Differential Equations

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Differential Equations

The topic is chemical reaction rate.

Two substances A and B are combined to form a product C. The formation of product is proportional to the time the reactants are combined. The final product is composed of two parts of B for every part of A. If initially A is 30 kg and B is 20 kg, and 5 kg of the product is formed after 30 mins., find the function of product formed at any given time.

Lorepersn (guest)

dx/dt = k(30-(1/3)x)(20-(2/3)x)
dx/dt = -1/9[k(x-90)(2x-60)
factoring out the negative and factor out the 1/3 gives -1/9

-9(dx/dt) = k(x-90)(2x-60)
-9(dx/dt)/((x-90)(2x-60) = k

-9dx/((x-90)(2x-60)) = kdt

Then use partial fraction to solve for -9dx/((x-90)(x-30))
Then we find that -> (3/20)/(x-30) + (-3/20)/(x-90) = kdt
Tranpose 3/20 to the right side, so that = (20/3)kdt
Integrating both sides give
-> ln |x-30| - ln |x-90| = (20/3)kt + C
Raising both sides to e gives to use log identities
-> (x-30)/(x-90) = e^(20kt/3)(e^C)
When x=0, t=0
-> (0-30)/(0-90) = e^((20k(0))/3)C ----> e^0 =1
-> (-30/-90) = C ----> thus c = 1/3
-> (x-30/x-90) = (1/3) e^(20kt/3) ----- (eq 420)
When x=5, t= 30
-> -25/-85 = (1/3)e^((20)(30)k/3)
We find that k = -0.000625
Substitute that to eq 420, we find that
-> (x-30)/(x-90) = (1/3)e^(-0.0125t/3) --- final answer

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