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

Given the expression

Factor out 0.02 from the xpression to have:

Separate the variables;

Integrate both sides of the eequation to have:

Apply the initial condition to get the value of C

Substitute the constant into the function A(t) to get the required proof

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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)

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step-by-step explanation:

correct answer is c. just did it

step-by-step explanation: