{"id":220571,"date":"2025-05-28T13:07:07","date_gmt":"2025-05-28T13:07:07","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=220571"},"modified":"2025-05-28T13:07:10","modified_gmt":"2025-05-28T13:07:10","slug":"find-a-linear-differential-operator-that-annihilates-the-given-function-7","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/28\/find-a-linear-differential-operator-that-annihilates-the-given-function-7\/","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 a function, we must find a differential operator L(D)L(D)<\/strong> such that: L(D)[f(x)]=0L(D)[f(x)] = 0<\/p>\n\n\n\n

We’ll treat each function separately.<\/p>\n\n\n\n

\n\n\n\n

1. Function: f1(x)=1+6x\u22122x3f_1(x) = 1 + 6x – 2x^3<\/strong><\/h3>\n\n\n\n

This is a polynomial function<\/strong> of degree 3. For polynomial functions:<\/p>\n\n\n\n

    \n
  • The (n+1)th derivative<\/strong> of a polynomial of degree n<\/strong> is 0.<\/li>\n<\/ul>\n\n\n\n

    Since f1(x)f_1(x) is a degree-3 polynomial, the fourth derivative will annihilate it.<\/p>\n\n\n\n

    Define the operator: L1(D)=D4L_1(D) = D^4<\/p>\n\n\n\n

    Then: D4[f1(x)]=0D^4[f_1(x)] = 0<\/p>\n\n\n\n

    \u2705 Annihilator<\/strong>: D4\\boxed{D^4}<\/p>\n\n\n\n

    \n\n\n\n

    2. Function: f2(x)=e\u2212x+2xex\u2212x2exf_2(x) = e^{-x} + 2xe^x – x^2e^x<\/strong><\/h3>\n\n\n\n

    This is a linear combination<\/strong> of terms involving exponentials multiplied by polynomials.<\/p>\n\n\n\n

    Break it into parts:<\/p>\n\n\n\n

      \n
    • e\u2212xe^{-x}<\/li>\n\n\n\n
    • 2xex2xe^x<\/li>\n\n\n\n
    • \u2212x2ex-x^2e^x<\/li>\n<\/ul>\n\n\n\n

      Let\u2019s identify annihilators of each term:<\/p>\n\n\n\n

        \n
      • e\u2212xe^{-x}: annihilated by (D+1)(D + 1)<\/li>\n\n\n\n
      • xexxe^x: annihilated by (D\u22121)2(D – 1)^2<\/li>\n\n\n\n
      • x2exx^2e^x: annihilated by (D\u22121)3(D – 1)^3<\/li>\n<\/ul>\n\n\n\n

        To annihilate the full function, we need the least common multiple (LCM)<\/strong> of the annihilators:<\/p>\n\n\n\n

          \n
        • LCM of (D+1),(D\u22121)2,(D\u22121)3(D + 1), (D – 1)^2, (D – 1)^3 is:<\/li>\n<\/ul>\n\n\n\n

          (D+1)(D\u22121)3(D + 1)(D – 1)^3<\/p>\n\n\n\n

          \u2705 Annihilator<\/strong>: (D+1)(D\u22121)3\\boxed{(D + 1)(D – 1)^3}<\/p>\n\n\n\n

          \n\n\n\n

          \ud83d\udcd8 Explanation)<\/h3>\n\n\n\n

          A linear differential operator<\/strong> is an operator involving derivatives, often denoted using D=ddxD = \\frac{d}{dx}. The process of annihilation<\/strong> means applying this operator to a function such that the result is zero. An annihilator<\/strong> effectively removes the function\u2014much like applying a fourth derivative to a cubic polynomial yields zero.<\/p>\n\n\n\n

          For f1(x)=1+6x\u22122x3f_1(x) = 1 + 6x – 2x^3:<\/h4>\n\n\n\n

          This is a polynomial of degree 3. The derivative of a constant is zero; each differentiation lowers the degree by one. Hence, after four differentiations:<\/p>\n\n\n\n

            \n
          • D[f1(x)]=6\u22126x2D[f_1(x)] = 6 – 6x^2<\/li>\n\n\n\n
          • D2[f1(x)]=\u221212xD^2[f_1(x)] = -12x<\/li>\n\n\n\n
          • D3[f1(x)]=\u221212D^3[f_1(x)] = -12<\/li>\n\n\n\n
          • D4[f1(x)]=0D^4[f_1(x)] = 0<\/li>\n<\/ul>\n\n\n\n

            So, D4D^4 annihilates this polynomial.<\/p>\n\n\n\n

            For f2(x)=e\u2212x+2xex\u2212x2exf_2(x) = e^{-x} + 2xe^x – x^2e^x:<\/h4>\n\n\n\n

            This function includes terms of the form xneaxx^n e^{ax}, common in differential equations. There is a systematic rule:<\/p>\n\n\n\n

              \n
            • eaxe^{ax}: annihilated by (D\u2212a)(D – a)<\/li>\n\n\n\n
            • xneaxx^n e^{ax}: annihilated by (D\u2212a)n+1(D – a)^{n+1}<\/li>\n<\/ul>\n\n\n\n

              So:<\/p>\n\n\n\n

                \n
              • e\u2212xe^{-x} is annihilated by (D+1)(D + 1)<\/li>\n\n\n\n
              • xexxe^x needs (D\u22121)2(D – 1)^2<\/li>\n\n\n\n
              • x2exx^2e^x needs (D\u22121)3(D – 1)^3<\/li>\n<\/ul>\n\n\n\n

                To annihilate the entire sum, we take the least common multiple: Annihilator=(D+1)(D\u22121)3\\text{Annihilator} = (D + 1)(D – 1)^3<\/p>\n\n\n\n

                This operator guarantees that all three terms vanish when the operator is applied.<\/p>\n\n\n\n

                \n\n\n\n

                \u2705 Final Answers:<\/h3>\n\n\n\n
                  \n
                • f1(x):D4f_1(x): \\boxed{D^4}<\/li>\n\n\n\n
                • f2(x):(D+1)(D\u22121)3f_2(x): \\boxed{(D + 1)(D – 1)^3}<\/li>\n<\/ul>\n\n\n\n
                  \"\"<\/figure>\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, we must find a differential operator L(D)L(D) such that: L(D)[f(x)]=0L(D)[f(x)] = 0 We’ll […]<\/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-220571","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220571","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=220571"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220571\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=220571"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=220571"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=220571"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}