Ok, not really... It's actually no harder than my Calc II class was, but only because it relies so heavily on Calc II and I suck so bad at methods of integration... Sigh... So obviously, when test time rolled around, I scored poorly on it.
But you try and solve 4(1/2)*Y'(t)+Y(t) = t(1/2) + t... That is no easy task. The only way to solve it apparently is by Solving for it's homogeneous solution and it's particular solution (We where given the inital states as well). I still don't get how to solve for the particular solution. The homogeneous one is really easy: Y'(t) + 4(-1/2)*Y(t) = 0; Y'(t) = 4(-1/2)*Y(t); Y'(t)/Y(t) = 4(-1/2); Ln |Y(t)| = 8(1/2); Y = e(8(1/2)). But without the particular solution, the formula Y = Yh+Yp can't be used. And originally, I thought the particular solution was the thing on the right hand side... But the book has this huge convoluted formula and a whole bunch of stuff I couldn't understand...
*Minutes Pass... Much reasearch occurs...*
Ok... So now I finally get it... I think the test had on it something like Y(0) = 1 & Y'(0) = 2... I should have seen it then. If they give us the derivitive for a value, then it should be obvious that I can just substitute everything together and get 4(1/2)*s+1 = 0, and then we solve for s to get our particular solution, 4(-1/2). Add the two together and get Y = e(8(1/2)) + 4(-1/2)... Or so I theorize.
However... It took me the whole time of the test and the infinite wealth of knowlege that is the Internet to derive that answer... What does this tell me?
*sigh* I gotta study more...
Study?! When do I sleep!!!
Mood: Tired and frustrated