{"id":36005,"date":"2025-06-25T01:18:32","date_gmt":"2025-06-25T01:18:32","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=36005"},"modified":"2025-06-25T01:18:34","modified_gmt":"2025-06-25T01:18:34","slug":"fracd3xdt3-3fracd2xdt2-2fracdxdt-6x-0-4","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/fracd3xdt3-3fracd2xdt2-2fracdxdt-6x-0-4\/","title":{"rendered":"\\frac{d^3x}{dt^3} – 3\\frac{d^2x}{dt^2} – 2\\frac{dx}{dt} + 6x = 0 4."},"content":{"rendered":"\n

. \\frac{d^3x}{dt^3} – 3\\frac{d^2x}{dt^2} – 2\\frac{dx}{dt} + 6x = 0 4. \\frac{d^3x}{dt^3} – 2\\frac{d^2x}{dt^2} – 10\\frac{dx}{dt} + 8x = 0<\/p>\n\n\n\n

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

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

Let’s solve the two third-order linear homogeneous differential equations using the characteristic equation method<\/strong> and then provide a clear explanation.<\/p>\n\n\n\n

\n\n\n\n

Problem 3<\/strong><\/h3>\n\n\n\n

d3xdt3\u22123d2xdt2\u22122dxdt+6x=0\\frac{d^3x}{dt^3} – 3\\frac{d^2x}{dt^2} – 2\\frac{dx}{dt} + 6x = 0dt3d3x\u200b\u22123dt2d2x\u200b\u22122dtdx\u200b+6x=0<\/p>\n\n\n\n

Step 1:<\/strong> Form the characteristic equation:r3\u22123r2\u22122r+6=0r^3 – 3r^2 – 2r + 6 = 0r3\u22123r2\u22122r+6=0<\/p>\n\n\n\n

Try rational roots. r=1r = 1r=1 and r=\u22121r = -1r=\u22121 do not work. Try r=3r = 3r=3:33\u22123(3)2\u22122(3)+6=27\u221227\u22126+6=03^3 – 3(3)^2 – 2(3) + 6 = 27 – 27 – 6 + 6 = 033\u22123(3)2\u22122(3)+6=27\u221227\u22126+6=0<\/p>\n\n\n\n

So r=3r = 3r=3 is a root. Divide the polynomial by (r\u22123)(r – 3)(r\u22123):r3\u22123r2\u22122r+6=(r\u22123)(r2\u22122)r^3 – 3r^2 – 2r + 6 = (r – 3)(r^2 – 2)r3\u22123r2\u22122r+6=(r\u22123)(r2\u22122)<\/p>\n\n\n\n

Solve r2\u22122=0\u21d2r=\u00b12r^2 – 2 = 0 \\Rightarrow r = \\pm \\sqrt{2}r2\u22122=0\u21d2r=\u00b12\u200b<\/p>\n\n\n\n

Solution:<\/strong>x(t)=C1e3t+C2e2t+C3e\u22122tx(t) = C_1 e^{3t} + C_2 e^{\\sqrt{2}t} + C_3 e^{-\\sqrt{2}t}x(t)=C1\u200be3t+C2\u200be2\u200bt+C3\u200be\u22122\u200bt<\/p>\n\n\n\n

\n\n\n\n

Problem 4<\/strong><\/h3>\n\n\n\n

d3xdt3\u22122d2xdt2\u221210dxdt+8x=0\\frac{d^3x}{dt^3} – 2\\frac{d^2x}{dt^2} – 10\\frac{dx}{dt} + 8x = 0dt3d3x\u200b\u22122dt2d2x\u200b\u221210dtdx\u200b+8x=0<\/p>\n\n\n\n

Step 1:<\/strong> Characteristic equation:r3\u22122r2\u221210r+8=0r^3 – 2r^2 – 10r + 8 = 0r3\u22122r2\u221210r+8=0<\/p>\n\n\n\n

Try r=4r = 4r=4:64\u221232\u221240+8=0\u21d2r=4 is a root64 – 32 – 40 + 8 = 0 \\Rightarrow r = 4 \\text{ is a root}64\u221232\u221240+8=0\u21d2r=4 is a root<\/p>\n\n\n\n

