To solve this initial-value problem using the Laplace transform, we need to take the Laplace transform of the differential equation and apply the initial conditions. Let’s break it down step by step.<\/p>\n\n\n\n
The given equation is:<\/h3>\n\n\n\n
y\u2032\u2032+4y=\u03b4(t\u22122)y” + 4y = \\delta(t – 2)y\u2032\u2032+4y=\u03b4(t\u22122)<\/p>\n\n\n\n
with initial conditions:y(0)=1,y\u2032(0)=0y(0) = 1, \\quad y'(0) = 0y(0)=1,y\u2032(0)=0<\/p>\n\n\n\n
where \u03b4(t\u22122)\\delta(t – 2)\u03b4(t\u22122) is the Dirac delta function, which represents an impulse at t=2t = 2t=2.<\/p>\n\n\n\n
Step 1: Apply the Laplace Transform to the differential equation<\/h3>\n\n\n\n
The Laplace transform of the second derivative y\u2032\u2032(t)y”(t)y\u2032\u2032(t) is given by:L{y\u2032\u2032(t)}=s2Y(s)\u2212sy(0)\u2212y\u2032(0)\\mathcal{L}\\{y”(t)\\} = s^2 Y(s) – sy(0) – y'(0)L{y\u2032\u2032(t)}=s2Y(s)\u2212sy(0)\u2212y\u2032(0)<\/p>\n\n\n\n
The Laplace transform of y(t)y(t)y(t) is Y(s)Y(s)Y(s), and the Laplace transform of the Dirac delta function \u03b4(t\u22122)\\delta(t – 2)\u03b4(t\u22122) is:L{\u03b4(t\u22122)}=e\u22122s\\mathcal{L}\\{\\delta(t – 2)\\} = e^{-2s}L{\u03b4(t\u22122)}=e\u22122s<\/p>\n\n\n\n
Substituting into the equation:(s2Y(s)\u2212s\u22c51\u22120)+4Y(s)=e\u22122s(s^2 Y(s) – s \\cdot 1 – 0) + 4Y(s) = e^{-2s}(s2Y(s)\u2212s\u22c51\u22120)+4Y(s)=e\u22122s<\/p>\n\n\n\n
Simplifying:(s2+4)Y(s)\u2212s=e\u22122s(s^2 + 4)Y(s) – s = e^{-2s}(s2+4)Y(s)\u2212s=e\u22122s<\/p>\n\n\n\n
Rearranging for Y(s)Y(s)Y(s):Y(s)=s+e\u22122ss2+4Y(s) = \\frac{s + e^{-2s}}{s^2 + 4}Y(s)=s2+4s+e\u22122s\u200b<\/p>\n\n\n\n
Step 2: Inverse Laplace Transform<\/h3>\n\n\n\n
Now, we need to take the inverse Laplace transform of Y(s)Y(s)Y(s). This expression can be split into two parts:Y(s)=ss2+4+e\u22122ss2+4Y(s) = \\frac{s}{s^2 + 4} + \\frac{e^{-2s}}{s^2 + 4}Y(s)=s2+4s\u200b+s2+4e\u22122s\u200b<\/p>\n\n\n\n
- \n
- Inverse Laplace Transform of ss2+4\\frac{s}{s^2 + 4}s2+4s\u200b:<\/strong> L\u22121{ss2+4}=cos\u2061(2t)\\mathcal{L}^{-1}\\left\\{\\frac{s}{s^2 + 4}\\right\\} = \\cos(2t)L\u22121{s2+4s\u200b}=cos(2t)<\/li>\n\n\n\n
- Inverse Laplace Transform of e\u22122ss2+4\\frac{e^{-2s}}{s^2 + 4}s2+4e\u22122s\u200b:<\/strong>
This is the Laplace transform of a shifted function: L\u22121{e\u22122ss2+4}=U(t\u22122)\u22c5sin\u2061[2(t\u22122)]\\mathcal{L}^{-1}\\left\\{\\frac{e^{-2s}}{s^2 + 4}\\right\\} = U(t – 2) \\cdot \\sin[2(t – 2)]L\u22121{s2+4e\u22122s\u200b}=U(t\u22122)\u22c5sin[2(t\u22122)] where U(t\u22122)U(t – 2)U(t\u22122) is the Heaviside step function.<\/li>\n<\/ol>\n\n\n\nStep 3: Combine the Results<\/h3>\n\n\n\n
Combining these results, we have:y(t)=cos\u2061(2t)+U(t\u22122)\u22c5sin\u2061[2(t\u22122)]y(t) = \\cos(2t) + U(t – 2) \\cdot \\sin[2(t – 2)]y(t)=cos(2t)+U(t\u22122)\u22c5sin[2(t\u22122)]<\/p>\n\n\n\n
This is the solution to the given initial-value problem. The term U(t\u22122)U(t – 2)U(t\u22122) shifts the sine wave to start at t=2t = 2t=2.<\/p>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
y(t)=cos\u2061(2t)+U(t\u22122)\u22c5sin\u2061[2(t\u22122)]y(t) = \\cos(2t) + U(t – 2) \\cdot \\sin[2(t – 2)]y(t)=cos(2t)+U(t\u22122)\u22c5sin[2(t\u22122)]<\/p>\n\n\n\n
This solution is correct because the cosine term satisfies the equation before the impulse at t=2t = 2t=2, and the shifted sine term handles the effect of the delta function at t=2t = 2t=2.<\/p>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-1.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"Use the Laplace transform to solve the given initial-value problem. y” + 4y = \\delta (t – 2), y(0) = 1, y'(0) = 0 y(t) = \\cos(2t) + U(t – 2) \\cdot \\frac{1}{2}\\sin[2(t – 2)] y(t) = \\frac{1}{2}U(t – 2)\\sin(2t) y(t) = \\frac{1}{2}U(t – 2)\\sin[2(t – 2)] y(t) = \\cos(2t) + U(t – 2) \\cdot […]<\/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-249253","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/249253","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=249253"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/249253\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=249253"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=249253"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=249253"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- Inverse Laplace Transform of e\u22122ss2+4\\frac{e^{-2s}}{s^2 + 4}s2+4e\u22122s\u200b:<\/strong>