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The section develops two key solution principles: linear combinations of solutions remain solutions, and the general homogeneous solution is built from two linearly independent solutions. It further explains how, when P, Q, and R are constant, solutions can be found using an exponential trial y=e^x, leading to an auxiliary (characteristic) equation and case analysis based on the discriminant.",{"@graph":69,"@context":122},[70,84,105],{"@type":71,"itemListElement":72},"BreadcrumbList",[73,77,79,82],{"item":74,"name":75,"@type":76,"position":8},"https://docshare.wps.com","Home","ListItem",{"item":78,"name":9,"@type":76,"position":14},"https://docshare.wps.com/document/",{"item":80,"name":20,"@type":76,"position":81},"https://docshare.wps.com/document/exam/",3,{"item":83,"name":65,"@type":76,"position":19},"https://docshare.wps.com/document/second-order-linear-equations-section-on-homogeneous-linear-differential-equations/441787/",{"url":83,"name":65,"@type":85,"image":86,"author":91,"headline":65,"publisher":94,"fileFormat":97,"inLanguage":63,"description":67,"dateModified":98,"datePublished":99,"encodingFormat":97,"isAccessibleForFree":100,"interactionStatistic":101},"DigitalDocument",{"url":87,"@type":88,"width":89,"height":90},"https://docshare.wps.com/thumbnails/second-order-linear-equations-section-on-homogeneous-linear-differential-equations/441787.png","ImageObject",300,407,{"name":92,"@type":93},"dawugda","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-10-20","2026-09-29",true,{"@type":102,"interactionType":103,"userInteractionCount":81},"InteractionCounter",{"@type":104},"ViewAction",{"@type":106,"mainEntity":107},"FAQPage",[108,114,118],{"name":109,"@type":110,"acceptedAnswer":111},"What is the homogeneous form of a second-order linear differential equation?","Question",{"text":112,"@type":113},"A second-order linear equation is homogeneous when the right-hand side G(x)=0 for all x. Its solution approach is then based on the homogeneous general solution framework.","Answer",{"name":115,"@type":110,"acceptedAnswer":116},"Why can the linear combination of two solutions still be a solution?",{"text":117,"@type":113},"If y1(x) and y2(x) solve the same linear homogeneous equation, then any linear combination y=c1y1+c2y2 also satisfies the equation due to the linearity of differentiation and substitution.",{"name":119,"@type":110,"acceptedAnswer":120},"How do you find solutions when the coefficients are constant?",{"text":121,"@type":113},"For constant coefficients, try y=e^x. Substituting into ay''+by'+cy=0 yields the auxiliary (characteristic) equation ar^2+br+c=0, and the roots determine the solution forms.","https://schema.org",{"og:url":83,"og:type":124,"og:title":65,"og:site_name":95,"og:description":67},"article",{"robots":126,"canonical":83},"index,follow",{"doc_id":128,"site_id":62},441787,1790826457,{"code":4,"msg":5,"data":131},{"doc_id":128,"user_id":132,"nickname":92,"user_avatar":133,"doc_module":4,"category_id":19,"category_name":20,"doc_title":65,"doc_description":67,"doc_content":134,"file_id":135,"file_url":136,"file_type":137,"file_size":138,"view_count":81,"is_deleted":4,"is_public":8,"is_downloadable":8,"audit_status":8,"page_count":139,"language":140,"language_code":63,"site_id":62,"html_lang":63,"table_of_contents":141,"faqs":142,"seo_title":143,"seo_description":67,"update_tm":144,"read_time":145},3985747870947,"https://ap-avatar.wpscdn.com/davatar_3d24733baf745e90a7e4bdd5f77d97b2","MORTH CAROLIMH STHTEINJVERS1TY  \nSECINI IIONALCULUSCOMCEPTS AND CONTEXTSJAMES STEWART  \nDigitized by the Internet Archivein 2022 with funding fromKahle/Austin Foundation  \nhttps://archive.org/details/isbn_9780534481025  \nCalculusSupplement  \nJames StewartMcMaster University  \n# BROOKS/COLE\n\n—TM来  \nTHOMSON LEARNING  \nPrinted in the United States of America  \nBrooks.Cole  \n511 Forest Lodge RoadPacific Grove,CA93950USA  \nFor information about our products,contact us:Thomson Learning Academic Resource Center1-800-423-0563http://www.brookscole.com  \nInternational HeadquartersThomson LearningInternational Division290 Harbor Drive,2dFloorStamford,CT 06902-7477USA  \n0-534-48102-7  \nCOPYRIGHT ◎2002 by the Wadsworth Group.Brooks/Cole is an imprint of the WadsworthGroup,a division of Thomson Learning Inc.Thomson LearningTM is a trademark used herein underlicense.  \nUK/Europe/Middle East/South AfricaThomson LearningBerkshire House168-173 High HolbornLondon WCIV7AA  \nAsia  \nThomson Learning60 Albert Street,\\#15-01Albert ComplexSingapore 189969  \nCanada  \nNelson Thomson Learning1120 Birchmount RoadToronto,Ontario MIK 5G4CanadaUnited Kingdom  \nALL RIGHTS RESERVED.No part of this work covered by the copyright hereon may bereproduced or used in any form or by any means—graphic,electronic,or mechanical,including photocopying,recording,taping,Web distribution,or information storage andretrieval systems—without the written permission of the publisher  \nThe Adaptable Courseware Program consists of products and additions to existing Brooks/Coleproducts that are produced from camera-ready copy.Peer review,class testing,and accuracy areprimarily the responsibility of the author(s).  \n# Second-Order Linear Equations\n\nA second-order linear differential equation has the form  \n1  \nwhere P,Q,R,and G are continuous functions.We saw in Section 10.1 that equations ofthis type arise in the study of the motion of a spring.In Section 18.3 we willfurther pur-sue this application as well as the application to electric circuits.  \nIn this section we study the case where G(x)=0,for allx,in Equation 1.Such equa-tions are called homogeneous linear equations.Thus,the form of a second-order linearhomogeneous differential equation is  \n2  \nIf G(x)≠0 for some x,Equation l is nonhomogeneous and is discussed in Section 18.2.  \nTwo basic facts enable us to solve homogeneous linear equations.The first of these saysthat if we know two solutions y and y₂of such an equation,then the linear combinationy=ciyi+C₂y2 is also a solution.  \n3 Theorem If y₁(x)and y₂(x)are both solutions of the linear homogeneous equa-tion (2)and c₁and c are any constants,then the function  \ny(x)=c₁y₁(x)+c₂y₂(x)  \nis also a solution of Equation 2.  \nProof Since y and y2 are solutions of Equation 2,we have  \nP(x)y\"+Q(x)yí+R(x)y=0andP(x)y2”+Q(x)yź+R(x)y2=0  \nTherefore,using the basic rules for differentiation,we have  \nP(x)y\"+Q(x)y'+R(x)y  \n=P(x)(c₁y₁+C₂y2)\"+Q(x)(c₁yi+c₂y2)'+R(x)(c₁yi+c₂y₂)=P(x)(c₁y\"+c₂y2)+Q(x)(c₁yí+c₂y2)+R(x)(cy1+C₂y₂)=c₁[P(x)y\"+Q(x)yí+R(x)y]+c₂[P(x)y2+Q(x)y+R(x)y₂2]=c₁(0)+c₂(0)=0  \nThus,y=CIyi+c₂y2 is a solution of Equation 2.  \nThe other fact we need is given by the following theorem,which is proved in moreadvanced courses.It says that the general solution is a linear combination of two linearlyindependent solutions y and y₂.This means that neither yi nor y₂is a constant multipleof the other.For instance,the functions f(x)=x²and g(x)=5x²are linearly dependent,but f(x)=e×and g(x)=xe*are linearly independent.  \n4 Theorem If y₁and y₂are linearly independent solutions of Equation 2,then thegeneral solution is given by  \ny(x)=ciy₁(x)+c₂y₂(x)  \nwhere c and cz are arbitrary constants.  \nTheorem 4 is very useful because it says that if we know two particular linearly inde-pendent solutions,then we know every solution.  \nIn general,it is not easy to discover particular solutions to a second-order linear equa-tion.But it is always possible to do so if the coefficient functions P,Q,and R are con","cbCaiklgGz14SFRT","https://ap.wps.com/l/cbCaiklgGz14SFRT","pdf",2831924,44,"English","# Second-Order Linear Equations\n## Homogeneous vs. nonhomogeneous form\n## Linear combination theorem\n## General solution from independent solutions\n## Constant-coefficient case and trial exponential\n## Auxiliary equation and discriminant cases","[{\"question\":\"What is the homogeneous form of a second-order linear differential equation?\",\"answer\":\"A second-order linear equation is homogeneous when the right-hand side G(x)=0 for all x. Its solution approach is then based on the homogeneous general solution framework.\"},{\"question\":\"Why can the linear combination of two solutions still be a solution?\",\"answer\":\"If y1(x) and y2(x) solve the same linear homogeneous equation, then any linear combination y=c1y1+c2y2 also satisfies the equation due to the linearity of differentiation and substitution.\"},{\"question\":\"How do you find solutions when the coefficients are constant?\",\"answer\":\"For constant coefficients, try y=e^x. Substituting into ay''+by'+cy=0 yields the auxiliary (characteristic) equation ar^2+br+c=0, and the roots determine the solution forms.\"}]","Second-Order Linear Equations - Section on Homogeneous Linear Differential Equations | PDF",1790697499,111]