{"id":226332,"date":"2025-06-05T06:10:14","date_gmt":"2025-06-05T06:10:14","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=226332"},"modified":"2025-06-05T06:10:17","modified_gmt":"2025-06-05T06:10:17","slug":"described-by-the-following-differential-equa-tions-with-the-specified-input-and-initial-conditions","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/05\/described-by-the-following-differential-equa-tions-with-the-specified-input-and-initial-conditions\/","title":{"rendered":"Described by the following differential equa- tions with the specified input and initial conditions"},"content":{"rendered":"\n

Described by the following differential equa- tions with the specified input and initial conditions: (a)<\/p>\n\n\n\n

y(0?) = -1, $\\frac{d}{dt}y(t)|_{t=0?} = 5, x(t) = e??u(t) (b) Determine the initial value for the forced and natural response individually. Do you know what is the difference between the natural response and forced response?
Determine the forced and natural responses for the LTI systems described by the following differential equa- tions with the specified input and initial conditions: (a)<\/p>\n\n\n\n

y(0?) = -1, $\\frac{d}{dt}y(t)|_{t=0?} = 5, x(t) = e??u(t) (b) Determine the initial value for the forced and natural response individually. Do you know what is the difference between the natural response and forced response?<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To determine the forced<\/strong> and natural responses<\/strong> of a linear time-invariant (LTI) system, we must first understand their meanings:<\/p>\n\n\n\n

\n\n\n\n

Natural Response vs Forced Response<\/strong><\/h3>\n\n\n\n
    \n
  • Natural Response<\/strong>: The system\u2019s output due to initial conditions only<\/strong> (i.e., assuming input x(t)=0x(t) = 0).<\/li>\n\n\n\n
  • Forced Response<\/strong>: The system\u2019s output due to external input only<\/strong> (i.e., assuming initial conditions are zero).<\/li>\n<\/ul>\n\n\n\n

    The total response<\/strong> is the sum<\/strong> of the natural and forced responses.<\/p>\n\n\n\n

    \n\n\n\n

    Given Information<\/strong><\/h3>\n\n\n\n

    We\u2019re given:<\/p>\n\n\n\n

      \n
    • Initial conditions:
      y(0\u2212)=\u22121y(0^-) = -1,
      dy(t)dt\u2223t=0\u2212=5\\left. \\frac{dy(t)}{dt} \\right|_{t = 0^-} = 5<\/li>\n\n\n\n
    • Input:
      x(t)=e\u2212tu(t)x(t) = e^{-t} u(t)
      (unit step input multiplied by exponential decay)<\/li>\n\n\n\n
    • The differential equation is not<\/strong> explicitly provided. But if we assume a standard second-order system of the form:<\/li>\n<\/ul>\n\n\n\n

      d2y(t)dt2+a1dy(t)dt+a0y(t)=x(t)\\frac{d^2y(t)}{dt^2} + a_1 \\frac{dy(t)}{dt} + a_0 y(t) = x(t)<\/p>\n\n\n\n

      Let\u2019s assume an example system: d2y(t)dt2+3dy(t)dt+2y(t)=x(t)\\frac{d^2y(t)}{dt^2} + 3 \\frac{dy(t)}{dt} + 2 y(t) = x(t)<\/p>\n\n\n\n

      \n\n\n\n

      Step-by-Step Solution<\/strong><\/h3>\n\n\n\n

      Step 1: Solve Homogeneous (Natural) Response<\/strong><\/p>\n\n\n\n

      Set input x(t)=0x(t) = 0: d2yn(t)dt2+3dyn(t)dt+2yn(t)=0\\frac{d^2y_n(t)}{dt^2} + 3 \\frac{dy_n(t)}{dt} + 2 y_n(t) = 0<\/p>\n\n\n\n

      Characteristic equation: r2+3r+2=0\u21d2r=\u22121,\u22122r^2 + 3r + 2 = 0 \\Rightarrow r = -1, -2<\/p>\n\n\n\n

      So, natural response: yn(t)=Ae\u2212t+Be\u22122ty_n(t) = A e^{-t} + B e^{-2t}<\/p>\n\n\n\n

      Apply initial conditions:<\/p>\n\n\n\n

        \n
      • y(0)=\u22121=A+By(0) = -1 = A + B<\/li>\n\n\n\n
      • y\u2032(t)=\u2212Ae\u2212t\u22122Be\u22122t\u21d2y\u2032(0)=\u2212A\u22122B=5y'(t) = -A e^{-t} – 2B e^{-2t} \\Rightarrow y'(0) = -A – 2B = 5<\/li>\n<\/ul>\n\n\n\n

