Can someone help me with this question?
Using as a trial solution solve the differential equation ...
given that x = 1 and at t = 0.
Use your solution to prove that for t>=0, x<=1
Thanks in advance
First solve to get the homogenous solution. Can you do this?
You're given that the particular solution has the form (you should think about why that is) so you need to get the value of A. Start by calculating and . Can you do this?
Note: Once you have the value of A and hence the particular solution you can construct the complete solution by adding together the homogenous and particular solutions. The boundary conditions x = 1 and at t = 0 are used to get the value of the arbitrary constants.
This is how i worked it out so far:
Using trial solution
therefore
comparing coefficients:
....Particular Integral
Considering
Let be a trial solution
therefore
u = -1(twice) therefore real equal roots
hence - Complimentary Function
x = Particular Integral + Complimentary Function
therefore ....[A]
putting x = 1, t = 0 in [A]
This is where i'm stuck, i don't know how to carry on from here.