{"id":238595,"date":"2025-06-18T08:51:17","date_gmt":"2025-06-18T08:51:17","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=238595"},"modified":"2025-06-18T08:51:19","modified_gmt":"2025-06-18T08:51:19","slug":"obtain-the-laplace-transform-of-the-function-5e-2t-3-_-2cos-2t","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/18\/obtain-the-laplace-transform-of-the-function-5e-2t-3-_-2cos-2t\/","title":{"rendered":"Obtain the Laplace transform of the function: 5e-2t + 3 _ 2cos 2t"},"content":{"rendered":"\n<p>Obtain the Laplace transform of the function: 5e-2t + 3 _ 2cos 2t<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To obtain the Laplace transform of the function<br>f(t)=5e\u22122t+3\u22122cos\u2061(2t),f(t) = 5e^{-2t} + 3 &#8211; 2\\cos(2t),f(t)=5e\u22122t+3\u22122cos(2t),<br>we will apply the basic Laplace transform rules to each term individually.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step-by-step Laplace Transform:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Laplace Transform of<\/strong> 5e\u22122t5e^{-2t}5e\u22122t: L{e\u2212at}=1s+a,soL{5e\u22122t}=5s+2\\mathcal{L}\\{e^{-at}\\} = \\frac{1}{s + a}, \\quad \\text{so} \\quad \\mathcal{L}\\{5e^{-2t}\\} = \\frac{5}{s + 2}L{e\u2212at}=s+a1\u200b,soL{5e\u22122t}=s+25\u200b<\/li>\n\n\n\n<li><strong>Laplace Transform of constant<\/strong> 333: L{3}=3s\\mathcal{L}\\{3\\} = \\frac{3}{s}L{3}=s3\u200b<\/li>\n\n\n\n<li><strong>Laplace Transform of<\/strong> \u22122cos\u2061(2t)-2\\cos(2t)\u22122cos(2t): L{cos\u2061(at)}=ss2+a2,soL{\u22122cos\u2061(2t)}=\u22122\u22c5ss2+4\\mathcal{L}\\{\\cos(at)\\} = \\frac{s}{s^2 + a^2}, \\quad \\text{so} \\quad \\mathcal{L}\\{-2\\cos(2t)\\} = -2 \\cdot \\frac{s}{s^2 + 4}L{cos(at)}=s2+a2s\u200b,soL{\u22122cos(2t)}=\u22122\u22c5s2+4s\u200b<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Laplace Transform:<\/h3>\n\n\n\n<p>Now, combine all the individual transforms: L{f(t)}=5s+2+3s\u22122ss2+4\\mathcal{L}\\{f(t)\\} = \\frac{5}{s + 2} + \\frac{3}{s} &#8211; \\frac{2s}{s^2 + 4}L{f(t)}=s+25\u200b+s3\u200b\u2212s2+42s\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>The Laplace transform is a powerful integral transform used to convert time-domain functions into s-domain expressions. This is especially useful in solving linear differential equations and analyzing systems in engineering and physics.<\/p>\n\n\n\n<p>The given function is: f(t)=5e\u22122t+3\u22122cos\u2061(2t)f(t) = 5e^{-2t} + 3 &#8211; 2\\cos(2t)f(t)=5e\u22122t+3\u22122cos(2t)<\/p>\n\n\n\n<p>We break this down into simpler parts and apply known Laplace transform formulas.<\/p>\n\n\n\n<p>First, for exponential functions like e\u2212ate^{-at}e\u2212at, the Laplace transform is: L{e\u2212at}=1s+a\\mathcal{L}\\{e^{-at}\\} = \\frac{1}{s + a}L{e\u2212at}=s+a1\u200b<\/p>\n\n\n\n<p>Multiplying by a constant (in this case 5) simply scales the result. So 5e\u22122t5e^{-2t}5e\u22122t becomes 5s+2\\frac{5}{s + 2}s+25\u200b.<\/p>\n\n\n\n<p>Next, for constants such as 3, the Laplace transform is: L{c}=cs\\mathcal{L}\\{c\\} = \\frac{c}{s}L{c}=sc\u200b<\/p>\n\n\n\n<p>Hence, the transform of 3 is 3s\\frac{3}{s}s3\u200b.<\/p>\n\n\n\n<p>For trigonometric functions, the Laplace transform of cosine is: L{cos\u2061(at)}=ss2+a2\\mathcal{L}\\{\\cos(at)\\} = \\frac{s}{s^2 + a^2}L{cos(at)}=s2+a2s\u200b<\/p>\n\n\n\n<p>In this case, cos\u2061(2t)\\cos(2t)cos(2t) transforms to ss2+4\\frac{s}{s^2 + 4}s2+4s\u200b, and multiplying by -2 gives \u22122ss2+4-\\frac{2s}{s^2 + 4}\u2212s2+42s\u200b.<\/p>\n\n\n\n<p>Adding all parts together, we get the total Laplace transform: 5s+2+3s\u22122ss2+4\\frac{5}{s + 2} + \\frac{3}{s} &#8211; \\frac{2s}{s^2 + 4}s+25\u200b+s3\u200b\u2212s2+42s\u200b<\/p>\n\n\n\n<p>This compact s-domain expression can now be used to analyze systems or solve differential equations more easily than working in the time domain.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1037.jpeg\" alt=\"\" class=\"wp-image-238596\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Obtain the Laplace transform of the function: 5e-2t + 3 _ 2cos 2t The Correct Answer and Explanation is: To obtain the Laplace transform of the functionf(t)=5e\u22122t+3\u22122cos\u2061(2t),f(t) = 5e^{-2t} + 3 &#8211; 2\\cos(2t),f(t)=5e\u22122t+3\u22122cos(2t),we will apply the basic Laplace transform rules to each term individually. Step-by-step Laplace Transform: Final Laplace Transform: Now, combine all the individual [&hellip;]<\/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-238595","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/238595","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=238595"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/238595\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=238595"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=238595"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=238595"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}