        Solve system:<\/p>\n\n\n\n

          \n
        • A+B=\u22121A + B = -1<\/li>\n\n\n\n
        • \u2212A\u22122B=5-A – 2B = 5<\/li>\n<\/ul>\n\n\n\n

          From first: A=\u22121\u2212BA = -1 – B
          Substitute into second: \u2212(\u22121\u2212B)\u22122B=5\u21d21+B\u22122B=5\u21d2\u2212B=4\u21d2B=\u22124,A=3-(-1 – B) – 2B = 5 \\Rightarrow 1 + B – 2B = 5 \\Rightarrow -B = 4 \\Rightarrow B = -4, A = 3<\/p>\n\n\n\n

          So, yn(t)=3e\u2212t\u22124e\u22122ty_n(t) = 3 e^{-t} – 4 e^{-2t}<\/p>\n\n\n\n

          \n\n\n\n

          Step 2: Solve Particular (Forced) Response<\/strong><\/p>\n\n\n\n

          Input: x(t)=e\u2212tu(t)x(t) = e^{-t} u(t)<\/p>\n\n\n\n

          Assume particular solution of form: yp(t)=Ke\u2212ty_p(t) = K e^{-t}<\/p>\n\n\n\n

          Plug into the equation: (Ke\u2212t)\u2032\u2032+3(Ke\u2212t)\u2032+2(Ke\u2212t)=e\u2212t(K e^{-t})” + 3(K e^{-t})’ + 2(K e^{-t}) = e^{-t} Ke\u2212t(1+3+2)=e\u2212t\u21d26Ke\u2212t=e\u2212t\u21d2K=16K e^{-t}(1 + 3 + 2) = e^{-t} \\Rightarrow 6K e^{-t} = e^{-t} \\Rightarrow K = \\frac{1}{6}<\/p>\n\n\n\n

          So, yp(t)=16e\u2212tu(t)y_p(t) = \\frac{1}{6} e^{-t} u(t)<\/p>\n\n\n\n

          \n\n\n\n

          Final Answer:<\/strong><\/p>\n\n\n\n

            \n
          • Natural response<\/strong>:
            yn(t)=3e\u2212t\u22124e\u22122ty_n(t) = 3 e^{-t} – 4 e^{-2t}<\/li>\n\n\n\n
          • Forced response<\/strong>:
            yf(t)=16e\u2212tu(t)y_f(t) = \\frac{1}{6} e^{-t} u(t)<\/li>\n\n\n\n
          • Total response<\/strong>:
            y(t)=yn(t)+yf(t)y(t) = y_n(t) + y_f(t)<\/li>\n<\/ul>\n\n\n\n
            \n\n\n\n

            Explanation<\/strong><\/h3>\n\n\n\n

            In Linear Time-Invariant (LTI) systems, analyzing system behavior involves decomposing the total response into two components: natural response<\/strong> and forced response<\/strong>. The natural response depends entirely on the system\u2019s initial conditions, with no external input<\/strong> applied. This behavior represents how the system’s stored energy dissipates over time due to internal dynamics (like resistance or damping). In contrast, the forced response is the system\u2019s output due to an external input<\/strong>, assuming the system starts from rest (i.e., zero initial conditions). This describes how the system reacts to the input function over time.<\/p>\n\n\n\n

            In the given example, the system is governed by a second-order linear differential equation with exponential input x(t)=e\u2212tu(t)x(t) = e^{-t} u(t), and initial conditions y(0\u2212)=\u22121y(0^-) = -1, y\u2032(0\u2212)=5y'(0^-) = 5. We first found the natural response by setting the input to zero and solving the homogeneous equation. The solution yielded exponential terms whose coefficients were determined by applying the initial conditions.<\/p>\n\n\n\n

            For the forced response, we assumed the system starts from rest and solved the differential equation with the input included. We found a particular solution that matched the input form e\u2212te^{-t}, yielding a steady-state behavior due to the input.<\/p>\n\n\n\n

            The total response is the sum<\/strong> of these components. Understanding this decomposition is crucial in system theory, as it helps engineers design systems that either suppress or enhance certain responses \u2014 for instance, in controlling vibrations or designing electrical filters<\/p>\n\n\n\n

            \"\"<\/figure>\n\n\n\n

            .<\/p>\n","protected":false},"excerpt":{"rendered":"

            Described by the following differential equa- tions with the specified input and initial conditions: (a) y(0?) = -1, $\\frac{d}{dt}y(t)|_{t=0?} = 5, x(t) = e??u(t) (b) Determine the initial value for the forced and natural response individually. Do you know what is the difference between the natural response and forced response?Determine the forced and natural responses […]<\/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-226332","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/226332","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=226332"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/226332\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=226332"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=226332"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=226332"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}