the velocity of a chemical reaction is given by dx/dx=k(a-x) where x is the amount transfered in time t,k is a constant and a is the condition at time t=0 when c=0 solve the equation and determine x in terms of t?
Plz show work plz plz
I think you might mean
dx/dt = k(a-x)
but what is c?
yes sir plz help
just keep on ...
dx/dt = k(a-x)
dx/(a-x) = kt
ln(a-x) = kt^2/2 or, using a new k
ln(a-x) = kt^2
a-x = e^(kt^2)
Now use your initial conditions to find a and k.
I am still not sure it is right
maybe
dx/dt = k(a-t) ??
if so try
x = c + p t - kt^2/2
then
dx/dt = p -kt
then p = ka and if c = 0 then
dx/dt = 0 +ka -kt = k(a-t)
so then
x = ka t - k t^2/2
Use Steve's solution if no typos in problem statement.
To solve the given differential equation, we need to integrate both sides. Let's solve it step by step:
Given equation: dx/dt = k(a - x)
Step 1: Separate the variables.
dx/(a - x) = k dt
Step 2: Integrate both sides with their respective variables.
∫ (1/(a - x)) dx = ∫ k dt
Step 3: Evaluate the integrals.
Let's evaluate the left-hand side integral first.
∫ (1/(a - x)) dx = -ln|a - x| + C1
∫ k dt = kt + C2
Step 4: Combine the integration results.
-ln|a - x| + C1 = kt + C2
Step 5: Combine the constants of integration into a single constant.
-ln|a - x| = kt + C
Step 6: Remove the logarithm using exponential notation.
|a - x| = e^(-kt - C)
Step 7: Simplify using properties of the absolute value.
Since the absolute value can be either positive or negative, we can remove it by considering both cases separately.
Case 1: a - x > 0
(a - x) = e^(-kt - C)
Case 2: a - x < 0
-(a - x) = e^(-kt - C)
Step 8: Solve each case.
Case 1:
(a - x) = e^(-kt - C)
x = a - e^(-kt - C)
Case 2:
-(a - x) = e^(-kt - C)
x - a = e^(-kt - C)
x = a + e^(-kt - C)
Therefore, the solution to the given differential equation is:
x = a - e^(-kt - C) for Case 1
x = a + e^(-kt - C) for Case 2
Note: The constants C1, C2, and C represent the constants of integration and can be determined using initial conditions or given values in the problem statement.