Zkirsche Posted December 6, 2011 Share Posted December 6, 2011 Ok, so I'm learning about ordinary differential equations for my interview at cambridge this week. (Which is why I'm inactive for mafia) However, I was just stumped/confused by a question. The question was, solve: (d^2y/dx^2)+4(dy/dx)+4y = 2cos^2(x) So I attempted to treat this as a normal linear inhomogeneous constant coefficient equation and began to use the method of variation of constants (or variation of parameters). However, when I reduced this equation (by setting 2cos^2(x) to 0), and tried to find the two variables y0 and y1 which were solutions to the reduced equation. However, the auxillary equation gives only one root. Meaning y0 = y1 = (A+Bx)e^(2x). Where A and B are arbitrary constants. For the complementary equation, which usually comes in the form of y = A0y0 + A1y1, do I simply multiple out the brackets? After doing so, do I simply replace A with V1(x) and B with V2(x) where V1(x) and V2(x) are functions of x and then work ahead normally, or is there a different route I need to take? Any help would be much appreciated. :) Link to comment Share on other sites More sharing options...
Aleph Posted December 6, 2011 Share Posted December 6, 2011 I don't remember this crap very well, but isn't it supposed to be scaled or something? Like "2 and 2t" or whatever? Also why 2? Wouldn't it be -2? lul I totally forgot DE. Also your syntax or whatever is way different from how we did it and not only do I not remember how we did inhomogeneous or that crap with matrices toward the end that made ungodly expressions look like common fucking sense, I almost don't even remember if that was a hazy dream or something that I actually went to school for... Link to comment Share on other sites More sharing options...
Zkirsche Posted December 7, 2011 Author Share Posted December 7, 2011 It would be -2 yes. I even wrote -2 so I don't know how that happened. Link to comment Share on other sites More sharing options...
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