> #OK to post Homework
> #Jeton Hida, September 20, 2021, Assignment 4
> #Number 1 #y''(t)+y(t)=0 #y1(t)=cos(t), y2(t)=sin(t) 
> diff(cos(t),$(t,2));
                                   -cos(t)
> diff(sin(t),$(t,2));
                                   -sin(t)
> #-cos(t)+cos(t)=0 #-sin(t)+sin(t)=0 
> diff(cos(t)+sin(t),$(t,2));
                               -cos(t) - sin(t)
> #(-cos(t)-sin(t))+(cos(t)+sin(t))=0 
#Conditions are met, both are solutions to the second order differential equation, and their sum is also a solution.
> #Number 2 #(y'(t))^2-4*y(t)=0 #y1(t)=t^2 diff(t^2,t)
                                     2 t
> evalf((2*t)^2-4*(t^2));
                                      0.
> #t^2 is a solution to the differential equation 
#Now we'll show why y2(t)=2*y1(t)=2t^2 is NOT a solution, a constant multiple of a solution is not also a solution because this is a nonlinear differential equation. If it was linear, this value would also be a solution
> diff(2*t^2,t);
                                     4 t
> evalf((4*t)^2-4*(2*t^2));
                                        2
                                    8. t 
> #Does not equal 0, so 2*t^2 is indeed NOT a solution.
> #Number 3 #a(n)=5*a(n-1)-6*a(n-2) #a1(n)=2^n, a2(n)=3^n
> evalf(2^2=5*2^(1)-6*2^(0))
                                   4. = 4.
> evalf(3^2=5*3^(1)-6*3^(0))
                                   9. = 9.
> evalf(2^2+3^2=5*(2^1+3^1)-6*(2^0+3^0))
                                  13. = 13.
> #Proven that 2^n+3^n is also a solution to the recurrence equation a(n)=5*a(n-1)-6*a(n-2)
> #Number 4
> #a(n)=a(n-1)^2, n>=0, a1(n)=2^(2^n), a2(n)=3^(2^n)
> seq(2^(2^n)=(2^(2^(n-1)))^2, n=0..5)
   2 = 2, 4 = 4, 16 = 16, 256 = 256, 65536 = 65536, 4294967296 = 4294967296
> seq(3^(2^n)=(3^(2^(n-1)))^2, n=0..5)
          3 = 3, 9 = 9, 81 = 81, 6561 = 6561, 43046721 = 43046721, 

            1853020188851841 = 1853020188851841
> seq((3^(2^n)+2^(2^n))=(3^(2^(n-1))+2^(2^(n-1)))^2, n=0..5)
                     2                                                       
    / (1/2)    (1/2)\                                                        
5 = \3      + 2     / , 13 = 25, 97 = 169, 6817 = 9409, 43112257 = 46471489, 

  1853024483819137 = 1858666703634049
> #Again, this is a nonlinear recurrence so we cannot add the two solutions
> together and have a new solution. As proven above.
> #Number 5
> #part ii rsolve({a(n)-7*a(n-1)+12*a(n-2)=6*n-11, a(0)=3, a(1)=9},a(n))
                                n    n        
                               3  + 4  + n + 1
> #part iv rsolve({a(n)-4*a(n-2)=-3*n+8, a(0)=2, a(1)=1},a(n))
                                 n       n    
                                2  + (-2)  + n
>