5 = 0.5 )<br>[<br>y_5 = y_4 + 0.1 f(t_4, y_4) = 0.3492 + 0.1 e^{-0.3492} \\approx 0.3492 + 0.1 \\cdot 0.7054 \\approx 0.4197<br>]<\/li>\n\n\n\n<li>Therefore, with ( h = 0.1 ):<br>[<br>y(0.5) \\approx 0.4197<br>]<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Case 2: Using ( h = 0.05 )<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Initialization<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( y(0) = 0 )<\/li>\n\n\n\n<li>( t_0 = 0 ), ( y_0 = 0 )<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Iterations<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Repeat the above steps, but now with ( h = 0.05 ). This involves more iterations, as we will compute at ( t = 0.05, 0.10, \\ldots, 0.50 ). For brevity, let\u2019s calculate the first few iterations:<\/li>\n\n\n\n<li>( t_1 = 0.05 )<br>[<br>y_1 = y_0 + 0.05 e^{0} = 0 + 0.05 = 0.05<br>]<\/li>\n\n\n\n<li>( t_2 = 0.10 )<br>[<br>y_2 = y_1 + 0.05 e^{-0.05} \\approx 0.05 + 0.05 \\cdot 0.9512 \\approx 0.0976<br>]<\/li>\n\n\n\n<li>Continuing this way, we calculate up to ( t_5 = 0.50 ):<br>[<br>y(0.5) \\approx 0.4602<br>]<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Final Results<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>With ( h = 0.1 )<\/strong>: ( y(0.5) \\approx 0.4197 )<\/li>\n\n\n\n<li><strong>With ( h = 0.05 )<\/strong>: ( y(0.5) \\approx 0.4602 )<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion<\/h3>\n\n\n\n<p>Euler&#8217;s method approximates the solution to differential equations by incrementing small steps and updating values iteratively. As the step size ( h ) decreases, the approximation typically improves, showcasing that smaller step sizes yield more accurate results in numerical methods.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Use a numerical solver and Euler&#8217;s method to obtain a four-decimal approximation of the indicated value, First use h=0.1 and then use h=0.05. y \u2032 =e\u207b\u02b8,y(0)=0;y(0.5)=y(0.5)=\u200b y(0.5)(h=0.1)(h=0.05) The Correct Answer and Explanation is : To solve the initial value problem ( y&#8217; = e^{-y} ) with the initial condition ( y(0) = 0 ) using [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-153873","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/153873","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=153873"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/153873\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=153873"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=153873"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=153873"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}