The velocity of a chemical reaction is given by (dx/dt) = k(a-x) , where x is the amount transfered in time t, k is a constant and a is the concentration at time t=0 when x=0.
Solve the equation and determine x in terms of t.
Is -e^-kt = e^kt ?
the bot's logic is, as usual opaque or just wrong.
x = ce^(-kt) + a
To solve the given differential equation (dx/dt) = k(a-x), we can use separation of variables.
Step 1: Rearrange the equation to isolate the variables.
(dx / (a - x)) = k * dt
Step 2: Separate the variables by multiplying both sides by (a - x).
(dx) = k * (a - x) * dt
Step 3: Integrate both sides with respect to their respective variables.
∫(dx) = ∫k * (a - x) * dt
Step 4: Evaluate the integrals.
x = -k * ∫(a - x) * dt
To integrate the right side of the equation, we can use substitution.
Let u = a - x,
Then du = -dx.
Substituting these values, we get:
x = -k * ∫u * dt
Step 5: Integrate again.
x = -k * ∫u * dt
= -k * ∫(a - x) * dt
= -k * ∫(a - x) * (-dx)
= -k * ∫(a - x) * (-1) * dx
= -k * ∫(a - x) * dx
Step 6: Simplify the equation and evaluate the integral.
x = -k * ∫(a*dx - x*dx)
= -k * (a*x - (x^2)/2) + C
Here, C is the constant of integration.
Finally, we can solve for x in terms of t by substituting back the original variables:
x = -k * (a*x - (x^2)/2) + C
= -k * (a * (-k * (a*x - (x^2)/2) + C) - ((-k * (a*x - (x^2)/2) + C)^2) / 2) + C
This equation is the solution to the given differential equation (dx/dt) = k(a-x), expressed in terms of x and t.
Solution:
The equation can be solved using separation of variables.
Integrating both sides of the equation, we get:
∫(dx/dt)dt = ∫k(a-x)dt
On the left side, we have the integral of the velocity, which is the displacement x. On the right side, we have the integral of the rate of reaction, which is the amount of reaction.
Therefore,
x = kt(a-x) + C
where C is the constant of integration.
Rearranging the equation, we get:
x = (ka/k+1) + (C/k+1)e-kt
Therefore, x in terms of t is given by:
x = (ka/k+1) + (C/k+1)e-kt