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posted by  Norby on 8/24/2008 9:22:09 PM  |  status: Closed  

Lifesaver! Please help!

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Verify that the given functions form a basis for the space of solutions of the given differential equation and find the solution given the initial conditions:




y(1)=0
y'(1)=0
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posted by Altair on 8/24/2008 9:38:24 PM  |  status: Live
Asker's Rating: Lifesaver   
Response Details:
if  are the solution of
 
  (1)
 
the general solution can be written as:
 
y=c1 x + c2 x2
 
y' = c1 + 2c2 x
 
y"=2c2
 
substitue in (1), we have:
 
2c2 x2 - 2x(c1 + 2c2x) + 2c1x + 2c2x2 = 4c2 x2 -2c1x - 4c2x2 +2c1x = 0
 
so we verify that are the solutions of (1)
 
y(1) = c1 + c2 = 0   => c1 = -c2
y'(1) = c1 + 2c2 = 0 -> c1 = -(1/2) c2
 
c1 = 0
c2 = 0
 
so the solution with the initial condition is not important one since y=0 in this case.
 
 
 
 
 

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posted by Ronan on 8/24/2008 9:42:59 PM  |  status: Live
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Response Details:
f1'(x) = 1,  f1''(x)  = 0
So, x^2 * 0 -2x*(1) +  2x = 0
 
f2'(x) = 2x,  f2''(x)  = 2
So , x^2 * 2 - 2x(2x) + 2x^2 = 0
 
So both given functions form a basis. 
 
General solution f(x) = a*x + b* x^2, where a & b are constants to be determined
So, f '(1) = a + 2* b(1) = 0 
and f(1) = a + b  = 0
 
So, unless I've made a mistake, a = b = 0
 
 
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