# The function x appears with no arguments.

Posted 9 years ago
7868 Views
|
3 Replies
|
2 Total Likes
|
 DSolve[{Derivative[x][t] == (k (-x[t] + y[t]))/Subscript[V, A], Derivative[y][t] == (k (x[t] - y[t]))/Subscript[V, B], x == Subscript[x, 0], y == Subscript[y, 0]}, {x[t], y[t]}, {t, 0, 200}]Given :dx/dt =k/Subscript[V, A] (y-x) dy/dt =k/Subscript[V, B] (x-y) where Subscript[V, A] and Subscript[V, B] are the volumes of the compartments and k > 0 is a permeability factor. Let x(0) = Subscript[x, 0] and y(0) = Subscript[y, 0] denote the initial conditions for the nutrient. Solely on the basis of equations in the system and the assumptions Subscript[x, 0] > Subscript[y, 0] > 0, sketch on the same set of coordinate axes possible solution curves of the system. Explain your reasoning. Discuss the behavior of the solutions over a long period of time. I want to know if i typed something wrong or if the problem is wrong
3 Replies
Sort By:
Posted 9 years ago
 Don't put your constants inside brackets. Also, Dsolve doesn't take a starting and ending value for the dependent variable. In:= Nutrient = {x'[t] == (k/Va) (y[t] - x[t]), y'[t] == (k/Vb) (x[t] - y[t]), x == x0, y == y0}; In:= Solution = DSolve[Nutrient, {x[t], y[t]}, {t}] Out= {{x[t] -> ( Va x0 + E^((k t (-Va - Vb))/(Va Vb)) Vb x0 + Vb y0 - E^((k t (-Va - Vb))/(Va Vb)) Vb y0)/(Va + Vb), y[t] -> -((-Va x0 + E^((k t (-Va - Vb))/(Va Vb)) Va x0 - E^((k t (-Va - Vb))/(Va Vb)) Va y0 - Vb y0)/(Va + Vb))}} 
Posted 9 years ago
 Nutrient = {x'[t] == (k/[Va]) (y[t] - x[t]), y'[t] == (k/[Vb]) (x[t] - y[t]), x == [x0], y == [y0]};Solution = DSolve[Nutrient, {x[t], y[t]}, {t, 0, 200}
Posted 9 years ago
 Suggest you put your code into a code sample block.