Depends on how you define "repetition." Repetition of a concept, not necessarily exact copies of equations where only one or two variables are changed in the structure, helps you learn. Like for instance, find the particular solution of this second order linear nonhomogeneous equation:
y"-3y'-4y = 3e^(2t)
to solve this, you would use these steps:
r^2-3r-4=0
r = 4 or -1
y = c_1e^(4t)+c_2e^(-t)+Y(t)
Y(t) = Ae^(2t)
Y'(t) = 2Ae^(2t)
Y"(t) = 4Ae^(2t)
Subsituting back into the original equation..
(4A-6A-4A)e^(2t) = 3e^(2t)
Hence A = -1/2 thus the solution is:
y = c_1e^(4t)+c_2e^(-t)-1/2e^(2t)
So after solving that and not doing anymore problems like this, can you therefore say that you know how to solve second order nonhomogeneous equations? No you can't, because depending on the situation of the equation, you will need to use different steps. If you try doing exactly what we did in the previous question for this one:
y"-3y'-4y = -8(e^t)cos2t
they look the same, but you can't exactly use the same series of steps to solve it. Instead you use these series of steps to find the general solution:
r^2-3r-4=0
r = 4 or -1
y = c_1e^(4t)+c_2e^(-t)+Y(t)
Y(t) = A(e^t)cos2t+B(e^t)sin2t
Y'(t) = (A+2B)(e^t)cos2t+(-2A+B)(e^t)sin2t
Y"(t) = (-3A+4B)(e^t)cos2t+(-4A-3B)(e^t)sin2t
substitute these into the original equation, we get:
10A+2B = 8, 2A-10B = 0
Hence A = 10/13 and B = 2/13 so therefore the answer is:
y = c_1e^(4t)+c_2e^(-t)+10/13(e^t)cos2t+2/13(e^t)sin2t
Both are same concept, just each using their own way of solving it. But then there can be some others that vary even more like:
y''+4y = 3csct
where the solution and the steps involved are:
y''+4y = 0
y_c(t) = c_1cos2t+c_2sin2t
y = u_1(t)cos2t+u_2(t)sin2t
y' = -2u_1(t)sin2t+2u_2(t)cos2t+u'_1(t)cos2t+u'_2(t)sin2t
we assume u'_1(t)cos2t+u'_2(t)sin2t = 0 so:
y' = -2u_1(t)sin2t+2u_2(t)cos2t
y" = -4u_1(t)cos2t-4u_2(t)sin2t-2u'_1(t)sin2t+2u'_2(t)cos2t
Then substituting back into the original equation, we get:
-2u'_1(t)sin2t+2u'_2(t)cos2t = 3csct
now using that and our assumption: u'_1(t)cos2t+u'_2(t)sin2t = 0, we get:
u'_1 = -3cost
u'_2 = 3/2csct-3sint
u_1 = -3sint+c_1
u_2 = 3/2ln|csct-cott|+3cost+c_2
now substituting those u's, we get:
y = -3sintcos2t+3/2ln|csct-cott|sin2t+3costsin2t+c_1cos2t+c_2sin2t
and using double-angle formulas, we get:
y = 3sint+3/2ln|csct-cott|sin2t+c_1cos2t+c_2sin2t
Again all are of which are the same concept. If you do alot more of these problems, you learn new methods of knowing how to solve them and more importantly, practical thinking and analytical skills when trying to mix, match, and apply these different methods of solving these equations. This is also a way of constant building; by learning more methods of how to solve these problems and being exposed to more and more different ways of applying methods into solving such equations helps one learn and understand the concept inside out. Repetition in this sense works.
Post has been edited 1 time(s), last time on Oct 8 2007, 4:38 am by MillenniumArmy.
None.