# Reduction of Order to find 2nd solution of DE

1. Feb 25, 2013

### zabumafu

1. The problem statement, all variables and given/known data
t^2y''-4ty'+6y=0 t>0, y1(t)=t^2 find the 2nd solution using method of reduction of order.

2. Relevant equations

3. The attempt at a solution

Let y=v(t)t^2
y'=2tv+v't^2
y''=2v+2tv'+v''t^2+2tv'

put in back into eq and solving reduces to (checked it twice)
t^4*v''=0
v''=0
Then I integrated with respect to T
v'=c (c is some constant)
v(t)=ct where c is a constant.

I substituted v(t) back into y=v(t)*t^2 and got y2=ct^3 where c is some constant so y2=t^3? I know thats the answer but did I correctly do all these steps? Also do I need to show the wronskian in order to prove that the solution is acceptable as y1 and y2 need to form a fundamental set of solutions? Thanks I just want to make sure I am doing this correctly.

--------Problem 2: same question different equation

(x-1)y''-xy'+y=0 x>1 y1=e^x

let y=v(t)e^x
y'=v'e^x +ve^x
y''=v''e^x +2v'e^x +ve^x

into the eq:
e^x[(x-1)(v''+2v'+v)-x(v'+v)+v]=0
(x-1)v''+(x-2)v'=0 then I get confused

Last edited: Feb 25, 2013
2. Feb 25, 2013

### HallsofIvy

Staff Emeritus
You have done problem 1 correctly.

For problem 2, after you have
e^x[(x-1)(v''+2v'+v)-x(v'+v)+v]=0
(x-1)v''+(x-2)v'=0
let u= v' so your differential equation is (x- 1)u'+ (x- 2)u= 0, a differential equation of order 1 (this was, after all, a "reduction of order" method).

That is, in fact, a separable equation- it can be written as (x- 1)u'= (2- x)u and then $$\frac{du}{u}= \frac{2- x}{x- 1}dx$$. Integrate both sides.