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

Step-by-step explanation:

dA/dt = 6 - 0.02A

dA/dt = - 0.02 (A - 300)

Separate the variables.

dA / (A - 300) = - 0.02 dt

Integrate.

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

Solve for A.

A - 300 = Ce^ (-0.02t)

A = 300 + Ce^ (-0.02t)

Use initial condition to find 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)