i Find the value of the constant k such that is a particular integral of the differential equation:
ii Find the solution of this differential equation for which y=1 and when x=0.
iii Use the differential equation to determine the value of when x=0. Hence prove that 0<=1 for x>0
I got this far:
Originally Posted by Mush
Stuck at this part now.
Trying to sum LHS together and I have no idea how to do that, while trying to expand the RHS to the LHS or get both sides to meet somewhere in the middle is still pretty much impossible for me :(
Okay, let's write the expressions for and in a different form, and see if that helps!
Originally Posted by Lonehwolf
Put them all into the original ODE:
Divide both sides by :
Now take out a factor of k:
For part two you must find the general solution of the equation (in part 1 you only found a particular solution!), and then implement the given conditions to find another particular solution. You do this using normal methods.
First of all, use the auxiliary equation:
Solve this for and you will get 1 solution! (Since the discriminant is 0!). That means that your complimentary function is
, where B and A are the arbitrary constants! You then need to consider the particular integral to account for the RHS of the ODE, to find . You should find, however, that the RHS is linearly dependent on your complimentary function, so you have to make adjustments! Do you know what basis function you use for when the RHS is , considering the linear dependency on the complimentary function?
Once you have found and , then you find the total function simply by adding these two together! Because your function is linear, then the solution can be superposed of lots of little solutions. Hence