Using separation of variables, solve the differential equation dB/dt = -k(B-24) for BB(0)=28

Answer:[tex]B(t) = 4\, e^{-k \cdot t} + 24[/tex]. There isn't enough information to determine the value of [tex]k[/tex]. It is thus treated as a non-zero constant. (In case that [tex]k = 0[/tex], [tex]B(t) = 28[/tex].)Step-by-step explanation:Start by separating the two differentials, [tex]dB[/tex] and [tex]dt[/tex]. Multiply both sides of this equation by the denominator [tex]dt[/tex] to obtain[tex]dB = - k (B -24) \, dt[/tex].Separate the variables. Move all terms that involve [tex]B[/tex] to the side with [tex]dB[/tex]; move all terms that involve [tex]t[/tex] to the side with [tex]dt[/tex].Divide both sides with [tex](B - 24)[/tex] assuming that it is not equal to zero.[tex]\displaystyle \frac{1}{B - 24} \,dB = -k \, dt[/tex]. Note that the first derivative of [tex](B - 24)[/tex] with respect to [tex]B[/tex] is just [tex]1[/tex]. Hence, leaving the constant [tex](-k)[/tex] on the side with [tex]dt[/tex] could potentially simplify the calculations.Integrate both sides. [tex]\displaystyle \int \frac{1}{B - 24}\, dB = \int -k \, dt[/tex].The left-hand side is similar to the first-derivative of a logarithm with [tex]e[/tex] as its base. [tex]\dfrac{d}{du} \ln |u| = \dfrac{1}{u}[/tex].Apply [tex]u[/tex]-substitution to the denominator [tex](B - 24)[/tex]. Let [tex]u = B - 24[/tex]. Then [tex]du = dB[/tex] since the expression [tex](B - 24)[/tex] is linear with a coefficient of [tex]1[/tex].[tex]\displaystyle \int \frac{1}{B - 24}\, dB = \int \dfrac{1}{u}\, du = \ln |u| = \ln |B - 24|[/tex]. The given condition [tex]B(0) = 28[/tex] implies that [tex]B - 24 > 0[/tex]. The right-hand side is constant. Only linear expressions can produce a first-derivative of a constant. [tex]\displaystyle \int -k\, dt = (-k) \int dt = -k\, t[/tex].Hence [tex]\ln (B - 24) = -k \, t + C[/tex].The constant [tex]C[/tex] only needs to appear on one side of the equation.[tex]\ln(x)[/tex] is the inverse of [tex]e^x[/tex]. Hence [tex]e^{\ln (B - 24)} = B - 24[/tex]. Place both sides of the equation as the exponent of [tex]e[/tex]:[tex]e^{\ln (B - 24)} = e^{-k \, t + C}[/tex].[tex]B - 24 = e^{-k \, t} \cdot e^{C}[/tex].Apply the given initial condition of [tex]t = 0[/tex] and [tex]B = 28[/tex] to find the value of [tex]e^{C}[/tex]:Left-hand side: [tex]28 - 24 = 4[/tex].Right-hand side: [tex]e^{-k \, t}\cdot e^{C} = e^{0} \cdot C = C[/tex].Hence [tex]C = 4[/tex]. Add [tex]24[/tex] to both sides of the equation to obtain[tex]B = 4 e^{-k \, t} + 24[/tex].Apparently, the value of [tex]B[/tex] is dependent on that of [tex]t[/tex]. Hence the function expression:[tex]B(t) = 4 e^{-k \, t} + 24[/tex].