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

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!