{"id":36010,"date":"2025-06-25T01:21:02","date_gmt":"2025-06-25T01:21:02","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=36010"},"modified":"2025-06-25T01:21:07","modified_gmt":"2025-06-25T01:21:07","slug":"what-is-the-order-of-the-differential-equation-d3x-dt3-t3-cos-2-dx-dt-convert-it-into-system-of-first-order-equations","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/what-is-the-order-of-the-differential-equation-d3x-dt3-t3-cos-2-dx-dt-convert-it-into-system-of-first-order-equations\/","title":{"rendered":"What is the order of the differential equation d3x dt3 t3 cos 2 dx dt Convert it into system of first order equations."},"content":{"rendered":"\n<p>What is the order of the differential equation d3x dt3 t3 cos 2 dx dt Convert it into system of first order equations. Do not attempt to find a solution [2 marks]<\/p>\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<h3 class=\"wp-block-heading\"><strong>Correct Answer:<\/strong><\/h3>\n\n\n\n<p>The given differential equation is:d3xdt3+t3cos\u2061(2)\u22c5dxdt=0\\frac{d^3x}{dt^3} + t^3 \\cos(2) \\cdot \\frac{dx}{dt} = 0dt3d3x\u200b+t3cos(2)\u22c5dtdx\u200b=0<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Order of the differential equation:<\/strong><br>The order of a differential equation is the highest derivative present. Here, the highest derivative is d3xdt3\\frac{d^3x}{dt^3}dt3d3x\u200b, so:<\/p>\n\n\n\n<p><strong>Order = 3<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>To convert it into a system of first-order equations:<\/strong><\/p>\n\n\n\n<p>Let:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>x1=xx_1 = xx1\u200b=x<\/li>\n\n\n\n<li>x2=dxdt=dx1dtx_2 = \\frac{dx}{dt} = \\frac{dx_1}{dt}x2\u200b=dtdx\u200b=dtdx1\u200b\u200b<\/li>\n\n\n\n<li>x3=d2xdt2=dx2dtx_3 = \\frac{d^2x}{dt^2} = \\frac{dx_2}{dt}x3\u200b=dt2d2x\u200b=dtdx2\u200b\u200b<\/li>\n<\/ul>\n\n\n\n<p>From the original equation:d3xdt3+t3cos\u2061(2)\u22c5dxdt=0\\frac{d^3x}{dt^3} + t^3 \\cos(2) \\cdot \\frac{dx}{dt} = 0dt3d3x\u200b+t3cos(2)\u22c5dtdx\u200b=0<\/p>\n\n\n\n<p>We substitute:d3xdt3=dx3dt=\u2212t3cos\u2061(2)\u22c5x2\\frac{d^3x}{dt^3} = \\frac{dx_3}{dt} = -t^3 \\cos(2) \\cdot x_2dt3d3x\u200b=dtdx3\u200b\u200b=\u2212t3cos(2)\u22c5x2\u200b<\/p>\n\n\n\n<p>So the system becomes:{dx1dt=x2dx2dt=x3dx3dt=\u2212t3cos\u2061(2)\u22c5x2\\begin{cases} \\frac{dx_1}{dt} = x_2 \\\\ \\frac{dx_2}{dt} = x_3 \\\\ \\frac{dx_3}{dt} = -t^3 \\cos(2) \\cdot x_2 \\end{cases}\u23a9\u23a8\u23a7\u200bdtdx1\u200b\u200b=x2\u200bdtdx2\u200b\u200b=x3\u200bdtdx3\u200b\u200b=\u2212t3cos(2)\u22c5x2\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 is a third-order ordinary differential equation (ODE) because the highest derivative involved is d3xdt3\\frac{d^3x}{dt^3}dt3d3x\u200b. Identifying the order of an ODE is important as it tells us how many initial conditions are required to determine a unique solution. In this case, three initial conditions would be necessary: the values of xxx, dxdt\\frac{dx}{dt}dtdx\u200b, and d2xdt2\\frac{d^2x}{dt^2}dt2d2x\u200b at some initial time.<\/p>\n\n\n\n<p>Transforming a higher-order ODE into a system of first-order equations is a standard technique used to make the equation more suitable for numerical or theoretical analysis. By defining new variables for each derivative up to the second order, we reduce the third-order equation to three coupled first-order equations. This approach is widely used in engineering, physics, and applied mathematics, especially for simulation and modeling using software that typically solves only first-order systems.<\/p>\n\n\n\n<p>In this conversion, we introduced three variables: x1x_1x1\u200b, x2x_2x2\u200b, and x3x_3x3\u200b, representing the function and its first two derivatives. This allowed us to express the original third derivative as a first-order derivative of x3x_3x3\u200b, and relate it to x2x_2x2\u200b and ttt using the original equation. The resulting system captures the full dynamics of the original third-order equation.<\/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-banner8-729.jpeg\" alt=\"\" class=\"wp-image-36011\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-729.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-729-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-729-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>What is the order of the differential equation d3x dt3 t3 cos 2 dx dt Convert it into system of first order equations. Do not attempt to find a solution [2 marks] The Correct Answer and Explanation is: Correct Answer: The given differential equation is:d3xdt3+t3cos\u2061(2)\u22c5dxdt=0\\frac{d^3x}{dt^3} + t^3 \\cos(2) \\cdot \\frac{dx}{dt} = 0dt3d3x\u200b+t3cos(2)\u22c5dtdx\u200b=0 Order of the [&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-36010","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36010","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=36010"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36010\/revisions"}],"predecessor-version":[{"id":36012,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36010\/revisions\/36012"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=36010"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=36010"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=36010"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}