in this problem we're given a second order linear differential equation and we're not asked to find
the general solution for this because it's a calculus 2 class but we're asked to verify the
general solution so remember a general solution for a second order differential equation should
have two arbitrary constants in it and that's part a and then in part b we're given two initial
conditions that should allow us to nail down those two constants and find the particular solution
so to show that something is a general solution you just run it through the differential equation
so i need to get y double prime and i need to get y prime and i need to show y double prime minus y
prime minus two times y all adds up to zero so i start taking derivatives and the chain rule
produces a factor of 2 in that first term and a factor of negative 1 in the second term then i
get y double prime again i produce a factor of 2 in the first term and a factor of negative 1
in the second term so now i have an expression for y
and y prime and y double prime and i put them all together and y double prime
that's four c one e to the two t plus c two e to the negative t minus y prime that's two c
one e to the two t minus c two e to the negative t minus twice the original y which is c1 e to the 2t
plus c2 e to the negative t now i'm really not sure how much work to show here because
the standard way to deal with it at this point is to do the bookkeeping one term at a time like i'm
going to look at all the terms with an e to the 2t in them there's one there one there and one there
and i immediately see that i have 4c1 e to the 2t but then i have the negative 2 c1 e to the 2t
and another negative 2 c 1 e to the 2 ts so i get a coefficient of 0 for that term
i guess i'll show that as my intermediate step and i do a similar type of bookkeeping
for the e to the negative t terms there's a c 2 e to the negative t
plus another c 2 e to the negative t just remembering that we had to distribute this minus
sign so now i've got 2 c 2 e to the negative t and then in my last term i see that i have a negative
2 c 2 e to the negative t when i distribute this negative 2. so that also adds up to 0
and i end up with a grand total of 0. so in this little proof i've started with
my original differential equation y double prime of t minus y prime of t minus two y of t
and then i went through the algebra and i showed that it's all zero
so this proves that that solution satisfies the original differential equation
in part b i'm asked for the particular solution corresponding to these initial conditions
so i'll start with y of 0 equals 1. so if i take my original function y
and i plug in 0 for t i'm supposed to get 1. so i have 1. is equal to c1 because e to the 0 is 1
plus c2 then my second initial condition is y prime of 0 equals 0. so i have an expression
for y prime written down already and if i replace t with zero i'm supposed to get zero again these
exponentials are simple when i plug in zero for t they just turn into one so i have two c one
minus c two now i have a little system of equations to solve and it's set up for c2 to
eliminate in one step so i'm just going to do that i'm going to add the two equations
and on the left hand side i get 1 and on the right hand side to get 3 c1 and i've immediately got c1
is one-third then i can plug back into the first equation one is equal to c1 that's one-third
plus c2 so this means c2 is two-thirds finally i have to actually write down what this particular
solution is so that's the general solution with the constants replaced with the values that are
going to satisfy the initial conditions so i get y of t is 1 third e to the two t
plus two thirds e to the negative t
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