Final answer:
To solve the differential equation dB/dt + 2B = 50, we can separate the variables and integrate both sides. Applying the initial condition B(1) = 60, we can solve for the constant of integration and find the solution to be B = 25 ± e^(-2(t - 1 - (1 + ln(±70) / 2))) / 2.
Step-by-step explanation:
To solve the differential equation dB/dt + 2B = 50, we can first rewrite it as dB = (50 - 2B)dt. Next, we can separate the variables by dividing both sides by (50 - 2B), giving us dB / (50 - 2B) = dt. We can then integrate both sides:
∫(1 / (50 - 2B)) dB = ∫dt
To evaluate the integrals, we can use a substitution. Let u = 50 - 2B. Then du = -2dB and dt = -du / 2. Substituting these back into the equation, we get:
∫(1 / u) (-1/2) du = ∫dt
By evaluating the integrals, we have -1/2 ln|u| = t + C, where C is the constant of integration.
Now, we can substitute back in for u: -1/2 ln|50 - 2B| = t + C. Next, we can solve for B:
ln|50 - 2B| = -2(t + C)
|50 - 2B| = e^(-2(t + C))
50 - 2B = ±e^(-2(t + C))
Solving for B, we get B = 25 ± (e^(-2(t + C)) / 2)
Now, we can apply the initial condition B(1) = 60. Since t = 1, we have B = 60. Plugging that in, we can solve for C:
60 = 25 ± (e^(-2(1 + C)) / 2)
Subtracting 25 and multiplying both sides by 2:
70 = ±e^(-2(1 + C))
e^(-2(1 + C)) = ±70
-2(1 + C) = ln(±70)
1 + C = -ln(±70) / 2
C = -1 - ln(±70) / 2
Plugging this value of C back into our earlier equation for B, we get:
B = 25 ± e^(-2(t - 1 - (1 + ln(±70) / 2))) / 2
Therefore, the solution to the differential equation with the given initial condition is B = 25 ± e^(-2(t - 1 - (1 + ln(±70) / 2))) / 2.