{"id":214308,"date":"2025-05-14T05:20:38","date_gmt":"2025-05-14T05:20:38","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=214308"},"modified":"2025-05-14T05:20:41","modified_gmt":"2025-05-14T05:20:41","slug":"find-a-linear-differential-operator-that-annihilates-the-given-function-3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/14\/find-a-linear-differential-operator-that-annihilates-the-given-function-3\/","title":{"rendered":"Find a linear differential operator that annihilates the given function"},"content":{"rendered":"\n

Find a linear differential operator that annihilates the given function. (Use D for the differential operator.)<\/p>\n\n\n\n

For, 1+6x – 2×3 and, e-x + 2xex – x2ex<\/p>\n\n\n\n

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

To find a linear differential operator<\/strong> that annihilates<\/em> a function means to find a differential operator $L(D)$, where $D = \\frac{d}{dx}$, such that:<\/p>\n\n\n\n

$$
L(D)[f(x)] = 0
$$<\/p>\n\n\n\n

We are given two functions:<\/p>\n\n\n\n

    \n
  1. $f_1(x) = 1 + 6x – 2x^3$<\/li>\n\n\n\n
  2. $f_2(x) = e^{-x} + 2x e^x – x^2 e^x$<\/li>\n<\/ol>\n\n\n\n

    We will find the annihilator for each function separately and then combine them.<\/p>\n\n\n\n

    \n\n\n\n

    \ud83d\udd39 Part 1: Polynomial Function<\/strong><\/h3>\n\n\n\n

    Function:<\/p>\n\n\n\n

    $$
    f_1(x) = 1 + 6x – 2x^3
    $$<\/p>\n\n\n\n

    This is a polynomial of degree 3<\/strong>. The annihilator for any polynomial of degree $n$ is:<\/p>\n\n\n\n

    $$
    D^{n+1}
    $$<\/p>\n\n\n\n

    So here, since the highest power is $x^3$, we use:<\/p>\n\n\n\n

    $$
    D^4
    $$<\/p>\n\n\n\n

    Why? Because differentiating a cubic polynomial four times gives zero:<\/p>\n\n\n\n

    $$
    D[1 + 6x – 2x^3] = 6 – 6x^2,\\quad D^2 = -12x,\\quad D^3 = -12,\\quad D^4 = 0
    $$<\/p>\n\n\n\n

    Thus, the annihilator is:<\/p>\n\n\n\n

    $$
    L_1(D) = D^4
    $$<\/p>\n\n\n\n

    \n\n\n\n

    \ud83d\udd39 Part 2: Exponential Function<\/strong><\/h3>\n\n\n\n

    Function:<\/p>\n\n\n\n

    $$
    f_2(x) = e^{-x} + 2x e^x – x^2 e^x
    $$<\/p>\n\n\n\n

    This is a linear combination of:<\/p>\n\n\n\n

      \n
    • $e^{-x}$<\/li>\n\n\n\n
    • $x e^x$<\/li>\n\n\n\n
    • $x^2 e^x$<\/li>\n<\/ul>\n\n\n\n

      \u2705 Step 1: Annihilator of $e^{-x}$<\/h4>\n\n\n\n

      Annihilator is:<\/p>\n\n\n\n

      $$
      (D + 1)
      $$<\/p>\n\n\n\n

      \u2705 Step 2: Annihilator of $x^2 e^x$<\/h4>\n\n\n\n

      The annihilator for $x^n e^{ax}$ is:<\/p>\n\n\n\n

      $$
      (D – a)^{n+1}
      $$<\/p>\n\n\n\n

      So for $x^2 e^x$, it is:<\/p>\n\n\n\n

      $$
      (D – 1)^3
      $$<\/p>\n\n\n\n

      \u2705 Combine Both<\/h4>\n\n\n\n

      To annihilate all terms in $f_2(x)$, we take the least common multiple (LCM)<\/strong> of the individual annihilators:<\/p>\n\n\n\n

      $$
      L_2(D) = (D + 1)(D – 1)^3
      $$<\/p>\n\n\n\n

      \n\n\n\n

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

      To annihilate the full expression:<\/p>\n\n\n\n

      $$
      f(x) = f_1(x) + f_2(x) = 1 + 6x – 2x^3 + e^{-x} + 2x e^x – x^2 e^x
      $$<\/p>\n\n\n\n

      We combine the two operators:<\/p>\n\n\n\n

        \n
      • From Part 1: $D^4$<\/li>\n\n\n\n
      • From Part 2: $(D + 1)(D – 1)^3$<\/li>\n<\/ul>\n\n\n\n

        So the final annihilator<\/strong> is:<\/p>\n\n\n\n

        $$
        \\boxed{D^4 (D + 1)(D – 1)^3}
        $$<\/p>\n\n\n\n

        This is the differential operator that, when applied to the full function, gives zero.<\/p>\n","protected":false},"excerpt":{"rendered":"

        Find a linear differential operator that annihilates the given function. (Use D for the differential operator.) For, 1+6x – 2×3 and, e-x + 2xex – x2ex The Correct Answer And Explanation is: To find a linear differential operator that annihilates a function means to find a differential operator $L(D)$, where $D = \\frac{d}{dx}$, such that: […]<\/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-214308","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/214308","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=214308"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/214308\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=214308"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=214308"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=214308"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}