# Diff EQ reduction of Order problem need help!

1. Aug 18, 2008

### koab1mjr

1. The problem statement, all variables and given/known data
One solution y1(t) of the differential equation is given
a) Use the method of reductions of order to obtain a second solution y2(t)

b) Compute the wronskian formed by y1(t) and y2(t)

2. Relevant equations

(t + 1)^2y'' - 4(t+1)y' + 6y = 0

y1(t) = (t+1)^2

3. The attempt at a solution

i assume that the second solutions is y1(t)*u(t) where u(t) is some function of t that statisfies the diff eq

y2(t) = y1(t)*u(t) = (t+1)^2*u(t)

y2(t)' = 2(t+1)*u(t) + (t+1)^2*u(t)'

y2(t)'' = 2u(t) + 2(t+1)*u(t)' + 2(t+1)*u(t)' + (t+1)^2*u(t)''
or
y2(t)'' = 2u(t) + 4(t+1)*u(t)' + (t+1)^2*u(t)''

I plug these into the eqaution and reduce and I wind up with

(t+1)^4*u(t)'' = 0

I do a reduction of order through v(t) = u(t)' and v(t)' = u(t)''

and the equations becomes (t+1)^4*v(t)' = 0

Here is where I may be going wrong, I divide the (t+1)^4 and just have
v(t)' = 0 which then becomes u(t) = C1t + C2

so y2(t) = y1(t)* C1t + C2]
or
y2(t) = C1*y1(t)*t + C2y1(t)

C1 can be folded into complimentary solution constant, so I assign 1 to that. C2 can be zero since its a multiple of the first solution.

My problem is the answer is wrong. The book has (t+1)^3, I checked both y1 and the books answer and they both work. So I am lost as to what I did wrong. I showed it to a couple of post docs and they got the answer I got. Any help is much appreciated.

2. Aug 18, 2008

### Dick

Everybody is correct. The book writes the general answer as D1*(t+1)^2+D2*(t+1)^3. That's the same as (t+1)^2*(D1+D2+D2*t). If you pick C1=D1+D2 and C2=D2, that's exactly your answer. They are both the same set of functions. There are many different choices for the 'other linearly independent function' y2. E.g. they also could have written y2=t*(t+1)^2.