{"id":36161,"date":"2025-06-25T03:55:13","date_gmt":"2025-06-25T03:55:13","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=36161"},"modified":"2025-06-25T03:55:15","modified_gmt":"2025-06-25T03:55:15","slug":"find-the-inverse-laplace-transform-of","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/find-the-inverse-laplace-transform-of\/","title":{"rendered":"Find the inverse Laplace transform of"},"content":{"rendered":"\n<pre id=\"preorder-ask-header-text\" class=\"wp-block-preformatted\"> Find the inverse Laplace transform of\u00a0<\/pre>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"638\" height=\"77\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-307.png\" alt=\"\" class=\"wp-image-36162\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-307.png 638w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-307-300x36.png 300w\" sizes=\"auto, (max-width: 638px) 100vw, 638px\" \/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>We are asked to find the <strong>inverse Laplace transform<\/strong> of the expression:L\u22121{slog\u2061(s\u22121s+1)}\\mathcal{L}^{-1}\\left\\{s \\log\\left(\\frac{s &#8211; 1}{s + 1}\\right)\\right\\}L\u22121{slog(s+1s\u22121\u200b)}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Solution:<\/strong><\/h3>\n\n\n\n<p>LetF(s)=slog\u2061(s\u22121s+1)F(s) = s \\log\\left(\\frac{s &#8211; 1}{s + 1}\\right)F(s)=slog(s+1s\u22121\u200b)<\/p>\n\n\n\n<p>We aim to find f(t)=L\u22121{F(s)}f(t) = \\mathcal{L}^{-1}\\{F(s)\\}f(t)=L\u22121{F(s)}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step-by-step Explanation:<\/strong><\/h3>\n\n\n\n<p>Let us denote:log\u2061(s\u22121s+1)=log\u2061(s\u22121)\u2212log\u2061(s+1)\\log\\left(\\frac{s &#8211; 1}{s + 1}\\right) = \\log(s &#8211; 1) &#8211; \\log(s + 1)log(s+1s\u22121\u200b)=log(s\u22121)\u2212log(s+1)<\/p>\n\n\n\n<p>Thus,F(s)=slog\u2061(s\u22121)\u2212slog\u2061(s+1)F(s) = s \\log(s &#8211; 1) &#8211; s \\log(s + 1)F(s)=slog(s\u22121)\u2212slog(s+1)<\/p>\n\n\n\n<p>Now, the Laplace transform of 1t(e\u2212at\u2212e\u2212bt)\\frac{1}{t} \\left(e^{-at} &#8211; e^{-bt}\\right)t1\u200b(e\u2212at\u2212e\u2212bt) is:L{1t(e\u2212at\u2212e\u2212bt)}=log\u2061(s+bs+a)\\mathcal{L}\\left\\{ \\frac{1}{t}(e^{-at} &#8211; e^{-bt}) \\right\\} = \\log\\left(\\frac{s + b}{s + a}\\right)L{t1\u200b(e\u2212at\u2212e\u2212bt)}=log(s+as+b\u200b)<\/p>\n\n\n\n<p>So by shifting variables and differentiating in the sss-domain, we can derive the inverse Laplace transform of slog\u2061(s\u22121s+1)s \\log\\left(\\frac{s &#8211; 1}{s + 1}\\right)slog(s+1s\u22121\u200b) using known results.<\/p>\n\n\n\n<p>Let\u2019s use known inverse Laplace transform results directly:<\/p>\n\n\n\n<p>There is a known Laplace transform identity:L\u22121{slog\u2061(s\u2212as+a)}=\u22122a2cosh\u2061(at)\u22121t\\mathcal{L}^{-1}\\left\\{ s \\log\\left(\\frac{s &#8211; a}{s + a} \\right) \\right\\} = -2a^2 \\frac{\\cosh(at) &#8211; 1}{t}L\u22121{slog(s+as\u2212a\u200b)}=\u22122a2tcosh(at)\u22121\u200b<\/p>\n\n\n\n<p>Set a=1a = 1a=1, so:L\u22121{slog\u2061(s\u22121s+1)}=\u22122\u22c5cosh\u2061(t)\u22121t\\mathcal{L}^{-1}\\left\\{ s \\log\\left(\\frac{s &#8211; 1}{s + 1} \\right) \\right\\} = -2 \\cdot \\frac{\\cosh(t) &#8211; 1}{t}L\u22121{slog(s+1s\u22121\u200b)}=\u22122\u22c5tcosh(t)\u22121\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Answer:<\/strong><\/h3>\n\n\n\n<p>f(t)=\u22122\u22c5cosh\u2061(t)\u22121t\\boxed{f(t) = -2 \\cdot \\frac{\\cosh(t) &#8211; 1}{t}}f(t)=\u22122\u22c5tcosh(t)\u22121\u200b\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation:<\/strong><\/h3>\n\n\n\n<p>This result is derived using the known Laplace transform pair involving logarithmic expressions. The hyperbolic cosine function cosh\u2061(t)=et+e\u2212t2\\cosh(t) = \\frac{e^t + e^{-t}}{2}cosh(t)=2et+e\u2212t\u200b arises naturally from the inverse transform of expressions involving log\u2061(s\u2212as+a)\\log\\left(\\frac{s &#8211; a}{s + a}\\right)log(s+as\u2212a\u200b). The factor 1t\\frac{1}{t}t1\u200b appears due to the s-multiplication in the Laplace domain, which corresponds to a differentiation operation or time division in the time domain.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-765.jpeg\" alt=\"\" class=\"wp-image-36163\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-765.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-765-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-765-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Find the inverse Laplace transform of\u00a0 The Correct Answer and Explanation is: We are asked to find the inverse Laplace transform of the expression:L\u22121{slog\u2061(s\u22121s+1)}\\mathcal{L}^{-1}\\left\\{s \\log\\left(\\frac{s &#8211; 1}{s + 1}\\right)\\right\\}L\u22121{slog(s+1s\u22121\u200b)} Solution: LetF(s)=slog\u2061(s\u22121s+1)F(s) = s \\log\\left(\\frac{s &#8211; 1}{s + 1}\\right)F(s)=slog(s+1s\u22121\u200b) We aim to find f(t)=L\u22121{F(s)}f(t) = \\mathcal{L}^{-1}\\{F(s)\\}f(t)=L\u22121{F(s)} Step-by-step Explanation: Let us denote:log\u2061(s\u22121s+1)=log\u2061(s\u22121)\u2212log\u2061(s+1)\\log\\left(\\frac{s &#8211; 1}{s + 1}\\right) = [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-36161","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36161","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=36161"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36161\/revisions"}],"predecessor-version":[{"id":36164,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36161\/revisions\/36164"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=36161"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=36161"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=36161"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}