Divide:r3\u22122r2\u221210r+8=(r\u22124)(r2+2r\u22122)r^3 – 2r^2 – 10r + 8 = (r – 4)(r^2 + 2r – 2)r3\u22122r2\u221210r+8=(r\u22124)(r2+2r\u22122)<\/p>\n\n\n\n

Solve quadratic:r=\u22122\u00b1(2)2+82=\u22122\u00b1122=\u22121\u00b13r = \\frac{-2 \\pm \\sqrt{(2)^2 + 8}}{2} = \\frac{-2 \\pm \\sqrt{12}}{2} = -1 \\pm \\sqrt{3}r=2\u22122\u00b1(2)2+8\u200b\u200b=2\u22122\u00b112\u200b\u200b=\u22121\u00b13\u200b<\/p>\n\n\n\n

Solution:<\/strong>x(t)=C1e4t+C2e(\u22121+3)t+C3e(\u22121\u22123)tx(t) = C_1 e^{4t} + C_2 e^{(-1 + \\sqrt{3})t} + C_3 e^{(-1 – \\sqrt{3})t}x(t)=C1\u200be4t+C2\u200be(\u22121+3\u200b)t+C3\u200be(\u22121\u22123\u200b)t<\/p>\n\n\n\n

\n\n\n\n

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

To solve higher-order linear differential equations with constant coefficients, we use the characteristic equation method<\/strong>. This technique transforms the differential equation into a polynomial by replacing derivatives with powers of a variable rrr. For example, d3xdt3\\frac{d^3x}{dt^3}dt3d3x\u200b becomes r3r^3r3, d2xdt2\\frac{d^2x}{dt^2}dt2d2x\u200b becomes r2r^2r2, and so on. Solving the resulting polynomial yields the roots, which guide the form of the solution.<\/p>\n\n\n\n

For Problem 3, we obtained the characteristic equation r3\u22123r2\u22122r+6=0r^3 – 3r^2 – 2r + 6 = 0r3\u22123r2\u22122r+6=0. Using factorization, we found the roots r=3,2,\u22122r = 3, \\sqrt{2}, -\\sqrt{2}r=3,2\u200b,\u22122\u200b. These real roots produce exponential solutions, leading to the general solution x(t)=C1e3t+C2e2t+C3e\u22122tx(t) = C_1 e^{3t} + C_2 e^{\\sqrt{2}t} + C_3 e^{-\\sqrt{2}t}x(t)=C1\u200be3t+C2\u200be2\u200bt+C3\u200be\u22122\u200bt.<\/p>\n\n\n\n

In Problem 4, the characteristic equation is r3\u22122r2\u221210r+8=0r^3 – 2r^2 – 10r + 8 = 0r3\u22122r2\u221210r+8=0. We found r=4r = 4r=4 and factored the remaining quadratic to get complex roots r=\u22121\u00b13r = -1 \\pm \\sqrt{3}r=\u22121\u00b13\u200b. These also lead to exponential solutions, resulting in the general solution x(t)=C1e4t+C2e(\u22121+3)t+C3e(\u22121\u22123)tx(t) = C_1 e^{4t} + C_2 e^{(-1+\\sqrt{3})t} + C_3 e^{(-1-\\sqrt{3})t}x(t)=C1\u200be4t+C2\u200be(\u22121+3\u200b)t+C3\u200be(\u22121\u22123\u200b)t.<\/p>\n\n\n\n

These solutions describe how the system evolves over time. The nature of the roots\u2014real or complex\u2014determines whether the behavior is exponential growth\/decay or oscillatory.<\/p>\n\n\n\n

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

. \\frac{d^3x}{dt^3} – 3\\frac{d^2x}{dt^2} – 2\\frac{dx}{dt} + 6x = 0 4. \\frac{d^3x}{dt^3} – 2\\frac{d^2x}{dt^2} – 10\\frac{dx}{dt} + 8x = 0 The Correct Answer and Explanation is: Let’s solve the two third-order linear homogeneous differential equations using the characteristic equation method and then provide a clear explanation. Problem 3 d3xdt3\u22123d2xdt2\u22122dxdt+6x=0\\frac{d^3x}{dt^3} – 3\\frac{d^2x}{dt^2} – 2\\frac{dx}{dt} + 6x […]<\/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-36005","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36005","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=36005"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36005\/revisions"}],"predecessor-version":[{"id":36008,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36005\/revisions\/36008"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=36005"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=36005"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=36005"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}