{"id":18399,"date":"2016-10-19T17:21:47","date_gmt":"2016-10-19T17:21:47","guid":{"rendered":"http:\/\/www.oracletutoring.ca\/blog\/?p=18399"},"modified":"2016-10-19T17:21:47","modified_gmt":"2016-10-19T17:21:47","slug":"how-to-solve-second-order-homogeneous-linear-differential-equations-w-constant-coefficients-basic-idea","status":"publish","type":"post","link":"https:\/\/www.oracletutoring.ca\/blog\/how-to-solve-second-order-homogeneous-linear-differential-equations-w-constant-coefficients-basic-idea\/","title":{"rendered":"How to solve second-order homogeneous linear differential equations w\/ constant coefficients: basic idea"},"content":{"rendered":"

The tutor shows the quadratic method (pka, characteristic equation) behind 2nd order LDEs w\/ constant coefficients.<\/h1>\n

An exponential function of the form<\/p>\n

y=ekt<\/sup><\/p>\n

has the first derivative<\/p>\n

y’=kekt<\/sup><\/p>\n

and second derivative<\/p>\n

y”=k2<\/sup>ekt<\/sup><\/p>\n

Seeing an equation like, for example,<\/p>\n

y” -3y’ +2y = 0<\/p>\n

one can suppose the solution to be of the form<\/p>\n

y=Cekt<\/sup> (C is some constant)<\/p>\n

so the equation becomes<\/p>\n

k2<\/sup>Cekt<\/sup> -3kCekt<\/sup> +2Cekt<\/sup> = 0<\/p>\n

Now we can factor out Cekt<\/sup>:<\/p>\n

Cekt<\/sup>(k2<\/sup> -3k +2)=0<\/p>\n

and then factor and solve for k:<\/p>\n

Cekt<\/sup>(k-2)(k-1)=0<\/p>\n

k=2 or k=1<\/p>\n

Therefore, a possible solution to the equation<\/p>\n

y” -3y’ +2y = 0<\/p>\n

is<\/p>\n

y=C1<\/sub>e1t<\/sup> + C2<\/sub>e2t<\/sup><\/p>\n

or just<\/p>\n

y=C1<\/sub>et<\/sup> + C2<\/sub>e2t<\/sup><\/p>\n

HTH:)<\/p>\n

Source:<\/p>\n

Boyce, William and Richard DiPrima. Elementary Differential Equations and Boundary Value Problems<\/u>. New York: John Wiley & Sons, Inc., 1986.<\/p>\n

Jack of Oracle Tutoring by Jack and Diane,<\/a> Campbell River, BC.<\/p>\n","protected":false},"excerpt":{"rendered":"

The tutor shows the quadratic method (pka, characteristic equation) behind 2nd order LDEs w\/ constant coefficients. An exponential function of the form y=ekt has the first derivative y’=kekt and second derivative y”=k2ekt Seeing an equation like, for example, y” -3y’ …<\/p>\n

How to solve second-order homogeneous linear differential equations w\/ constant coefficients: basic idea<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1960,3],"tags":[],"class_list":["post-18399","post","type-post","status-publish","format-standard","hentry","category-differential-equations","category-math"],"_links":{"self":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/18399","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/comments?post=18399"}],"version-history":[{"count":17,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/18399\/revisions"}],"predecessor-version":[{"id":18416,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/18399\/revisions\/18416"}],"wp:attachment":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/media?parent=18399"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/categories?post=18399"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/tags?post=18399"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}