We are asked to solve two initial value problems (IVPs) using Laplace Transforms<\/strong>.<\/p>\n\n\n\n d2ydt2+6dydt+13y=13u4(t),y(0)=3,y\u2032(0)=1\\frac{d^2y}{dt^2} + 6\\frac{dy}{dt} + 13y = 13u_4(t), \\quad y(0) = 3, \\quad y'(0) = 1dt2d2y\u200b+6dtdy\u200b+13y=13u4\u200b(t),y(0)=3,y\u2032(0)=1<\/p>\n\n\n\n Step 1: Take Laplace Transforms of both sides<\/strong><\/p>\n\n\n\n Let Y(s)Y(s)Y(s) be the Laplace Transform of y(t)y(t)y(t). Apply the Laplace Transform:s2Y(s)\u22123s\u22121+6(sY(s)\u22123)+13Y(s)=13\u22c5e\u22124sss^2Y(s) – 3s – 1 + 6(sY(s) – 3) + 13Y(s) = 13 \\cdot \\frac{e^{-4s}}{s}s2Y(s)\u22123s\u22121+6(sY(s)\u22123)+13Y(s)=13\u22c5se\u22124s\u200b<\/p>\n\n\n\n Simplify:s2Y(s)\u22123s\u22121+6sY(s)\u221218+13Y(s)=13e\u22124sss^2Y(s) – 3s – 1 + 6sY(s) – 18 + 13Y(s) = \\frac{13e^{-4s}}{s}s2Y(s)\u22123s\u22121+6sY(s)\u221218+13Y(s)=s13e\u22124s\u200b<\/p>\n\n\n\n Combine like terms:(s2+6s+13)Y(s)=13e\u22124ss+3s+19(s^2 + 6s + 13)Y(s) = \\frac{13e^{-4s}}{s} + 3s + 19(s2+6s+13)Y(s)=s13e\u22124s\u200b+3s+19<\/p>\n\n\n\n Solve for Y(s)Y(s)Y(s):Y(s)=13e\u22124ss(s2+6s+13)+3s+19s2+6s+13Y(s) = \\frac{13e^{-4s}}{s(s^2 + 6s + 13)} + \\frac{3s + 19}{s^2 + 6s + 13}Y(s)=s(s2+6s+13)13e\u22124s\u200b+s2+6s+133s+19\u200b<\/p>\n\n\n\n Now take the inverse Laplace of each term separately using known formulas and shift theorems.<\/p>\n\n\n\n d2ydt2+4dydt+9y=20u2(t)sin\u2061(t),y(0)=1,y\u2032(0)=2\\frac{d^2y}{dt^2} + 4\\frac{dy}{dt} + 9y = 20u_2(t)\\sin(t), \\quad y(0) = 1, \\quad y'(0) = 2dt2d2y\u200b+4dtdy\u200b+9y=20u2\u200b(t)sin(t),y(0)=1,y\u2032(0)=2<\/p>\n\n\n\n Step 1: Take Laplace Transforms<\/strong>L{y\u2032\u2032(t)}+4L{y\u2032(t)}+9L{y(t)}=20L{u2(t)sin\u2061(t)}\\mathcal{L}\\{y”(t)\\} + 4\\mathcal{L}\\{y'(t)\\} + 9\\mathcal{L}\\{y(t)\\} = 20\\mathcal{L}\\{u_2(t)\\sin(t)\\}L{y\u2032\u2032(t)}+4L{y\u2032(t)}+9L{y(t)}=20L{u2\u200b(t)sin(t)}<\/p>\n\n\n\n Use:L{sin\u2061(t)u2(t)}=e\u22122s\u22c51s2+1\\mathcal{L}\\{\\sin(t)u_2(t)\\} = e^{-2s} \\cdot \\frac{1}{s^2 + 1}L{sin(t)u2\u200b(t)}=e\u22122s\u22c5s2+11\u200b<\/p>\n\n\n\n So:20\u22c5e\u22122ss2+120 \\cdot \\frac{e^{-2s}}{s^2 + 1}20\u22c5s2+1e\u22122s\u200b<\/p>\n\n\n\n Now, apply Laplace to LHS:s2Y(s)\u2212sy(0)\u2212y\u2032(0)+4(sY(s)\u2212y(0))+9Y(s)=20e\u22122ss2+1s^2Y(s) – sy(0) – y'(0) + 4(sY(s) – y(0)) + 9Y(s) = \\frac{20e^{-2s}}{s^2 + 1}s2Y(s)\u2212sy(0)\u2212y\u2032(0)+4(sY(s)\u2212y(0))+9Y(s)=s2+120e\u22122s\u200b<\/p>\n\n\n\n Substitute initial values:s2Y(s)\u2212s\u22122+4sY(s)\u22124+9Y(s)=20e\u22122ss2+1s^2Y(s) – s – 2 + 4sY(s) – 4 + 