{"id":256735,"date":"2025-07-16T12:56:54","date_gmt":"2025-07-16T12:56:54","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=256735"},"modified":"2025-07-16T12:56:56","modified_gmt":"2025-07-16T12:56:56","slug":"use-the-laplace-transform-to-solve-ivp","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/16\/use-the-laplace-transform-to-solve-ivp\/","title":{"rendered":"Use the Laplace Transform to solve IVP"},"content":{"rendered":"\n<p>Use the Laplace Transform to solve IVP: y&#8221; + y = f(t). y(0) = 0, y'(0) = 1, where 0 &lt; t &lt; 6, 1 &lt; t. f(t) =<\/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 solve the initial value problem (IVP) using the <strong>Laplace Transform<\/strong>, we are given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Differential equation:<br><strong>y&#8221; + y = f(t)<\/strong><\/li>\n\n\n\n<li>Initial conditions:<br><strong>y(0) = 0<\/strong>, <strong>y'(0) = 1<\/strong><\/li>\n\n\n\n<li>For <strong>t > 1<\/strong>, the forcing function <strong>f(t) = 1<\/strong>, and for <strong>0 &lt; t &lt; 1<\/strong>, <strong>f(t) = 0<\/strong><\/li>\n<\/ul>\n\n\n\n<p>This means:<br><strong>f(t) = u\u2081(t)<\/strong> where <strong>u\u2081(t)<\/strong> is the <strong>unit step function<\/strong> starting at <strong>t = 1<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Take the Laplace Transform of both sides<\/h3>\n\n\n\n<p>Use the Laplace transform properties:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>L{y&#8221;} = s\u00b2Y(s) &#8211; sy(0) &#8211; y'(0) = s\u00b2Y(s) &#8211; 1<\/strong><\/li>\n\n\n\n<li><strong>L{y} = Y(s)<\/strong><\/li>\n\n\n\n<li><strong>L{f(t)} = L{u\u2081(t)} = e^{-s}\/s<\/strong><\/li>\n<\/ul>\n\n\n\n<p>Now apply the Laplace transform to the differential equation:<\/p>\n\n\n\n<p><strong>s\u00b2Y(s) &#8211; 1 + Y(s) = e^{-s}\/s<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Solve for Y(s)<\/h3>\n\n\n\n<p>Combine like terms:<\/p>\n\n\n\n<p><strong>(s\u00b2 + 1)Y(s) = 1 + e^{-s}\/s<\/strong><\/p>\n\n\n\n<p>Solve for Y(s):<\/p>\n\n\n\n<p><strong>Y(s) = [1 + e^{-s}\/s] \/ (s\u00b2 + 1)<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Take the inverse Laplace Transform<\/h3>\n\n\n\n<p>Break it into two terms:<\/p>\n\n\n\n<p><strong>Y(s) = 1\/(s\u00b2 + 1) + e^{-s}\/[s(s\u00b2 + 1)]<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>L\u207b\u00b9{1\/(s\u00b2 + 1)} = sin(t)<\/strong><\/li>\n\n\n\n<li><strong>L\u207b\u00b9{e^{-s}F(s)} = u\u2081(t) * f(t &#8211; 1)<\/strong> (time-shifting property)<\/li>\n<\/ul>\n\n\n\n<p>Let <strong>F(s) = 1\/[s(s\u00b2 + 1)]<\/strong><br>Use partial fractions:<\/p>\n\n\n\n<p><strong>1\/[s(s\u00b2 + 1)] = A\/s + (Bs + C)\/(s\u00b2 + 1)<\/strong><br>Solve:<br>1 = A(s\u00b2 + 1) + (Bs + C)(s)<br>\u2192 A = 1, B = -1, C = 0<\/p>\n\n\n\n<p>So:<\/p>\n\n\n\n<p><strong>F(s) = 1\/s &#8211; s\/(s\u00b2 + 1)<\/strong><br>Thus:<br><strong>L\u207b\u00b9{F(s)} = 1 &#8211; cos(t)<\/strong><\/p>\n\n\n\n<p>Now apply the time shift:<\/p>\n\n\n\n<p><strong>L\u207b\u00b9{e^{-s}F(s)} = u\u2081(t)(1 &#8211; cos(t &#8211; 1))<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p><strong>y(t) = sin(t) + u\u2081(t)(1 &#8211; cos(t &#8211; 1))<\/strong><\/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>We solved the second-order linear differential equation using the Laplace transform method. This method is powerful because it turns differential equations into algebraic equations, which are easier to handle.<\/p>\n\n\n\n<p>We began by transforming the original equation <strong>y&#8221; + y = f(t)<\/strong>. Using Laplace transform rules, we converted the left-hand side into terms of <strong>Y(s)<\/strong>, the transform of <strong>y(t)<\/strong>. For the right-hand side, we recognized <strong>f(t)<\/strong> as the <strong>unit step function u\u2081(t)<\/strong> because the forcing term is zero before t = 1 and one afterward. The Laplace transform of <strong>u\u2081(t)<\/strong> is <strong>e^{-s}\/s<\/strong>.<\/p>\n\n\n\n<p>After applying the Laplace transform to both sides and substituting initial conditions <strong>y(0) = 0<\/strong> and <strong>y'(0) = 1<\/strong>, we solved for <strong>Y(s)<\/strong>. The solution was a sum of two terms. The first, <strong>1\/(s\u00b2 + 1)<\/strong>, corresponds to the inverse transform <strong>sin(t)<\/strong>. The second term, involving <strong>e^{-s}<\/strong>, required the <strong>second shifting theorem<\/strong> (also called the time-delay property).<\/p>\n\n\n\n<p>We rewrote <strong>1\/[s(s\u00b2 + 1)]<\/strong> using partial fractions and took its inverse Laplace transform to get <strong>1 &#8211; cos(t)<\/strong>. Applying the shift gave us <strong>u\u2081(t)(1 &#8211; cos(t &#8211; 1))<\/strong>.<\/p>\n\n\n\n<p>Combining both parts, the solution to the original IVP is:<br><strong>y(t) = sin(t) + u\u2081(t)(1 &#8211; cos(t &#8211; 1))<\/strong><br>This piecewise behavior reflects the fact that the forcing function <strong>f(t)<\/strong> turns on at <strong>t = 1<\/strong>, affecting the system starting from that time.<\/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-banner6-680.jpeg\" alt=\"\" class=\"wp-image-256736\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Use the Laplace Transform to solve IVP: y&#8221; + y = f(t). y(0) = 0, y'(0) = 1, where 0 &lt; t &lt; 6, 1 &lt; t. f(t) = The Correct Answer and Explanation is: To solve the initial value problem (IVP) using the Laplace Transform, we are given: This means:f(t) = u\u2081(t) where u\u2081(t) [&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 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