- #1
Kevin Qi
- 3
- 0
Hello,
I noticed that the solution of a homogeneous linear second order DE can be interpreted as the kernel of a linear transformation.
It can also be easily shown that the general solution, Ygeneral, of a nonhomogenous DE is given by:
Ygeneral = Yhomogeneous + Yparticular
My question: Is it possible to arrive at the above result by using arguments involving the relationship between the kernel, column space, row space, etc of the linear differential operator of a DE?
My current attempt: The solution space of a non-homogeneous DE is isomorphic to the solution space (kernel) of the corresponding homogeneous DE, and is "displaced" from the kernel by the vector, Yparticular. Hence Ygeneral could be constructing by adding any Yparticular back to the Yparticular.
My solution is probably pretty flawed, and I have no idea how I can justify it. Could someone please enlighten me (or tell me that my question is dumb)? :D
Thanks in advance!
I noticed that the solution of a homogeneous linear second order DE can be interpreted as the kernel of a linear transformation.
It can also be easily shown that the general solution, Ygeneral, of a nonhomogenous DE is given by:
Ygeneral = Yhomogeneous + Yparticular
My question: Is it possible to arrive at the above result by using arguments involving the relationship between the kernel, column space, row space, etc of the linear differential operator of a DE?
My current attempt: The solution space of a non-homogeneous DE is isomorphic to the solution space (kernel) of the corresponding homogeneous DE, and is "displaced" from the kernel by the vector, Yparticular. Hence Ygeneral could be constructing by adding any Yparticular back to the Yparticular.
My solution is probably pretty flawed, and I have no idea how I can justify it. Could someone please enlighten me (or tell me that my question is dumb)? :D
Thanks in advance!