{"id":217666,"date":"2025-05-22T10:07:42","date_gmt":"2025-05-22T10:07:42","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=217666"},"modified":"2025-05-22T10:07:44","modified_gmt":"2025-05-22T10:07:44","slug":"find-a-linear-differential-operator-that-annihilates-the-given-function-5","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/22\/find-a-linear-differential-operator-that-annihilates-the-given-function-5\/","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 means to find an operator $L(D)$, expressed in terms of the differential operator $D = \\frac{d}{dx}$, such that:<\/p>\n\n\n\n

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

We will analyze each function separately.<\/p>\n\n\n\n

\n\n\n\n

Part 1: Function<\/strong> $f(x) = 1 + 6x – 2x^3$<\/h3>\n\n\n\n

This is a polynomial function<\/strong>. A general rule is:<\/p>\n\n\n\n

    \n
  • A polynomial of degree $n$ is annihilated by the $(n+1)^\\text{th}$ derivative.<\/li>\n<\/ul>\n\n\n\n

    Here, the highest degree term is $x^3$, so:<\/p>\n\n\n\n

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

    Therefore, the linear differential operator is:<\/p>\n\n\n\n

    $$
    \\boxed{D^4}
    $$<\/p>\n\n\n\n

    \n\n\n\n

    Part 2: Function<\/strong> $g(x) = e^{-x} + 2x e^x – x^2 e^x$<\/h3>\n\n\n\n

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

      \n
    • $e^{-x}$: exponential<\/li>\n\n\n\n
    • $2x e^x$, $x^2 e^x$: polynomial \u00d7 exponential<\/li>\n<\/ul>\n\n\n\n

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

      The function $e^{-x}$ is annihilated by:<\/p>\n\n\n\n

      $$
      (D + 1)
      \\quad \\text{because} \\quad (D + 1)[e^{-x}] = De^{-x} + e^{-x} = -e^{-x} + e^{-x} = 0
      $$<\/p>\n\n\n\n

      Step 2: Annihilate $2x e^x – x^2 e^x$<\/h4>\n\n\n\n

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

        \n
      • $x e^x$ and $x^2 e^x$<\/li>\n<\/ul>\n\n\n\n

        Let\u2019s denote a general form:<\/p>\n\n\n\n

          \n
        • $e^x$, $x e^x$, $x^2 e^x$ \u21d2 annihilated by $(D – 1)^3$<\/li>\n<\/ul>\n\n\n\n

          This is because:<\/p>\n\n\n\n

            \n
          • $(D – 1)[e^x] = 0$<\/li>\n\n\n\n
          • $(D – 1)^2[x e^x] = 0$<\/li>\n\n\n\n
          • $(D – 1)^3[x^2 e^x] = 0$<\/li>\n<\/ul>\n\n\n\n

            Thus, $(D – 1)^3$ annihilates the entire $(2x – x^2) e^x$ part.<\/p>\n\n\n\n

            Step 3: Annihilator for entire function<\/h4>\n\n\n\n

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

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

            We combine annihilators using the least common multiple (LCM)<\/strong> of the two operators:<\/p>\n\n\n\n

              \n
            • $(D + 1)$ and $(D – 1)^3$<\/li>\n<\/ul>\n\n\n\n

              Since these are distinct, the combined annihilator is:<\/p>\n\n\n\n

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

              \n\n\n\n

              \u2705 Final Answers:<\/h3>\n\n\n\n
                \n
              1. For<\/strong> $f(x) = 1 + 6x – 2x^3$
                Annihilator<\/strong>:\u2003\u2003$\\boxed{D^4}$<\/li>\n\n\n\n
              2. For<\/strong> $g(x) = e^{-x} + 2x e^x – x^2 e^x$
                Annihilator<\/strong>:\u2003\u2003$\\boxed{(D + 1)(D – 1)^3}$<\/li>\n<\/ol>\n\n\n\n
                \n\n\n\n

                \ud83d\udcd8 Explanation (Approx. 300 words):<\/h3>\n\n\n\n

                In differential equations, a linear differential operator<\/strong> is an operator involving derivatives that acts linearly on functions. For example, $D = \\frac{d}{dx}$, $D^2 = \\frac{d^2}{dx^2}$, etc. If a differential operator $L(D)$ applied to a function $f(x)$ yields zero, we say that $L(D)$ annihilates<\/strong> $f(x)$.<\/p>\n\n\n\n

                In the first function $f(x) = 1 + 6x – 2x^3$, we have a polynomial of degree 3. It\u2019s a known rule that a polynomial of degree $n$ is annihilated by $D^{n+1}$. Here, $D^4 f(x) = 0$, so $D^4$ is the annihilator.<\/p>\n\n\n\n

                In the second function $g(x) = e^{-x} + 2x e^x – x^2 e^x$, we deal with exponentials and polynomial-exponential terms. The function $e^{-x}$ is annihilated by $D + 1$, because differentiating $e^{-x}$ gives $-e^{-x}$, and adding back $e^{-x}$ yields zero.<\/p>\n\n\n\n

                For terms like $x e^x$, $x^2 e^x$, we use a standard rule: $x^n e^{ax}$ is annihilated by $(D – a)^{n+1}$. So $x^2 e^x$ is annihilated by $(D – 1)^3$, which also annihilates $x e^x$ and $e^x$.<\/p>\n\n\n\n

                To annihilate the full function, we combine these operators. Since they operate on independent parts of the function, their least common multiple gives the full annihilator: $(D + 1)(D – 1)^3$.<\/p>\n\n\n\n

                This method is fundamental in solving differential equations using the method of annihilators, especially when handling nonhomogeneous equations<\/p>\n\n\n\n

                \"\"<\/figure>\n\n\n\n

                .<\/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 given function means to find an operator $L(D)$, expressed in terms of the 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-217666","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217666","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=217666"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217666\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217666"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217666"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217666"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}