To find a linear differential operator<\/strong> that annihilates<\/em> a given function means to find a differential operator L(D)L(D), written in terms of D=ddxD = \\frac{d}{dx}, such that: L(D)[f(x)]=0L(D)[f(x)] = 0<\/p>\n\n\n\n We are given two separate functions to analyze:<\/p>\n\n\n\n This is a polynomial function<\/strong> of degree 3.<\/p>\n\n\n\n Any polynomial of degree nn is annihilated by the differential operator Dn+1D^{n+1}. So, for a polynomial of degree 3: D4[1+6x\u22122×3]=0D^4[1 + 6x – 2x^3] = 0<\/p>\n\n\n\n The linear differential operator that annihilates this function is: D4\\boxed{D^4}<\/p>\n\n\n\n This is a linear combination<\/strong> of exponential and exponential-polynomial terms. Let’s break it down:<\/p>\n\n\n\n In general, xneaxx^n e^{ax} is annihilated by (D\u2212a)n+1(D – a)^{n+1}<\/p>\n<\/blockquote>\n\n\n\n So, we take the least common multiple (LCM)<\/strong> of all individual annihilators:<\/p>\n\n\n\n The full differential operator is: (D+1)(D\u22121)3\\boxed{(D + 1)(D – 1)^3}<\/p>\n\n\n\n D4\\boxed{D^4}<\/p>\n\n\n\n (D+1)(D\u22121)3\\boxed{(D + 1)(D – 1)^3}<\/p>\n\n\n\n In differential equations, a linear differential operator<\/strong> like D=ddxD = \\frac{d}{dx} acts on a function to produce its derivative. An operator annihilates<\/strong> a function when applying it results in zero. To find such an operator, we look at the structure of the function.<\/p>\n\n\n\n For the first function, 1+6x\u22122×31 + 6x – 2x^3, it\u2019s a polynomial of degree 3. Each differentiation reduces the degree by 1. After four derivatives, the result is zero. Hence, D4D^4 is the smallest differential operator that annihilates it.<\/p>\n\n\n\n For the second function, e\u2212x+2xex\u2212x2exe^{-x} + 2x e^x – x^2 e^x, each term is exponential or exponential times a polynomial. The term e\u2212xe^{-x} is annihilated by D+1D + 1 because the derivative of e\u2212xe^{-x} is \u2212e\u2212x-e^{-x}. The terms 2xex2x e^x and \u2212x2ex-x^2 e^x are of the form xnexx^n e^x, which are annihilated by (D\u22121)n+1(D – 1)^{n+1}. Specifically, xexx e^x is annihilated by (D\u22121)2(D – 1)^2, and x2exx^2 e^x by (D\u22121)3(D – 1)^3.<\/p>\n\n\n\n To annihilate the whole function, we need a common operator that eliminates all terms. That means we need to take the product of their annihilators: (D+1)(D + 1) and (D\u22121)3(D – 1)^3. This product ensures that each term individually is sent to zero.<\/p>\n\n\n\n Therefore, the second function is annihilated by (D+1)(D\u22121)3(D + 1)(D – 1)^3.<\/p>\n\n\n\n Match each function name with its equation. 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 a 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-219790","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219790","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=219790"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219790\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=219790"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=219790"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=219790"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nFunction 1: f(x)=1+6x\u22122x3f(x) = 1 + 6x – 2x^3<\/strong><\/h3>\n\n\n\n
Step 1: Recall the rule<\/h4>\n\n\n\n
\u2705 Answer:<\/strong><\/h4>\n\n\n\n
\n\n\n\nFunction 2: f(x)=e\u2212x+2xex\u2212x2exf(x) = e^{-x} + 2x e^x – x^2 e^x<\/strong><\/h3>\n\n\n\n
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\u2705 Answer:<\/strong><\/h4>\n\n\n\n
\n\n\n\n\u2705 Final Answers:<\/h3>\n\n\n\n
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\n\n\n\n\u270d\ufe0f Explanation <\/h3>\n\n\n\n
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