Treadstone Posted October 22, 2004 Share Posted October 22, 2004 due tomorrow, oh how i hate procrastination.. need to solve y''=2y+2tan^3x anywho, i know about the parameter thing, but how do i solve y''=2y....just use the charactoristic equation r^2=0? seemed to easy... edit**figured it out Link to comment Share on other sites More sharing options...
Asimov Pupil Posted April 30, 2005 Share Posted April 30, 2005 I wanna try please correct me if i'm wrong. [math]y=2y+2(tan(x))^3[/math] [math]\frac{dy}{dx}=2y\frac{dy}{dx}-6((tan(x))^2)sec(x)[/math] [math]6((tan(x))^2)sec(x)=2y\frac{dy}{dx}-\frac{dy}{dx}[/math] [math]\frac{6((tan(x))^2)sec(x)}{2y-1}=\frac{dy}{dx}[/math] am i right so far? [math](\frac{dy}{dx})'=\frac{-((2y-1)((cos(x))^2)(12(tan(x))((sec(x))^2)-6((tan(x))^2)(2y-1)sin2x+((2y-1)((cos(x))^2)(12(tan(x))(sec(x))^2}{((2y-1)((cos(x))^2)^2}[/math] uh i think the last part is to long! Link to comment Share on other sites More sharing options...
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