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steposvetlana
1 month ago
10

Show that A(t)=300−250e0.2−0.02t satisfies the differential equation ⅆAⅆt=6−0.02A with initial condition A(10)=50 .

Mathematics
1 answer:
Leona [12.6K]1 month ago
8 0

Detailed derivation:

dA/dt = 6 - 0.02A

dA/dt = -0.02 (A - 300)

Rearranging terms.

dA / (A - 300) = -0.02 dt

Integrate both sides.

ln(A - 300) = -0.02t + C

Isolate A.

A - 300 = Ce^(-0.02t)

A = 300 + Ce^(-0.02t)

Apply initial condition to determine C.

50 = 300 + Ce^(-0.02 × 10)

50 = 300 + Ce^(-0.2)

-250 = Ce^(-0.2)

C = -250e^(0.2)

A = 300 - 250e^(0.2)e^(-0.02t)

A = 300 - 250e^(0.2 - 0.02t)

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