{"id":186892,"date":"2025-01-27T05:35:04","date_gmt":"2025-01-27T05:35:04","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=186892"},"modified":"2025-01-27T05:35:07","modified_gmt":"2025-01-27T05:35:07","slug":"the-auxiliary-equation-of-y-4y-4y-16y-0-is-r%c2%b3-4r%c2%b2-4r-16-0","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/27\/the-auxiliary-equation-of-y-4y-4y-16y-0-is-r%c2%b3-4r%c2%b2-4r-16-0\/","title":{"rendered":"The auxiliary equation of y”” \u2013 4y” \u2013 4y’ + 16y = 0 is r\u00b3 \u2013 4r\u00b2 \u2013 4r + 16 = 0"},"content":{"rendered":"\n

The auxiliary equation of y”” \u2013 4y” \u2013 4y’ + 16y = 0 is r\u00b3 \u2013 4r\u00b2 \u2013 4r + 16 = 0. The auxiliary equation has a root in the interval [-0.75, 3.75]. Let f(r) = r\u00b3 – 4r\u00b2 – 4r + 16 so that the roots ” of the auxiliary quation can be determined by solving f(r) = 0. Then, applying four iterations of the Bisection Method to f(r), using the 55<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

To solve ( f(r) = r^3 – 4r^2 – 4r + 16 = 0 ) using the Bisection Method<\/strong> in the interval ([-0.75, 3.75]), we iterate as follows:<\/p>\n\n\n\n

Steps:<\/h3>\n\n\n\n
    \n
  1. Initial interval<\/strong>: ([a, b] = [-0.75, 3.75]).<\/li>\n<\/ol>\n\n\n\n
      \n
    • Compute ( f(a) = f(-0.75) = (-0.75)^3 – 4(-0.75)^2 – 4(-0.75) + 16 = 15.078125 > 0 ).<\/li>\n\n\n\n
    • Compute ( f(b) = f(3.75) = (3.75)^3 – 4(3.75)^2 – 4(3.75) + 16 = -6.828125 < 0 ).<\/li>\n\n\n\n
    • Since ( f(a) \\cdot f(b) < 0 ), the root lies in this interval.<\/li>\n<\/ul>\n\n\n\n
        \n
      1. First iteration<\/strong>:<\/li>\n<\/ol>\n\n\n\n
          \n
        • Midpoint: ( c_1 = \\frac{a + b}{2} = \\frac{-0.75 + 3.75}{2} = 1.5 ).<\/li>\n\n\n\n
        • Compute ( f(c_1) = f(1.5) = (1.5)^3 – 4(1.5)^2 – 4(1.5) + 16 = 1.375 > 0 ).<\/li>\n\n\n\n
        • Update: Since ( f(c_1) > 0 ), set ( a = c_1 = 1.5 ), ( b = 3.75 ).<\/li>\n<\/ul>\n\n\n\n
            \n
          1. Second iteration<\/strong>:<\/li>\n<\/ol>\n\n\n\n
              \n
            • Midpoint: ( c_2 = \\frac{a + b}{2} = \\frac{1.5 + 3.75}{2} = 2.625 ).<\/li>\n\n\n\n
            • Compute ( f(c_2) = f(2.625) = (2.625)^3 – 4(2.625)^2 – 4(2.625) + 16 = -3.78515625 < 0 ).<\/li>\n\n\n\n
            • Update: Since ( f(c_2) < 0 ), set ( a = 1.5 ), ( b = c_2 = 2.625 ).<\/li>\n<\/ul>\n\n\n\n
                \n
              1. Third iteration<\/strong>:<\/li>\n<\/ol>\n\n\n\n
                  \n
                • Midpoint: ( c_3 = \\frac{a + b}{2} = \\frac{1.5 + 2.625}{2} = 2.0625 ).<\/li>\n\n\n\n
                • Compute ( f(c_3) = f(2.0625) = (2.0625)^3 – 4(2.0625)^2 – 4(2.0625) + 16 = -0.31494140625 < 0 ).<\/li>\n\n\n\n
                • Update: Since ( f(c_3) < 0 ), set ( a = 1.5 ), ( b = c_3 = 2.0625 ).<\/li>\n<\/ul>\n\n\n\n
                    \n
                  1. Fourth iteration<\/strong>:<\/li>\n<\/ol>\n\n\n\n
                      \n
                    • Midpoint: ( c_4 = \\frac{a + b}{2} = \\frac{1.5 + 2.0625}{2} = 1.78125 ).<\/li>\n\n\n\n
                    • Compute ( f(c_4) = f(1.78125) = (1.78125)^3 – 4(1.78125)^2 – 4(1.78125) + 16 = 0.41693115234375 > 0 ).<\/li>\n\n\n\n
                    • Update: Since ( f(c_4) > 0 ), set ( a = c_4 = 1.78125 ), ( b = 2.0625 ).<\/li>\n<\/ul>\n\n\n\n

                      Result:<\/h3>\n\n\n\n

                      After 4 iterations, the root lies in ([1.78125, 2.0625]). The approximate root is the midpoint ( c_4 = 1.78125 ).<\/p>\n\n\n\n

                      \n\n\n\n

                      Explanation:<\/h3>\n\n\n\n

                      The Bisection Method<\/strong> is a numerical approach to find a root of a function by iteratively halving the interval where the root lies. At each step:<\/p>\n\n\n\n

                        \n
                      1. Compute the midpoint of the interval.<\/li>\n\n\n\n
                      2. Evaluate the function at the midpoint.<\/li>\n\n\n\n
                      3. Update the interval based on the sign of the function at the midpoint.<\/li>\n<\/ol>\n\n\n\n

                        Since the function ( f(r) ) is continuous, and ( f(a) \\cdot f(b) < 0 ), the Intermediate Value Theorem guarantees a root exists in the interval. Repeated halving reduces the interval\u2019s size, narrowing closer to the root.<\/p>\n","protected":false},"excerpt":{"rendered":"

                        The auxiliary equation of y”” \u2013 4y” \u2013 4y’ + 16y = 0 is r\u00b3 \u2013 4r\u00b2 \u2013 4r + 16 = 0. The auxiliary equation has a root in the interval [-0.75, 3.75]. Let f(r) = r\u00b3 – 4r\u00b2 – 4r + 16 so that the roots ” of the auxiliary quation can be […]<\/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-186892","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186892","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=186892"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186892\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186892"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186892"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186892"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}