[QUOTE=nepmma;585567] dy= -2(x- 3)dx.
"Change in y" over what interval? $\displaystyle \Delta t= 1$?b) change in y at time t = -2(y at time (t-1))+12
i am stuck on part b as it includes the time element
Then $\displaystyle y(t)- y(t-1)= -1y(t-1)+ 12$ so that math]y(t)= -2y(t-1)+ 12[/tex].
In order to solve that you would need to know y(t) over some initial interval, say t= 0 to t= 1.
Given that y(t)= f(t) for t= 0 to t= 1, y(t)= 12- 2f(t) for t= 1 to 2, y(t)= 12- 24- 4f(t)= -12- 4f(t) for t= 2 to t= 3, y(t)= 12- (-24- 8f(t))= 36+ 4f(t) for t= 3 to 4, etc.