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The function x appears with no arguments.

Posted 10 years ago

DSolve[{Derivative[1][x][t] == (k (-x[t] + y[t]))/Subscript[V, A], Derivative[1][y][t] == (k (x[t] - y[t]))/Subscript[V, B], x[0] == Subscript[x, 0], y[0] == 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

Don't put your constants inside brackets. Also, Dsolve doesn't take a starting and ending value for the dependent variable.

In[5]:= Nutrient = {x'[t] == (k/Va) (y[t] - x[t]), 
   y'[t] == (k/Vb) (x[t] - y[t]), x[0] == x0, y[0] == y0};

In[6]:= Solution = DSolve[Nutrient, {x[t], y[t]}, {t}]

Out[6]= {{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 BY: Frank Kampas

Nutrient = {x'[t] == (k/[Va]) (y[t] - x[t]), y'[t] == (k/[Vb]) (x[t] - y[t]), x[0] == [x0], y[0] == [y0]};

Solution = DSolve[Nutrient, {x[t], y[t]}, {t, 0, 200}

Suggest you put your code into a code sample block.

POSTED BY: Frank Kampas
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