2) I would try
I have 3 non-homogeneous differential equations. The first i have done myself and believe to be correct, but can someone just confirm. The second and third i just need some guidance on what form the particular solution will take and the rest i will try to do myself:
1) y" - y' - 12y = e^6x here i used yp = ce^6x and got the general solution as yh = Ae^-3x + Be^4x + e^6x/18
2) y" - y' -6y = e^-x + 12x and yp = ???? y(0)=1, y'(0)=-2
3) y" + 2y' + 5y = cos3x and yp = ???? y(0)=1, y'(0)=0
surely is the constant not e^6x/18 as
>>yp=ce^6x so y'p=6e^6x and y"p=36e^6x
>>sub into the diff. equ. to get (36c - 6c - 12c)e^6x=e^6x
>> so 18c=1 hence c=1/18 therefore yh = Ae^-3x + Be^4x + e^6x/18
so how is it e^6x/144??
Your solution to the first problem is correct, I believe, though I would encourage you to use parentheses around the arguments of functions. That is, write it as y = Ae^-3x + Be^4x + e^(6x)/18, or better yet use LaTeX.
ok so im having difficulty on solving the 2nd, below is what i have done:
diff equa. >> y" - y' -6y = e^-x + 12x
particular equa.>> yp = Ae^-x + Bx + C so y'p= -Ae^-x + B and y"p= Ae^-x
sub into diff eq >> Ae^-x -(-Ae^-x + B) - 6(Ae^-x + Bx + C) = e^-x + 12x
simplify >> -4e^-x - 6Bx - 6C - B = e^-x + 12x
how would i solve from here??
subbed into to get general solution as >> yh= Ae^3x + Be^-2x - 1/4e^-x - 2x - 1/3
subbed in initial conditions to get A= 19/12 and B=0
so initial value problem is yh = 19/12e^3x + 0e^-2x - 1/4e^-x - 2x - 1/3
simplified >>yh = 19/12e^3x- 1/4e^-x - 2x - 1/3
can sum1 confirm if that is correct??