A chemical compound decays over time when exposed to air, at a rate proportional to its concentration to the power of 3/2. At the same time, the compound is produced by another process. The differential equation for its instantaneous concentration is: dn(t) dt =-0.8 n³/2 + 10 n₁(1-e-³t) = 2000 is the initial concentration. where n(t) is the instantaneous concentration and n at t = 0. Solve the differential equation to find the concentration as a function of time from t=0 until t = 0.5 s, using Taylor series method. Use a step size of h = 0.125 s.
A chemical compound decays over time when exposed to air, at a rate proportional to its concentration to the power of 3/2. At the same time, the compound is produced by another process. The differential equation for its instantaneous concentration is: dn(t) dt =-0.8 n³/2 + 10 n₁(1-e-³t) = 2000 is the initial concentration. where n(t) is the instantaneous concentration and n at t = 0. Solve the differential equation to find the concentration as a function of time from t=0 until t = 0.5 s, using Taylor series method. Use a step size of h = 0.125 s.
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Transcribed Image Text:A chemical compound decays over time when exposed to air, at a rate proportional to its
concentration to the power of 3/2. At the same time, the compound is produced by another
process. The differential equation for its instantaneous concentration is:
dn(t)
-0.8 n³/² + 10 n₁(1-e-3t)
dt
where n(t) is the instantaneous concentration and n₁ = 2000 is the initial concentration
at t = 0. Solve the differential equation to find the concentration as a function of time
from t = 0 until t = 0.5 s, using Taylor series method. Use a step size of h = 0.125 s.
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