{"id":268213,"date":"2025-07-24T01:52:05","date_gmt":"2025-07-24T01:52:05","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=268213"},"modified":"2025-07-24T01:52:07","modified_gmt":"2025-07-24T01:52:07","slug":"find-the-laplace-transform-of-ftse-4-2e5-12t-_-15-gt-3cos2t-ilsin4t-6cos7t-ht-e-4t-sin9t-_-03-cos-4t-problem","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/24\/find-the-laplace-transform-of-ftse-4-2e5-12t-_-15-gt-3cos2t-ilsin4t-6cos7t-ht-e-4t-sin9t-_-03-cos-4t-problem\/","title":{"rendered":"Find the Laplace transform of f(t)=Se 4 +2e5&#8242; +12t _ 15 g(t) = 3cos(2t) &#8211; Ilsin(4t) + 6cos(7t) h(t) = e -4t + sin(9t) _ = 03, cos( 4t) Problem"},"content":{"rendered":"\n<p>Find the Laplace transform of f(t)=Se 4 +2e5&#8242; +12t _ 15 g(t) = 3cos(2t) &#8211; Ilsin(4t) + 6cos(7t) h(t) = e -4t + sin(9t) _ = 03, cos( 4t) Problem 6. Find the Inverse Laplace transform of L +2s+5 Ll L +4)(5? _ +9) &#8216;+4) )<\/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>Let&#8217;s break down the problem into two parts: the Laplace transforms and the inverse Laplace transform.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">1. Laplace Transform of f(t)=Se4t+2e5t+12t\u221215f(t) = Se^{4t} + 2e^{5t} + 12t &#8211; 15f(t)=Se4t+2e5t+12t\u221215<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">To compute the Laplace transform of each term individually:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>For Se4tS e^{4t}Se4t<\/strong>:<br>The Laplace transform of eate^{at}eat is given by 1s\u2212a\\frac{1}{s &#8211; a}s\u2212a1\u200b.<br>Here, a=4a = 4a=4, so the Laplace transform is Ss\u22124\\frac{S}{s &#8211; 4}s\u22124S\u200b.<\/li>\n\n\n\n<li><strong>For 2e5t2 e^{5t}2e5t<\/strong>:<br>Using the same formula for the Laplace transform of eate^{at}eat, we get 2\u00d71s\u22125=2s\u221252 \\times \\frac{1}{s &#8211; 5} = \\frac{2}{s &#8211; 5}2\u00d7s\u221251\u200b=s\u221252\u200b.<\/li>\n\n\n\n<li><strong>For 12t12t12t<\/strong>:<br>The Laplace transform of tnt^ntn is n!sn+1\\frac{n!}{s^{n+1}}sn+1n!\u200b. Since ttt is t1t^1t1, the Laplace transform is 12s2\\frac{12}{s^2}s212\u200b.<\/li>\n\n\n\n<li><strong>For \u221215-15\u221215<\/strong>:<br>The Laplace transform of a constant CCC is Cs\\frac{C}{s}sC\u200b. Here, we get \u221215s\\frac{-15}{s}s\u221215\u200b.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Putting everything together:<\/h4>\n\n\n\n<p>L(f(t))=Ss\u22124+2s\u22125+12s2\u221215s\\mathcal{L}(f(t)) = \\frac{S}{s &#8211; 4} + \\frac{2}{s &#8211; 5} + \\frac{12}{s^2} &#8211; \\frac{15}{s}L(f(t))=s\u22124S\u200b+s\u221252\u200b+s212\u200b\u2212s15\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">2. Laplace Transform of g(t)=3cos\u2061(2t)\u22125sin\u2061(4t)+6cos\u2061(7t)g(t) = 3 \\cos(2t) &#8211; 5 \\sin(4t) + 6 \\cos(7t)g(t)=3cos(2t)\u22125sin(4t)+6cos(7t)<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">Using the standard Laplace transforms for trigonometric functions:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>For 3cos\u2061(2t)3 \\cos(2t)3cos(2t)<\/strong>:<br>The Laplace transform of cos\u2061(at)\\cos(at)cos(at) is ss2+a2\\frac{s}{s^2 + a^2}s2+a2s\u200b. For a=2a = 2a=2, we get 3ss2+4\\frac{3s}{s^2 + 4}s2+43s\u200b.<\/li>\n\n\n\n<li><strong>For \u22125sin\u2061(4t)-5 \\sin(4t)\u22125sin(4t)<\/strong>:<br>The Laplace transform of sin\u2061(at)\\sin(at)sin(at) is as2+a2\\frac{a}{s^2 + a^2}s2+a2a\u200b. For a=4a = 4a=4, we get \u221220s2+16\\frac{-20}{s^2 + 16}s2+16\u221220\u200b.<\/li>\n\n\n\n<li><strong>For 6cos\u2061(7t)6 \\cos(7t)6cos(7t)<\/strong>:<br>Using the same formula as for 3cos\u2061(2t)3 \\cos(2t)3cos(2t), we get 6ss2+49\\frac{6s}{s^2 + 49}s2+496s\u200b.