show that the solutions of the differential equation u'''- 3u''+4u = 0
form a vector space . find a basis for it.

madukar-ga,

First of all, let me give you the definition of a vector space from
mathworld.com (http://mathworld.wolfram.com/VectorSpace.html)

In terms of our question we can simplify the requirements as follows
(because functions in general already satisfy most of the rules) :

1) 0 must be a solution to the differential equation (Additive
Identity)

2) If u=f(x) and u=g(x) are solutions to the differential equation
then a*f(x)+b*g(x) must also be a solution for all constants a and b.

So first, is u=0 a solution

If u = 0, then u' = 0, u'' = 0 and u''' = 0.

So u''' - 3u'' + 4u = 0 - 3*0 + 4*0 = 0. Hence u = 0 is a solution.

Secondly, let f and g be solutions.

ie. f''' - 3f'' + 4f = 0
and g''' - 3g'' + 4g = 0

Consider h(x) = af + bg.

h'   = af' + bg'
h''  = af'' + bg''
h''' = af''' + bg'''

So h''' - 3h'' + 4h = af''' + bg''' - 3(af'' + bg'') + 4(af + bg)
= a(f''' - 3f'' + 4f) + b(g''' - 3g'' + 4g) = a*0 + b*0 = 0 for all
a,b.

So both requirements 1 and 2 are satisfied. The other requirements
follow from the vector properties of functions in general.

Hence the solutions to u''' - 3u'' + 4u = 0 form a vector space.

Hope this helps.

calebu2-ga

Search strategy : "Vector space" within mathworld.com

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Clarification of Answer by calebu2-ga on 21 Jan 2003 03:12 PST

A good point is made below. kennyh-ga points out that it is important
to show that the      two solutions span the space of solutions (by
showing that they form a asis and are linearly independent). So
consider his text as part of my answer. I'll write it out here for
clarity :

Given u'''- 3u''+4u = 0, by solving its characteristic equation :

m^3 - 3 m^2 + 4 = 0 or (m + 1) (m - 2)^2 = 0. 
So m = -1, -2 -2 and we obtain 3 functions e^(-x), e^(2x) & xe^(2x)
which form a basis of a 3-dim vector space.
 
Let's call them f1,f2, & f3. 

Testing {f1,f2,f3} are indep. : 

Consider af1 + bf2 +cf3= 0, That is: 

a e^(-x) + be^(2x) + cxe^(2x) = 0...(1) 

Set x = 0, we have a + b = 0 ..(1) 

[ Since e^(-x) <> 0 for all x, cancel it(by mult. e^x)] 

Multiply e^x to (1): a + b e^(3x) + c x e^(3x) = 0..(2) 

take dev. wrt x of (2) 

3b e^(3x) + c e^(3x) + 3cx e^(3x) = 0 

Cancel e^(3x): 3b + c + 3cx = 0 

Set x = 0, 3b + c = 0 ..(3) 

Set x = 1, 3b + 4c = 0...(4) 
 
(3)&(4) imply b=c=0 goto(1) a = 0 

This shows {f1, f2, f3} are linearly indep and therefore they form a
basis of the solution space of the given diff. eq. (aasume we know
that the sol. space of a k-degree linear diff eq. is of dim k)

calebu2-ga

It seems that the answer is not complete. Since, it did not show the
basis and independency.
 Given u'''- 3u''+4u = 0, by solving its characteristic equation 
say m^3 - 3 m^2 + 4 = 0 or (m + 1) (m - 2)^2 = 0.
 So m = -1, -2 -2 and we obtain 3 functions e^(-x), e^(2x) & xe^(2x)
 which form a basis of a 3-dim vector space.
 Let's call them f1,f2, & f3.
 Testing {f1,f2,f3} are indep. :
 Consider af1 + bf2 +cf3= 0, That is:
 a e^(-x) + be^(2x) + cxe^(2x) = 0...(1)
 Set x = 0, we have a + b = 0 ..(1)
[ Since e^(-x) <> 0 for all x, cancel it(by mult. e^x)]
 Multiply e^x to (1): a + b e^(3x) + c x e^(3x) = 0..(2)
 take dev. wrt x of (2)
 3b e^(3x) + c e^(3x) + 3cx e^(3x) = 0
 Cancel e^(3x): 3b + c + 3cx = 0
 Set x = 0, 3b + c = 0 ..(3)
 Set x = 1, 3b + 4c = 0...(4)
 (3)&(4) imply b=c=0 goto(1) a = 0

 This shows {f1, f2, f3} are linearly indep and therefore they form a
basis of the solution space of the given diff. eq. (aasume we know
that the sol. space of a k-degree linear diff eq. is of dim k)

 Kenny