I need help with a question I had a few hours ago on a test. Using the method of undetermined coefficients find the solution to
y'' - y' = 4te^(2t) y(0)=0 y'(0)=0

Question

I need help with a question I had a few hours ago on a test. Using the method of undetermined coefficients find the solution to 
y'' - y' = 4te^(2t) y(0)=0 y'(0)=0

nubeer · Answer

ok have you solved the left side?

anonymous · Answer

that's what the problem is asking. The left side is the differential equation.

nubeer · Answer

Y=Ae^(mx)
this is a general form
find Y', Y''
and plug them in your equation.. and remember when u r working on left side, the right side is equal to 0
so Y''-Y' = 0
your equation now should be 
m^2-m = 0
please solve and see if u get this or not.

anonymous · Answer

yeah so 
y''-y'=0
$$r^2-r = 0$$$$r(r-1)=0$$$$r _{1} = 0$$$$r _{2} = 1$$$$y_{c}=C _{1}e^0 + C _{2}e^t$$$$y_{c}=C_{1} + C_{2}e^t$$Carrying on from that: $$y _{c} = Ae ^{2t}(Bt+C)$$ I think.. that's where I got confused.

nubeer · Answer

ya its yp 
now find yp ' and yp'' ... now plug this derivative in equation
so 
yp''-yp' = Ae^(2t)(Bt+C)

anonymous · Answer

ok cool thanks. Yeah i meant to type yp but I didn't. thanks I got it now.

nubeer · Answer

ok good :)