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

[nemma14778]

Homework Statement

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.)

Homework Equations

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

 

[LCKurtz]

Homework Statement

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.)

Homework Equations

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
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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]

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.