Question

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

B(0)=28

Answer 1

`B(t) = 4\, e^(-k \cdot t) + 24`.

There isn't enough information to determine the value of
`k`. It is thus treated as a non-zero constant. (In case that
`k = 0`,
`B(t) = 28`.)

Explanation:

Start by separating the two differentials,
`dB` and
`dt`. Multiply both sides of this equation by the denominator
`dt` to obtain

`dB = - k (B -24) \, dt`.

Separate the variables. Move all terms that involve
`B` to the side with
`dB`; move all terms that involve
`t` to the side with
`dt`.

Divide both sides with
`(B - 24)` assuming that it is not equal to zero.

`\displaystyle (1)/(B - 24) \,dB = -k \, dt`.

Note that the first derivative of
`(B - 24)` with respect to
`B` is just
`1`. Hence, leaving the constant
`(-k)` on the side with
`dt` could potentially simplify the calculations.

Integrate both sides.

`\displaystyle \int (1)/(B - 24)\, dB = \int -k \, dt`.

The left-hand side is similar to the first-derivative of a logarithm with
`e` as its base.
`(d)/(du) \ln |u| = (1)/(u)`.

Apply
`u`-substitution to the denominator
`(B - 24)`. Let
`u = B - 24`. Then
`du = dB` since the expression
`(B - 24)` is linear with a coefficient of
`1`.

`\displaystyle \int (1)/(B - 24)\, dB = \int (1)/(u)\, du = \ln |u| = \ln |B - 24|`.

The given condition
`B(0) = 28` implies that
`B - 24 > 0`.

The right-hand side is constant. Only linear expressions can produce a first-derivative of a constant.
`\displaystyle \int -k\, dt = (-k) \int dt = -k\, t`.

Hence
`\ln (B - 24) = -k \, t + C`.

The constant
`C` only needs to appear on one side of the equation.

`\ln(x)` is the inverse of
`e^x`. Hence
`e^(\ln (B - 24)) = B - 24`. Place both sides of the equation as the exponent of
`e`:

`e^(\ln (B - 24)) = e^(-k \, t + C)`.

`B - 24 = e^(-k \, t) \cdot e^(C)`.

Apply the given initial condition of
`t = 0` and
`B = 28` to find the value of
`e^(C)`:

• Left-hand side:
  `28 - 24 = 4`.
• Right-hand side:
  `e^(-k \, t)\cdot e^(C) = e^(0) \cdot C = C`.

Hence
`C = 4`. Add
`24` to both sides of the equation to obtain

`B = 4 e^(-k \, t) + 24`.

Apparently, the value of
`B` is dependent on that of
`t`. Hence the function expression:

`B(t) = 4 e^(-k \, t) + 24`.