T, thus changes as dT/dt = ebt − kT. Solve the differential equation using this technique. (Let T(0) = T0.)

Question

mRNA is an intermediate molecule that arises in the decoding of DNA. mRNA is produced by a process called transcription and it eventually decays. Suppose that the rate of transcription is changing exponentially according to the expression e^bt

 and mRNA has a constant per capita decay rate of k. The number of mRNA transcript molecules, T, thus changes as

dT/dt = ebt − kT.

The first term on the right side is time-varying. As a result, the differential equation is not separable. However, the equation can be solved using the change of variables y(t) = be^ktT(t).

 Solve the differential equation using this technique. (Let T(0) = T₀.)

Answer 1

Using the suggested substitution y = be^ktT, we have

dy/dt = bke^ktT + be^kt * dT/dt (product rule)

Looking back at the original equation, you can substitute e^bt - kT for dT/dt in the equation above.

dy/dt = bke^ktT + be^kt * (e^bt - kT) = bke^ktT + bke^(k+b)t - bke^ktT

The first and last terms on the right hand side cancel, so we are left with

dy/dt = bke^(k+b)t

Therefore y = be^(k+b)t/(k+b) [Note: will deal with constant of integration later.]

To convert back to T, fill in be^ktT for y:

be^ktT = be^(k+b)t/(k+b)

Dividing both sides by be^kt gives

T = e^bt/(k+b) + C [I've now included the constant of integration]

T₀ = T(0) = 1/(k+b) + C. Therefore C = T₀ - 1/(k+b)

Plugging that in and simplifying gives us the final answer:

T = (e^bt - 1)/(k+b) + T₀

Answer 2

It is a first order linear ODE, which is solved by multiplying thru by integrating factor

dT/dt + kT = exp(bt)

the integrating factor is exp( integral(k)) = exp ( kt)

Multiplies both sides by exp(kt)

dT/dt * exp(kt) + kT ( exp(kt)) = exp( (b+k)t)

d/dt [ exp(kt) * T ] = exp( (b+k)t)

integrating both sides w.r.t. time t:

exp(kt) * T = integral ( exp((b+k)t) dt <---- u=(b+k)t

du = (b+k) dt

du / (b+k) = dt

exp(kt) * T = 1/(b+k) * integral [exp(u) du]

exp(kt) * T = 1/(b+k) * [exp(u) + c]

exp(kt) * T = 1/(b+k) * [ exp((b+k)t + c ] = 1/(b+k) * [ exp(bt+kt) + c ]

mutiplies both sides by exp(-kt) , which causes the exponents to be added:

T = 1/(b+k) *[ exp(bt) + c * exp(-kt) ]

CHECK:

dT/dt = 1/(b+k) * [ b * exp(bt) - ck * exp(-kt) ]

plugging into the ODE:

dT/dt + kT =

1/(b+k) * [ b * exp(bt) - ck * exp(-kt) ] + k/(b+k) [ exp(bt) + c*exp(-kt) ]

coefficients of exp(bt)

b/(b+k) + k/(b+k) = (b+k)/(b+k) = 1

coefficients of exp(-kt)

(-ck)/(b+k) + (ck)/(b+k) = 0, so exp(-kt) vanishes

the final result is exp(bt) which proves the solution is correct.

The solution is:

T = 1/(b+k) *[ exp(bt) + c * exp(-kt) ]

finally, resolving the initial conidition: T(0) = T0

T0 = 1/(b+k) * [ exp(0) + c*exp(0) ]

T0 = 1/(b+k) * [1 + c]

T0 * (b+k) = 1+c

c = T0(b+k)-1