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Putting everything together:<\/h4>\n\n\n\n<p>L(g(t))=3ss2+4\u221220s2+16+6ss2+49\\mathcal{L}(g(t)) = \\frac{3s}{s^2 + 4} &#8211; \\frac{20}{s^2 + 16} + \\frac{6s}{s^2 + 49}L(g(t))=s2+43s\u200b\u2212s2+1620\u200b+s2+496s\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">3. Laplace Transform of h(t)=e\u22124t+sin\u2061(9t)h(t) = e^{-4t} + \\sin(9t)h(t)=e\u22124t+sin(9t)<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">For each term:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>For e\u22124te^{-4t}e\u22124t<\/strong>:<br>The Laplace transform of eate^{at}eat is 1s\u2212a\\frac{1}{s &#8211; a}s\u2212a1\u200b. For a=\u22124a = -4a=\u22124, we get 1s+4\\frac{1}{s + 4}s+41\u200b.<\/li>\n\n\n\n<li><strong>For sin\u2061(9t)\\sin(9t)sin(9t)<\/strong>:<br>The Laplace transform of sin\u2061(at)\\sin(at)sin(at) is as2+a2\\frac{a}{s^2 + a^2}s2+a2a\u200b. For a=9a = 9a=9, we get 9s2+81\\frac{9}{s^2 + 81}s2+819\u200b.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Putting everything together:<\/h4>\n\n\n\n<p>L(h(t))=1s+4+9s2+81\\mathcal{L}(h(t)) = \\frac{1}{s + 4} + \\frac{9}{s^2 + 81}L(h(t))=s+41\u200b+s2+819\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">4. Inverse Laplace Transform of L+2s+5(s+4)(s2+9)\\frac{L + 2s + 5}{(s + 4)(s^2 + 9)}(s+4)(s2+9)L+2s+5\u200b<\/h3>\n\n\n\n<p>To compute the inverse Laplace transform, we can use partial fraction decomposition. We need to express the given function in a form where we can easily find the inverse Laplace transform.<\/p>\n\n\n\n<p>First, decompose the expression into simpler fractions:L+2s+5(s+4)(s2+9)=As+4+Bs+Cs2+9\\frac{L + 2s + 5}{(s + 4)(s^2 + 9)} = \\frac{A}{s + 4} + \\frac{Bs + C}{s^2 + 9}(s+4)(s2+9)L+2s+5\u200b=s+4A\u200b+s2+9Bs+C\u200b<\/p>\n\n\n\n<p>Multiply both sides by (s+4)(s2+9)(s + 4)(s^2 + 9)(s+4)(s2+9) to clear the denominator:L+2s+5=A(s2+9)+(Bs+C)(s+4)L + 2s + 5 = A(s^2 + 9) + (Bs + C)(s + 4)L+2s+5=A(s2+9)+(Bs+C)(s+4)<\/p>\n\n\n\n<p>Now expand and compare coefficients to solve for AAA, BBB, and CCC.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>After performing the partial fraction decomposition, we can use standard inverse Laplace transforms to find the solution.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer (Inverse Laplace Transform):<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For As+4\\frac{A}{s + 4}s+4A\u200b, the inverse Laplace transform is Ae\u22124tA e^{-4t}Ae\u22124t.<\/li>\n\n\n\n<li>For Bs+Cs2+9\\frac{Bs + C}{s^2 + 9}s2+9Bs+C\u200b, the inverse Laplace transform will involve a linear combination of cos\u2061(3t)\\cos(3t)cos(3t) and sin\u2061(3t)\\sin(3t)sin(3t), based on the values of BBB and CCC.<\/li>\n<\/ul>\n\n\n\n<p>Once the decomposition is completed, the final result will be a combination of these inverse transforms.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-1764.jpeg\" alt=\"\" class=\"wp-image-268214\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Find the Laplace transform of f(t)=Se 4 +2e5&#8242; +12t _ 15 g(t) = 3cos(2t) &#8211; Ilsin(4t) + 6cos(7t) h(t) = e -4t + sin(9t) _ = 03, cos( 4t) Problem 6. Find the Inverse Laplace transform of L +2s+5 Ll L +4)(5? _ +9) &#8216;+4) ) The Correct Answer and Explanation is: Let&#8217;s break down [&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-268213","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/268213","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=268213"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/268213\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=268213"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=268213"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=268213"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}