Proving solutions of an ODE of the form y''+by'+cy=0


by nemma14778
Tags: axiom, differential eqn, ode, vector spaces
nemma14778
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#1
Mar8-11, 09:55 PM
P: 1
1. The problem statement, all variables and given/known data
The set of solutions of an ODE of the form y''+by'+cy=0 forms a vector space. To convince yourself of that, prove that axioms 1,4,5 and 9 of the definition of a vector space hold for this set of solutions. (You may want to check the others as well, but no need to present their proof.)


2. Relevant equations



3. The attempt at a solution
I started to prove the first axiom as follows:


Axiom #1: If u and v are objects in V, then u + v is in V.
If we let c1u and c2v be two solutions to our second order differential equation, then c1u + c2v must be a solution for the second order differential equation for any constants c1 and c2.

F(y) = 0 = (c1u+c2v)’’ + b(c1u+c2v)’ + c(c1u+c2v)
0 = c1u’’+c2v’’ + bc1u’+ bc2v’ + cc1u+cc2v
0 = c1u’’ + bc1u’ + cc1u + c2v’’ + bc2v’ + cc2v
0 = c1(u’’ + bu’ + cu) + c2(v’’ + bv’ + cv)
0 = c1F(u) + c2F(v)
0 = c1(0) + c2(0) = 0 + 0 = 0
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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LCKurtz
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#2
Mar9-11, 10:24 PM
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Quote Quote by nemma14778 View Post
1. The problem statement, all variables and given/known data
The set of solutions of an ODE of the form y''+by'+cy=0 forms a vector space. To convince yourself of that, prove that axioms 1,4,5 and 9 of the definition of a vector space hold for this set of solutions. (You may want to check the others as well, but no need to present their proof.)


2. Relevant equations



3. The attempt at a solution
I started to prove the first axiom as follows:


Axiom #1: If u and v are objects in V, then u + v is in V.
If we let c1u and c2v be two solutions to our second order differential equation, then c1u + c2v must be a solution for the second order differential equation for any constants c1 and c2.

F(y) = 0 = (c1u+c2v)’’ + b(c1u+c2v)’ + c(c1u+c2v)
0 = c1u’’+c2v’’ + bc1u’+ bc2v’ + cc1u+cc2v
0 = c1u’’ + bc1u’ + cc1u + c2v’’ + bc2v’ + cc2v
0 = c1(u’’ + bu’ + cu) + c2(v’’ + bv’ + cv)
0 = c1F(u) + c2F(v)
0 = c1(0) + c2(0) = 0 + 0 = 0
[
I don't see any question here. Also I don't know what axioms 1,4,5,9 are in your text. However, in your argument above you want to start with u and v, not c1u and c2v. I think you will find that your argument will show that u and v are solutions then so is c1u + c2v, which may solve more than one of your required properties.
lurflurf
lurflurf is offline
#3
Mar10-11, 12:44 AM
HW Helper
P: 2,168
I do not understand any of that stuff. Any way if one wanted to verify existence uniqueness without apeal to general theorems a standard method is to apply variation of parameters to deduce that y is a solution if and only if it is a linear combination of a basis of the solution space. This also makes it easy to show te solution space is a vector space.


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