9Y(s) = \\frac{20e^{-2s}}{s^2 + 1}s2Y(s)\u2212s\u22122+4sY(s)\u22124+9Y(s)=s2+120e\u22122s\u200b<\/p>\n\n\n\n Simplify:(s2+4s+9)Y(s)=20e\u22122ss2+1+s+6(s^2 + 4s + 9)Y(s) = \\frac{20e^{-2s}}{s^2 + 1} + s + 6(s2+4s+9)Y(s)=s2+120e\u22122s\u200b+s+6<\/p>\n\n\n\n Solve for Y(s)Y(s)Y(s):Y(s)=20e\u22122s(s2+1)(s2+4s+9)+s+6s2+4s+9Y(s) = \\frac{20e^{-2s}}{(s^2 + 1)(s^2 + 4s + 9)} + \\frac{s + 6}{s^2 + 4s + 9}Y(s)=(s2+1)(s2+4s+9)20e\u22122s\u200b+s2+4s+9s+6\u200b<\/p>\n\n\n\n Take inverse Laplace term by term using partial fraction decomposition and known transforms. The shifting property applies due to the e\u22122se^{-2s}e\u22122s factor, which shifts the function by 2 units in time.<\/p>\n\n\n\n These solutions represent the response of the systems to a unit step or shifted sine input, filtered through a second-order differential equation. Laplace Transform simplifies solving such linear time-invariant differential equations by converting them into algebraic equations in the frequency domain.<\/p>\n\n\n\n Find the solution of the IVP using Laplace Transform a) \\frac{d^2y}{dt^2} + 6\\frac{dy}{dt} + 13y = 13u_4(t), y(0) = 3, y'(0) = 1 b) \\frac{d^2y}{dt^2} + 4\\frac{dy}{dt} + 9y = 20u_2(t)sin(t), y(0) = 1, y'(0) = 2 The Correct Answer and Explanation is: We are asked to solve two initial value problems (IVPs) using Laplace […]<\/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-24491","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/24491","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=24491"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/24491\/revisions"}],"predecessor-version":[{"id":24499,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/24491\/revisions\/24499"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=24491"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=24491"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=24491"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nProblem (a):<\/h3>\n\n\n\n
Use known transforms:L{y\u2032\u2032(t)}=s2Y(s)\u2212sy(0)\u2212y\u2032(0)\\mathcal{L}\\left\\{y”(t)\\right\\} = s^2Y(s) – sy(0) – y'(0)L{y\u2032\u2032(t)}=s2Y(s)\u2212sy(0)\u2212y\u2032(0)L{y\u2032(t)}=sY(s)\u2212y(0)\\mathcal{L}\\left\\{y'(t)\\right\\} = sY(s) – y(0)L{y\u2032(t)}=sY(s)\u2212y(0)L{u4(t)}=e\u22124ss\\mathcal{L}\\left\\{u_4(t)\\right\\} = \\frac{e^{-4s}}{s}L{u4\u200b(t)}=se\u22124s\u200b<\/p>\n\n\n\n
\n\n\n\nProblem (b):<\/h3>\n\n\n\n
\n\n\n\n![\"\"](\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-1.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"