{"id":214180,"date":"2025-05-13T17:49:42","date_gmt":"2025-05-13T17:49:42","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=214180"},"modified":"2025-05-13T17:49:45","modified_gmt":"2025-05-13T17:49:45","slug":"find-a-linear-differential-operator-that-annihilates-the-given-function-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/13\/find-a-linear-differential-operator-that-annihilates-the-given-function-2\/","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<\/strong>. (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 solve this problem, we are asked to find linear differential operators<\/strong> that annihilate<\/strong> given functions. That means we are to find differential operators L=anDn+an\u22121Dn\u22121+\u2026+a0L = a_n D^n + a_{n-1} D^{n-1} + \\ldots + a_0, such that L[f(x)]=0L[f(x)] = 0. Here, DD denotes the differential operator ddx\\frac{d}{dx}.<\/p>\n\n\n\n

\n\n\n\n

Function 1:<\/h3>\n\n\n\n

f(x)=1+6x\u22122x3f(x) = 1 + 6x – 2x^3<\/strong><\/p>\n\n\n\n

This is a polynomial<\/strong> of degree 3. We know that the (n+1)(n+1)th derivative of a degree nn polynomial is zero.<\/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[1+6x\u22122×3]=\u221212xD^2[1 + 6x – 2x^3] = -12x<\/li>\n\n\n\n
  • D3[1+6x\u22122×3]=\u221212D^3[1 + 6x – 2x^3] = -12<\/li>\n\n\n\n
  • D4[1+6x\u22122×3]=0D^4[1 + 6x – 2x^3] = 0<\/li>\n<\/ul>\n\n\n\n

    Thus, the operator D4D^4 annihilates<\/strong> the function:<\/p>\n\n\n\n

    \n

    \u2705 Answer<\/strong>: D4D^4<\/p>\n<\/blockquote>\n\n\n\n

    \n\n\n\n

    Function 2:<\/h3>\n\n\n\n

    f(x)=e\u2212x+2xex\u2212x2exf(x) = e^{-x} + 2x e^x – x^2 e^x<\/strong><\/p>\n\n\n\n

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

      \n
    • e\u2212xe^{-x}<\/li>\n\n\n\n
    • xexx e^x<\/li>\n\n\n\n
    • x2exx^2 e^x<\/li>\n<\/ul>\n\n\n\n

      We use a method called annihilator method<\/strong>:<\/p>\n\n\n\n

        \n
      • The annihilator of eaxe^{ax} is D\u2212aD – a<\/li>\n\n\n\n
      • The annihilator of xneaxx^n e^{ax} is (D\u2212a)n+1(D – a)^{n+1}<\/li>\n<\/ul>\n\n\n\n

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

          \n
        • e\u2212x\u21d2e^{-x} \\Rightarrow annihilator: D+1D + 1<\/li>\n\n\n\n
        • xex\u21d2x e^x \\Rightarrow annihilator: (D\u22121)2(D – 1)^2<\/li>\n\n\n\n
        • x2ex\u21d2x^2 e^x \\Rightarrow annihilator: (D\u22121)3(D – 1)^3<\/li>\n<\/ul>\n\n\n\n

          Now, to annihilate all<\/strong> terms in the function, we must take the least common multiple (LCM)<\/strong> of all these operators:<\/p>\n\n\n\n

            \n
          • LCM[(D+1),(D\u22121)3]=(D+1)(D\u22121)3\\text{LCM}[(D + 1), (D – 1)^3] = (D + 1)(D – 1)^3<\/li>\n<\/ul>\n\n\n\n
            \n

            \u2705 Answer<\/strong>: (D+1)(D\u22121)3(D + 1)(D – 1)^3<\/p>\n<\/blockquote>\n\n\n\n

            \n\n\n\n

            \u2728 Explanation (300+ words)<\/h3>\n\n\n\n

            To solve the problem of finding a linear differential operator that annihilates<\/strong> a given function, we need to understand how different types of functions behave under repeated differentiation.<\/p>\n\n\n\n

            For the first function:<\/h4>\n\n\n\n

            f(x)=1+6x\u22122x3f(x) = 1 + 6x – 2x^3<\/strong>
            This is a polynomial. The highest power is x3x^3, so its fourth derivative will be zero. This is because:<\/p>\n\n\n\n

              \n
            • The first derivative reduces the degree of each term by 1.<\/li>\n\n\n\n
            • Repeating differentiation will eventually remove all terms.<\/li>\n<\/ul>\n\n\n\n

              For example:<\/p>\n\n\n\n

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

                So, D4D^4 is the simplest operator that annihilates this polynomial.<\/p>\n\n\n\n

                For the second function:<\/h4>\n\n\n\n

                f(x)=e\u2212x+2xex\u2212x2exf(x) = e^{-x} + 2x e^x – x^2 e^x<\/strong>
                This is a linear combination of exponential functions and exponential-polynomial products.<\/p>\n\n\n\n

                To handle such functions, we apply the annihilator method<\/strong>:<\/p>\n\n\n\n

                  \n
                • The annihilator for eaxe^{ax} is D\u2212aD – a<\/li>\n\n\n\n
                • For xneaxx^n e^{ax}, the annihilator is (D\u2212a)n+1(D – a)^{n+1}<\/li>\n<\/ul>\n\n\n\n

                  So, analyze each term:<\/p>\n\n\n\n

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

                    The operator that can annihilate all three terms must handle both e\u2212xe^{-x} and x2exx^2 e^x. This leads to taking the least common multiple:<\/p>\n\n\n\n

                      \n
                    • LCM[(D+1),(D\u22121)3]=(D+1)(D\u22121)3\\text{LCM}[(D + 1), (D – 1)^3] = (D + 1)(D – 1)^3<\/li>\n<\/ul>\n\n\n\n

                      This operator, when applied to the function, will give zero. Hence, it’s the required linear differential operator.<\/p>\n\n\n\n

                      \n\n\n\n

                      \u2705 Final Answers:<\/h3>\n\n\n\n
                        \n
                      • For f(x)=1+6x\u22122x3f(x) = 1 + 6x – 2x^3: D4D^4<\/strong><\/li>\n\n\n\n
                      • For f(x)=e\u2212x+2xex\u2212x2exf(x) = e^{-x} + 2x e^x – x^2 e^x: (D+1)(D\u22121)3(D + 1)(D – 1)^3<\/strong><\/li>\n<\/ul>\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 solve this problem, we are asked to find linear differential operators that annihilate given functions. That means we are to find differential […]<\/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-214180","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/214180","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=214180"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/214180\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=214180"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=214180"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=214180"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}