{"id":219200,"date":"2025-05-25T19:53:47","date_gmt":"2025-05-25T19:53:47","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=219200"},"modified":"2025-05-25T19:53:49","modified_gmt":"2025-05-25T19:53:49","slug":"find-a-linear-differential-operator-that-annihilates-the-given-function-6","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/25\/find-a-linear-differential-operator-that-annihilates-the-given-function-6\/","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<\/strong> a given function, we look for a differential operator L=p(D)L = p(D) such that L[f(x)]=0L[f(x)] = 0, where D=ddxD = \\frac{d}{dx}.<\/p>\n\n\n\n

\n\n\n\n

Given Functions:<\/strong><\/h3>\n\n\n\n
    \n
  1. f1(x)=1+6x\u22122x3f_1(x) = 1 + 6x – 2x^3<\/li>\n\n\n\n
  2. f2(x)=e\u2212x+2xex\u2212x2exf_2(x) = e^{-x} + 2x e^x – x^2 e^x<\/li>\n<\/ol>\n\n\n\n
    \n\n\n\n

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

    This is a polynomial of degree 3.<\/p>\n\n\n\n

    A standard result:<\/p>\n\n\n\n

    \n

    The annihilator of a polynomial of degree nn is Dn+1D^{n+1}, because successive derivatives of a polynomial eventually reduce it to zero.<\/p>\n<\/blockquote>\n\n\n\n

    Here, f1(x)f_1(x) is degree 3. So its annihilator is: D4\\boxed{D^4}<\/p>\n\n\n\n

    Because:<\/p>\n\n\n\n

      \n
    • D(1+6x\u22122×3)=6\u22126x2D(1 + 6x – 2x^3) = 6 – 6x^2<\/li>\n\n\n\n
    • D2=\u221212xD^2 = -12x<\/li>\n\n\n\n
    • D3=\u221212D^3 = -12<\/li>\n\n\n\n
    • D4=0D^4 = 0<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

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

      Let\u2019s break it down:<\/p>\n\n\n\n

        \n
      • e\u2212xe^{-x} \u2192 annihilator: D+1D + 1<\/li>\n\n\n\n
      • 2xex2x e^x, x2exx^2 e^x \u2192 both are of the form P(x)exP(x)e^x, where P(x)P(x) is a polynomial.<\/li>\n<\/ul>\n\n\n\n

        A general rule:<\/p>\n\n\n\n

        \n

        For xneaxx^n e^{ax}, the annihilator is (D\u2212a)n+1(D – a)^{n+1}<\/p>\n<\/blockquote>\n\n\n\n

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

          \n
        • 2xex2x e^x \u2192 annihilator: (D\u22121)2(D – 1)^2<\/li>\n\n\n\n
        • x2exx^2 e^x \u2192 annihilator: (D\u22121)3(D – 1)^3<\/li>\n<\/ul>\n\n\n\n

          So for the full function:<\/p>\n\n\n\n

            \n
          • e\u2212x+(2x\u2212x2)exe^{-x} + (2x – x^2) e^x<\/li>\n<\/ul>\n\n\n\n

            The total annihilator is the least common multiple (LCM)<\/strong> of:<\/p>\n\n\n\n

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

              So the annihilator is: (D+1)(D\u22121)3\\boxed{(D + 1)(D – 1)^3}<\/p>\n\n\n\n

              \n\n\n\n

              Final Answer:<\/strong><\/h3>\n\n\n\n
                \n
              • Annihilator of 1+6x\u22122×31 + 6x – 2x^3:\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003D4\\boxed{D^4}<\/li>\n\n\n\n
              • Annihilator of e\u2212x+2xex\u2212x2exe^{-x} + 2x e^x – x^2 e^x:\u2003\u2003\u2003(D+1)(D\u22121)3\\boxed{(D + 1)(D – 1)^3}<\/li>\n<\/ul>\n\n\n\n
                \n\n\n\n

                Explanation <\/strong><\/h3>\n\n\n\n

                Annihilators are differential operators that send a function to zero. These are especially useful in solving differential equations using the method of annihilators.<\/p>\n\n\n\n

                The first function, f1(x)=1+6x\u22122x3f_1(x) = 1 + 6x – 2x^3, is a polynomial of degree 3. Derivatives of a polynomial reduce its degree by 1 at each step. Thus, applying DD (i.e., taking the derivative) four times will reduce this function to zero. Therefore, D4D^4 is the minimal annihilator. That is, D4[f1(x)]=0D^4[f_1(x)] = 0.<\/p>\n\n\n\n

                The second function, f2(x)=e\u2212x+2xex\u2212x2exf_2(x) = e^{-x} + 2x e^x – x^2 e^x, is a combination of exponential terms, some of which are multiplied by polynomials. For such functions, the annihilator depends on both the exponential part and the polynomial degree.<\/p>\n\n\n\n

                  \n
                • For e\u2212xe^{-x}, the annihilator is D+1D + 1, since D[e\u2212x]=\u2212e\u2212xD[e^{-x}] = -e^{-x}.<\/li>\n\n\n\n
                • For xnexx^n e^x, the annihilator is (D\u22121)n+1(D – 1)^{n+1}. Hence:\n
                    \n
                  • 2xex2x e^x is annihilated by (D\u22121)2(D – 1)^2<\/li>\n\n\n\n
                  • x2exx^2 e^x is annihilated by (D\u22121)3(D – 1)^3<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n

                    Since the function contains both e\u2212xe^{-x} and terms with exe^x, we need a differential operator that annihilates all components. The least common multiple (LCM) of all the individual annihilators gives the annihilator for the entire function.<\/p>\n\n\n\n

                    Thus, the annihilator of f2(x)f_2(x) is (D+1)(D\u22121)3(D + 1)(D – 1)^3, which when applied to f2(x)f_2(x), gives zero.<\/p>\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 given function, we look for a differential operator L=p(D)L = p(D) such that L[f(x)]=0L[f(x)] […]<\/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-219200","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219200","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=219200"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219200\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=219200"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=219200"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=219200"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}