Link to test: http://www.math.ucla.edu/~cmarshak/GRE2.pdf
#18  I figure this ultimately comes down to knowing power series expansions for some common mathematical expressions, but I am weak in this area. How would you approach this? What should I brush up on? Please explain it like I'm 5.
#19  "Which of the following is a general solution of the differential equation:
y'''  3y'' + 3y'  y = 0
And then each option is a bunch of e^t terms.
This is a type of problem where I'd probably waste a lot of time trying to find the derivatives and plugging things together. It's easy to make a mistake going that route. Are there any facts or approaches to this problem which make it less mechanical or more obvious what *isnt* the answer?
Thanks!
GR9367 #18, 19

 Posts: 39
 Joined: Mon May 30, 2011 10:18 pm
Re: GR9367 #18, 19
For #18, notice that the expression inside the sum can be written as (x^2)^n so the infinite sum is a geometric series with rate r=x^2. Then using the formula for the infinite series here http://en.wikipedia.org/wiki/Geometric_series#Formula should get you the solution.
For #19, I just solved the r^3  3r^2 + 3r  1 = 0 and noticed r=1 was a solution. After using that, I figured out that r=1 was a solution with multiplicity three. So the solution is of the form c1*e^t + c2*t*e^t + c3*t^2*e^t. Not sure of any other way to do it.
For #19, I just solved the r^3  3r^2 + 3r  1 = 0 and noticed r=1 was a solution. After using that, I figured out that r=1 was a solution with multiplicity three. So the solution is of the form c1*e^t + c2*t*e^t + c3*t^2*e^t. Not sure of any other way to do it.

 Posts: 157
 Joined: Sun Oct 14, 2012 12:15 pm
Re: GR9367 #18, 19
for #19, use the coefficients of t in the power of e in every solution and check which one worksTheYoungin wrote: For #19, I just solved the r^3  3r^2 + 3r  1 = 0 and noticed r=1 was a solution. After using that, I figured out that r=1 was a solution with multiplicity three. So the solution is of the form c1*e^t + c2*t*e^t + c3*t^2*e^t. Not sure of any other way to do it.
Re: GR9367 #18, 19
18) Its nothing fancy when it comes to series its a geometric series, as @TheYoungin points out.
So its probably easier to sum and then take the derivative. Sometimes its easier to do the termbyterm derivative and then to sum the series, but not in this case.
f(x) = 1/(1+x^2)
f'(x) = 2x/(1+x^2)^2
19) Let D be the derivative, considered as an operator, and then factor the equation they way you would a polynomial
D^3 y  3D^2 y +3 Dy  y =0
or (D^3  3D^2 + 3D  1)y = 0
or (D1)^3 y = 0
So one solution is given by solving (D1)y = 0, which is yields y=e^t (in general, a factor of (Dr) gives a solution of e^rt )
So given a solution of e^t and a repeated factor, the next is given by solving (D1)y = e^t, which happens to a have a solution te^t.
Given a solution of the form te^t and a factor which appears three times, the next
is given by solving (D1)y = te^t, which happens to have a solution 1/2 t^2 e^t
In general, if you have k+1 repeated factors like (Dr) in the equation, then you can iteratively solve starting with e^rt to get a general solution like 1/k! t^k e^rt (though the leading constant factor doesn't really matter). You can see this taking the solution e^rt and using the technique of variation of parameters to try a solution of the form y=u(t) e^rt to get an equation like u'(t) = t^(k1)/(k1)!, which you can directly integrate.
So its probably easier to sum and then take the derivative. Sometimes its easier to do the termbyterm derivative and then to sum the series, but not in this case.
f(x) = 1/(1+x^2)
f'(x) = 2x/(1+x^2)^2
19) Let D be the derivative, considered as an operator, and then factor the equation they way you would a polynomial
D^3 y  3D^2 y +3 Dy  y =0
or (D^3  3D^2 + 3D  1)y = 0
or (D1)^3 y = 0
So one solution is given by solving (D1)y = 0, which is yields y=e^t (in general, a factor of (Dr) gives a solution of e^rt )
So given a solution of e^t and a repeated factor, the next is given by solving (D1)y = e^t, which happens to a have a solution te^t.
Given a solution of the form te^t and a factor which appears three times, the next
is given by solving (D1)y = te^t, which happens to have a solution 1/2 t^2 e^t
In general, if you have k+1 repeated factors like (Dr) in the equation, then you can iteratively solve starting with e^rt to get a general solution like 1/k! t^k e^rt (though the leading constant factor doesn't really matter). You can see this taking the solution e^rt and using the technique of variation of parameters to try a solution of the form y=u(t) e^rt to get an equation like u'(t) = t^(k1)/(k1)!, which you can directly integrate.