I think you are "overworking" this. For any linear transformation, L, L(0)= 0 so that, if u is in the kernel of L, Lu= 0, the . That is, for any linear transformation, L, the kernel of L is a subspace of the kernel of .
Let denote the vector space of twice differentiable functions on . Define a linear map on by the formula
Suppose that is a basis for the solution space of . Find a basis for the solution space of the fourth order differential equation . What can you say about the kernals of and ?
This is my working out so far:
Since is a basis for the solution space of , then is a basis for the kernal of .
Therefore we can write
So since is spanned by two linearly independent vectors.
Now, consider . Since every solution of can be represented as ,
the equation becomes
, which then becomes
Let denote the homogenous solution and the particular solution respectively.
Obviously as is a basis for the solution space of .
Now this is where I got stuck. I did’t know what the particular solution is. I let
and tried to solve for the constants . It was a mess and I don’t think I was on the right track.
I was thinking that if I can find , then would be the basis of the fourth order differential equation . Am I correct?
Any help will be appreciated.
Thanks. Yep I understand what you mean by is a subspace of '.
I think because I have read somewhere that the solution space of an nth order homogeneous differential equation forms an nth dimensional vector space. Is a fourth order homogeneous equation? How would you find the basis?