Question

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

Answer 1

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)