{"id":218658,"date":"2025-05-24T07:18:59","date_gmt":"2025-05-24T07:18:59","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=218658"},"modified":"2025-05-24T07:19:01","modified_gmt":"2025-05-24T07:19:01","slug":"determine-the-longest-interval-in-which-the-given-initial-value-problem-is-certain-to-have-a-unique-twice-differentiable-solution","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/24\/determine-the-longest-interval-in-which-the-given-initial-value-problem-is-certain-to-have-a-unique-twice-differentiable-solution\/","title":{"rendered":"Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution."},"content":{"rendered":"\n

Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution. Do not attempt to find the solution. (Enter your answer using interval notation.) (t – 1)y” – 3ty’ + 3y = sin t, y(-8) = 8, y'(-8) = 1 1 GO Tutorial
Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution. Do not attempt to find the slotuion. (Enter your answer using interval notation.)<\/p>\n\n\n\n

(t-1)y”-3ty’+3y=sin t<\/p>\n\n\n\n

y(-8)=8<\/p>\n\n\n\n

y'(-8)=<\/p>\n\n\n\n

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

We are given a second-order linear nonhomogeneous differential equation<\/strong>: (t\u22121)y\u2032\u2032\u22123ty\u2032+3y=sin\u2061t,(t – 1)y” – 3t y’ + 3y = \\sin t,<\/p>\n\n\n\n

with initial conditions: y(\u22128)=8,y\u2032(\u22128)=1.y(-8) = 8, \\quad y'(-8) = 1.<\/p>\n\n\n\n

\n\n\n\n

Step 1: Standard Form<\/h3>\n\n\n\n

To apply the existence and uniqueness theorem<\/strong> for second-order linear differential equations, rewrite the equation in standard form<\/strong> by dividing the entire equation by t\u22121t – 1 (assuming t\u22601t \\neq 1): y\u2032\u2032\u22123tt\u22121y\u2032+3t\u22121y=sin\u2061tt\u22121.y” – \\frac{3t}{t – 1} y’ + \\frac{3}{t – 1} y = \\frac{\\sin t}{t – 1}.<\/p>\n\n\n\n

\n\n\n\n

Step 2: Identify Points of Discontinuity<\/h3>\n\n\n\n

The functions:<\/p>\n\n\n\n

    \n
  • p(t)=\u22123tt\u22121p(t) = -\\frac{3t}{t – 1},<\/li>\n\n\n\n
  • q(t)=3t\u22121q(t) = \\frac{3}{t – 1},<\/li>\n\n\n\n
  • g(t)=sin\u2061tt\u22121g(t) = \\frac{\\sin t}{t – 1},<\/li>\n<\/ul>\n\n\n\n

    must all be continuous<\/strong> for the existence and uniqueness theorem to guarantee a unique, twice-differentiable solution.<\/p>\n\n\n\n

    All of these functions are discontinuous at t=1t = 1<\/strong> (division by zero). So, the equation is not defined<\/strong> at t=1t = 1.<\/p>\n\n\n\n

    \n\n\n\n

    Step 3: Determine the Interval of Validity<\/h3>\n\n\n\n

    The initial value is given at t=\u22128t = -8, which lies to the left<\/strong> of the singular point t=1t = 1.<\/p>\n\n\n\n

    Therefore, the largest interval<\/strong> centered at t=\u22128t = -8 and not containing the singular point t=1t = 1 is: (\u2212\u221e,1)(-\\infty, 1)<\/p>\n\n\n\n

    On this interval:<\/p>\n\n\n\n

      \n
    • The coefficients p(t)p(t), q(t)q(t), and the nonhomogeneous term g(t)g(t) are continuous,<\/li>\n\n\n\n
    • The initial point t=\u22128\u2208(\u2212\u221e,1)t = -8 \\in (-\\infty, 1),<\/li>\n<\/ul>\n\n\n\n

      So by the existence and uniqueness theorem<\/strong>, there is a unique, twice-differentiable solution<\/strong> on the interval (\u2212\u221e,1)(-\\infty, 1).<\/p>\n\n\n\n

      \n\n\n\n

      \u2705 Final Answer:<\/h3>\n\n\n\n

      (\u2212\u221e,1)\\boxed{(-\\infty, 1)}<\/p>\n\n\n\n

      <\/h3>\n\n\n\n

      To determine where a unique, twice-differentiable solution to a second-order differential equation exists, we rely on the existence and uniqueness theorem<\/strong> for linear differential equations. This theorem states that if the equation is written in standard form: y\u2032\u2032+p(t)y\u2032+q(t)y=g(t),y” + p(t)y’ + q(t)y = g(t),<\/p>\n\n\n\n

      and if the functions p(t)p(t), q(t)q(t), and g(t)g(t) are continuous on an interval containing the initial value point, then a unique twice-differentiable solution exists on that interval.<\/p>\n\n\n\n

      Our equation, (t\u22121)y\u2032\u2032\u22123ty\u2032+3y=sin\u2061t(t – 1)y” – 3t y’ + 3y = \\sin t, becomes: y\u2032\u2032\u22123tt\u22121y\u2032+3t\u22121y=sin\u2061tt\u22121,y” – \\frac{3t}{t – 1} y’ + \\frac{3}{t – 1} y = \\frac{\\sin t}{t – 1},<\/p>\n\n\n\n

      after dividing by t\u22121t – 1. Here, all coefficients become undefined at t=1t = 1, creating a singularity.<\/p>\n\n\n\n

      The initial condition is given at t=\u22128t = -8, well to the left of this singularity. Therefore, the largest open interval containing t=\u22128t = -8 that excludes the singularity is (\u2212\u221e,1)(-\\infty, 1). Within this interval, all functions in the standard form are continuous, and hence the solution is guaranteed to exist and be unique.<\/p>\n\n\n\n

      Thus, the longest interval in which the solution is certain to exist and be unique is (\u2212\u221e,1)(-\\infty, 1).<\/p>\n\n\n\n

      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution. Do not attempt to find the solution. (Enter your answer using interval notation.) (t – 1)y” – 3ty’ + 3y = sin t, y(-8) = 8, y'(-8) = 1 1 GO TutorialDetermine the longest interval in […]<\/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-218658","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218658","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=218658"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218658\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218658"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218658"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218658"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}